\section{Introduction} \label{sec:intro}
Inspired by a series of works by Ebisu \cite{Ebisu1,Ebisu2,Ebisu3}, 
Iwasaki \cite{Iwasaki} set out to develop a theoretical study of 
non-terminating hypergeometric summations with one free parameter.  
In the present article we continue this study and settle some of the open 
problems posed there, where two new symmetries which we 
call {\sl duality} and {\sl reciprocity} will play central roles. 
\par
The origin of these symmetries is very simple.  
The Gauss hypergeometric equation 
\begin{equation} \label{eqn:hge}
z(1-z) \dfrac{d u}{d z^2} + \{\gamma - (\alpha+\beta+1) z \} 
\dfrac{d u}{d z} - \alpha \beta \, u = 0 
\end{equation} 
has the Riemann scheme: 
\begin{equation} \label{eqn:rs}
\left\{\begin{matrix}
z = 0    & z = 1               & z = \infty \\[1mm]
0        & 0                   & \alpha     \\[1mm]
1-\gamma & \gamma-\alpha-\beta & \beta 
\end{matrix}\right\}, 
\end{equation}
in which the hypergeometric series $\hgF(\alpha, \beta; \gamma; z)$ is 
just the solution of local exponent $0$ at the origin $z = 0$. 
Now the idea is this: exchanging the solution of exponent $0$ at 
$z = 0$ with that of exponent $1-\gamma$ at the same point $z = 0$  
produces duality, while exchanging the solution of exponent $0$ at 
$z = 0$ with that of exponent $\gamma-\alpha-\beta$ at $z = 1$ gives 
rise to reciprocity. 
\par
The purpose of this article is to explain how these two symmetries 
can be used not only to extend our previous results to a larger domain, 
but also to strengthen themselves substantially in the original domain. 
In particular we are able to show that they are powerful enough to 
settle the {\sl rationality} and {\sl finite-cardinality} 
conjectures for the numbers $a$ and $b$ in our notation below 
(see Theorem \ref{thm:ab-rf} among other results).          
\par
We recall the main issues of our study.   
For a data $\lambda = (p,q,r;a,b,c;x) \in \mathbb{R}^6 \times 
(-1, \, 1)$, let     
\[  
f(w;\lambda) := \hgF(pw+a, \, qw+b; \, rw + c; \, x).           
\] 
Putting the case $r = 0$ aside, we assume $r \neq 0$.  
Then we may assume $r > 0$, for otherwise a simultaneous 
change of signs $(p,q,r;w) \mapsto (-p,-q,-r;-w)$ makes $r > 0$.  
Moreover we may assume $c = 0$ by a shift of variable 
$w \mapsto w - c/r$.   
Thus from the outset we have only to consider 
\begin{equation} \label{eqn:f}  
f(w;\lambda) := \hgF(pw+a, \, qw+b; \, rw; \, x),           
\end{equation} 
where the revised data $\lambda = (p,q,r;a,b;x)$ ranges over the domain    
\begin{equation} \label{eqn:real}
p, \, q \in \R; \qquad r > 0; \qquad a, \, b \in \R; \qquad  
-1 < x < 0 \quad \mbox{or} \quad 0 < x < 1,    
\end{equation}
with the trivial case $x = 0$ being excluded. 
We are interested in the following problems. 
\begin{problemgpf} \label{prob:gpf} 
Find a data $\lambda = (p,q,r;a,b;x)$ for which $f(w; \lambda)$ 
has a {\sl gamma product formula}: 
\begin{equation} \label{eqn:gpf}
f(w; \lambda) = C \cdot d^w \cdot 
\dfrac{\varGamma(w+u_1)\cdots\varGamma(w+u_m)}{\varGamma(w+v_1)\cdots\varGamma(w+v_n)},      
\end{equation}
for some $C, \, d \in \C^{\times}$; $m, \, n \in \Z_{\ge 0}$; 
$u_1, \dots, u_m \in \C$; and $v_1, \dots, v_n \in \C$.    
\end{problemgpf}
\par
Problem $\rI$ is something like the peak of a high mountain, 
which is difficult to climb up directly.  
We need a base camp for attacking it and this role is 
played by the following.  
\begin{problemocf} \label{prob:ocf} 
Find a data $\lambda = (p,q,r;a,b;x)$ for which  
$f(w; \lambda)$ is {\sl of closed form}, that is,  
\begin{equation} \label{eqn:ocf} 
\dfrac{f(w+1; \lambda)}{f(w; \lambda)} =: R(w; \lambda) 
\in \mathbb{C}(w) \,\,:\,\, \mbox{a rational function of $w$}.  
\end{equation}
\end{problemocf}
\par
It is evident from recursion formula $\varGamma(w+1) = w \, \varGamma(w)$ 
that any solution to Problem $\rI$ leads to a solution to 
Problem $\rII$, where the corresponding rational function is 
given by 
\begin{equation} \label{eqn:R}
R(w; \lambda) = 
d \cdot \dfrac{(w+u_1) \cdots (w+u_m)}{(w+v_1) \cdots (w+v_n)}. 
\end{equation}  
However the converse question is by no means trivial. 
We say that a solution \eqref{eqn:ocf} to Problem $\rII$ with 
rational function \eqref{eqn:R} leads back to a solution \eqref{eqn:gpf} 
to Problem $\rI$, if there exists a constant $C \in \C^{\times}$ 
such that formula \eqref{eqn:gpf} holds true with $d$; $u_1,\dots,u_m$; 
and $v_1,\dots,v_n$ coming from \eqref{eqn:R}.  
\begin{problem} \label{prob:back}
When does a solution to Problem $\rII$ lead back to a solution 
to Problem $\rI$ ?   
\end{problem}
\par
A solution $\lambda$ to Problem $\rI$ or $\rII$ is said to be 
{\sl elementary}, if the function $f(w; \lambda)$ has at most 
finitely many poles in $\C_w$; otherwise, 
it is said to be {\sl non-elementary}.  
\begin{problem} \label{prob:elem}
Enumerate all elementary solutions and determine them explicitly. 
\end{problem} 
\par
For a non-elementary solution the existence of infinitely many poles 
allows us to discuss asymptotics and dynamics of their residues as 
the poles tends to infinity. 
This brings us significant information about the solution under 
consideration (see Iwasaki \cite[\S 7--\S 10]{Iwasaki}). 
\begin{definition} \label{def:trichtomy} 
A data $\lambda = (p,q,r;a,b;x)$ is said to be {\sl integral} if 
$\bp := (p,q;r) \in \Z^3$; {\sl rational} if $\bp \in \Q^3$;  
and {\sl irrational} if $\bp \not\in \Q^3$.  
One can speak of an integral, rational or irrational 
solution to Problem $\rI$ or $\rII$. 
Note that the component $(a,b;x)$ has nothing to do with this definition. 
\end{definition}
\par
There is an efficient method to find {\sl integral} solutions to 
Problem $\rII$. 
To describe it we simplify the notation by writing $\hgF(\ba;z) = 
\hgF(\alpha,\beta;\gamma;z)$ with $\ba = (a_1,a_2;a_3) = 
(\alpha,\beta;\gamma)$.   
A {\sl three-term relation} for $\hgF(\ba;z)$ is an identity of 
the form: 
\begin{equation} \label{eqn:3trF}
\hgF(\ba+\bp;z) = r(\ba; z) \, \hgF(\ba;z) + q(\ba; z) \, \hgF(\ba+\1;z), 
\end{equation}
where $\bp = (p,q;r) \in \mathbb{Z}^3$ is an integer vector;  
$\1 := (1,1;1)$; and $q(\ba;z)$ and $r(\ba;z)$ are rational 
functions of $(\ba;z)$ uniquely determined by $\bp$. 
Relation \eqref{eqn:3trF} can be obtained by composing a finite 
sequence of fifteen contiguous relations of Gauss (see Andrews et al. 
\cite[\S 2.5]{AAR}).  
Ebisu \cite{Ebisu1} found an efficient formula for the coefficients 
$q(\ba;z)$ and $r(\ba;z)$ (see also Vidunas \cite{Vidunas1}).   
\par
For an integral data $\lambda = (p,q,r;a,b;x)$, substituting 
$\ba = \bal(w) := (p w+a, q w +b; r w)$ and $z = x$ into equation  
\eqref{eqn:3trF} we have  
\begin{equation} \label{eqn:3trf}
f(w+1; \lambda) = R(w; \lambda) \, f(w; \lambda) + 
Q(w; \lambda) \, \tilde{f}(w; \lambda),  
\end{equation}
with $\tilde{f}(w; \lambda) := \hgF(\bal(w)+\1;x)$, where 
$Q(w;\lambda) := q(\bal(w);x)$ and $R(w;\lambda):= r(\bal(w);x)$ are 
rational functions of $w$ depending uniquely on $\lambda$. 
If $\lambda$ happens to be such a data that  
\begin{equation} \label{eqn:cfcr}
Q(w;\lambda) = 0 \qquad \mbox{in} \quad \C(w), 
\end{equation}
then three-term relation \eqref{eqn:3trf} reduces to a two-term one 
\eqref{eqn:ocf}, yielding a solution to Problem $\rII$. 
An integral solution to Problem $\rII$ so obtained is said to 
{\sl come from contiguous relations}.  
Finding a solution in this manner is called the {\sl method of 
contiguous relations}.  
Condition \eqref{eqn:cfcr} yields an overdetermined system of algebraic 
equations for $(a,b;x)$ with a given triple $(p, q, r) \in \Z^3$.    
\begin{problem} \label{prob:cfcr} 
When does an integral solution to Problem $\rII$ come from contiguous 
relations ? 
\end{problem}
\par
Given a data $\lambda = (p,q,r; a,b; x)$ and a positive integer 
$k \in \N$, the new data $k \lambda := (k p, k q, kr; a,b; x)$ is 
referred to as the {\sl multiplication} of $\lambda$ by $k$.  
It is said to be {\it nontrivial} if $k \ge 2$, in particular  
$2 \lambda$ is the {\sl duplication} of $\lambda$.  
In view of definition \eqref{eqn:f} we have  
\begin{equation} \label{eqn:f-mult}
f(w; k \lambda) = f(k w; \lambda) \qquad (k \in \N). 
\end{equation}
Gauss's multiplication formula for the gamma function 
\cite[Theorem 1.5.2]{AAR}:  
\begin{equation} \label{eqn:mult}
\varGamma(k w) = (2 \pi)^{(1-k)/2} \cdot k^{k w-1/2} \cdot 
\prod_{j=0}^{k-1} 
\varGamma\left(w + \textstyle \frac{j}{k}\right) \qquad (k \in \N)   
\end{equation}
implies that if $\lambda$ is a solution to Problem $\rI$ with gamma 
product formula \eqref{eqn:gpf} then $k \lambda$ is also a solution to 
the same problem with the ``multiplied" gamma product formula 
\begin{equation} \label{eqn:gpf-mult}
f(w; k \lambda) = C_k \cdot d_k^w \cdot 
\prod_{j=0}^{k-1} 
\dfrac{\varGamma\left(w+ \frac{u_1 + j}{k}\right) \cdots 
\varGamma\left(w+ \frac{u_m +j}{k}\right)}{\varGamma\left(w+ \frac{v_1 + j}{k}\right) 
\cdots \varGamma\left(w+ \frac{v_n +j}{k}\right)}, 
\end{equation}
where $C_k := C \cdot (2 \pi)^{(k-1)(n-m)/2} \cdot 
k^{u-v+(n-m)/2}$ with $u := u_1+\cdots+u_m$, $v := v_1+\cdots+v_n$, and 
$d_k := d^k \cdot k^{k(m-n)}$.  
In a similar manner, if $\lambda$ is a solution to Problem $\rII$ with 
rational function $R(w; \lambda)$ as in condition \eqref{eqn:ocf} then 
$k \lambda$ is also a solution to the same problem with  
\begin{equation} \label{eqn:R-mult}
R(w; k \lambda) = \prod_{j=0}^{k-1} R(k w + j; \lambda)
= d_k \cdot 
\prod_{j=0}^{k-1} 
\dfrac{\left(w+ \frac{u_1 + j}{k}\right) \cdots 
\left(w+ \frac{u_m +j}{k}\right)}{\left(w+ \frac{v_1 + j}{k}\right) 
\cdots \left(w+ \frac{v_n +j}{k}\right)}. 
\end{equation}
\par 
For any rational solution $\lambda = (p,q,r;a,b;x)$ to Problem $\rII$ 
there exists a positive integer $k$ such that $k \bp := (k p, k q; k r) 
\in \Z^3$, so it makes sense to ask whether the integral solution 
$k \lambda$ comes from contiguous relations.  
If the answer is ``yes" for some $k \in \N$, we say that the rational 
solution $\lambda$ {\sl essentially comes from contiguous relations}. 
As a variant of Problem \ref{prob:cfcr} we pose:   
\begin{problem} \label{prob:cfcr-ess}
When does a rational solution to Problem $\rII$ {\sl essentially} come from 
contiguous relations ? 
\end{problem}
\par
It is easy to see that formula \eqref{eqn:gpf} can be recovered from formula 
\eqref{eqn:gpf-mult} via relation \eqref{eqn:f-mult}.   
Indeed, we have only to replace $w$ by $w/k$ in \eqref{eqn:gpf-mult} and 
use the multiplication formula \eqref{eqn:mult} in the other way round 
(see Lemma \ref{lem:divisible} for a relevant discussion).   
However, it is totally unclear whether formula \eqref{eqn:R} can be 
recovered from formula \eqref{eqn:R-mult}, because with respect to the 
multiplication $R(w; \lambda)$ has no such simple structure as the 
relation \eqref{eqn:f-mult} for $f(w; \lambda)$, so we have to establish 
a bilateral link between Problems $\rI$ and $\rII$.  
This is one of several reasons for the importance of 
discussing Problem \ref{prob:back}. 
We refer to Remark \ref{rem:kummer} for another reason.        
\section{Symmetries} \label{sec:sym}
There are some transformations indispensable in discussing Problems 
$\rI$ and $\rII$. 
They consist of well-known classical symmetries in 
\S\ref{subsec:classical}, as well as new ones we shall introduce 
in \S\ref{subsec:d-r}.    
\subsection{Classical Symmetries} \label{subsec:classical} 
The hypergeometric series $\hgF(\alpha, \beta; \gamma; z)$ admits the 
following well-known symmetries: 
\begin{subequations} \label{eqn:transf}
\begin{alignat}{2}
\hgF(\alpha, \, \beta; \, \gamma; \, z)  
&= \hgF(\beta, \, \alpha; \, \gamma; \, z)  
& \qquad & \mbox{(trivial)} \label{eqn:transf0} \\ 
&= (1-z)^{\gamma-\alpha-\beta} \hgF(\gamma-\alpha, \, \gamma-\beta; \, \gamma; \, z)  
& \qquad & \mbox{(Euler)} \label{eqn:transf1} \\
&= (1-z)^{-\alpha} \hgF(\alpha, \, \gamma-\beta; \, \gamma; \, z/(z-1))  
&\qquad & \mbox{(Pfaff)} \label{eqn:transf2} \\
&= (1-z)^{-\beta} \hgF(\gamma-\alpha, \, \beta; \, \gamma; \, z/(z-1)).   
&\qquad & \mbox{(Pfaff)} \label{eqn:transf3}    
\end{alignat}
\end{subequations}
Compatible with Problems $\rI$ and $\rII$, they induce symmetries on  
solutions to those problems:  
\begin{subequations} \label{eqn:sym}
\begin{alignat}{2}
\lambda = (p,q,r; a,b; x)  
& \mapsto (q,p,r; b,a; x) 
& \qquad & \mbox{(trivial)} \label{eqn:sym0} \\
& \mapsto (r-p, \, r-q, \, r; \, -a, \, -b; \, x)
& \qquad & \mbox{(Euler)} \label{eqn:sym1} \\
& \mapsto (p, \, r-q, \, r; \, a, \, -b; \,x/(x-1))
& \qquad & \mbox{(Pfaff)}  \label{eqn:sym2} \\
& \mapsto (r-p, \, q, \, r; \, -a, \, b; \, x/(x-1)). 
& \qquad & \mbox{(Pfaff)} \label{eqn:sym3}   
\end{alignat}
\end{subequations} 
The symmetries \eqref{eqn:sym} were already introduced and discussed 
in Iwasaki \cite[\S 1]{Iwasaki}. 
\subsection{Division into Several Components } \label{subsec:compo}
As to the argument $x$ in domain \eqref{eqn:real}, Pfaff's 
transformation \eqref{eqn:sym2} or \eqref{eqn:sym3} takes the negative 
unit interval $-1 < x < 0$ to a positive one $0 < x < 1/2$, so we have 
only to work on the positive unit interval $0 < x < 1$. 
Thus domain \eqref{eqn:real} can be reduced to the following. 
\begin{equation} \label{eqn:real2}
p, \, q \in \R; \qquad r > 0; \qquad a, \, b \in \R; \qquad 0 < x < 1.   
\end{equation} 
\par
According to the location of $(p, q)$ relative to $r$ the domain 
\eqref{eqn:real2} is divided into several components, the figures of 
which are depicted upon projected to the $(p, q)$ plane with a fixed 
value of $r > 0$ as in Figures \ref{fig:D} and \ref{fig:E}.   
First we consider the ``square" domain
\[ 
\cD := \{\, \lambda \in \R^6\,:\,  
0 < p < r, \quad 0 < q < r; \quad a, \, b \in \R; \quad 0 < x < 1 \,\},       
\]     
\begin{figure}[t]
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\label{fig:D}
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\put(14.9000,-18.8000){\makebox(0,0)[lb]{$\cI^{-*}$}
\put(20.5000,-18.9000){\makebox(0,0)[lb]{$\cI^{+*}$}
\put(18.3000,-22.3000){\makebox(0,0)[lb]{$\cI^{*-}$}
\put(18.3000,-15.6000){\makebox(0,0)[lb]{$\cI^{*+}$}
\end{picture
\end{center}
\caption{The domain $\cE$ and the region $\cI$.}
\label{fig:E}
\end{minipage} 
\end{figure}
which is further decomposed into three components  
$\cD = \cD^- \cup \cD^0 \cup \cD^+$, where     
\begin{align*}  
\cD^- &:= \{\, \lambda \in \R^6 \,:\,\, 
p > 0, \quad q > 0, \quad p+q < r; 
\quad a, \, b \in \R; \quad 0 < x < 1 \, \}, 
\\
\cD^+ &:= \{\, \lambda \in \R^6 \,:\,\, 
p < r, \quad q < r, \quad p+q > r; 
\quad a, \, b \in \R; \quad 0 < x < 1 \, \},  
\\
\cD^0 &:= \{\, \lambda \in \R^6 \,:\,\, 
p > 0, \quad q > 0, \quad p+q = r; 
\quad a, \, b \in \R; \quad 0 < x < 1 \, \}.        
\end{align*}     
Euler's transformation \eqref{eqn:sym1} swaps $\cD^-$ and $\cD^+$   
while preserving the diagonal $\cD^0$.  
Thus we have only to deal with $\cD^- \cup \cD^0$. 
Next we define 
$\cE := \cE^{*-} \cup \cE^{*+} \cup \cE^{-*} \cup \cE^{+*}$ by   
\begin{align*} 
\cE^{*-} &:= \{\, \lambda \in \R^6\,:\,\,  
0 < p < r, \quad q < 0; \quad a, \, b \in \R; \quad 0 < x < 1 \, \}, 
\\
\cE^{*+} &:= \{\, \lambda \in \R^6\,:\,\,  
0 < p < r, \quad q > r; \quad a, \, b \in \R; \quad 0 < x < 1 \, \}, 
\\
\cE^{-*} &:= \{\, \lambda \in \R^6\,:\,\, 
p < 0, \quad 0 < q < r; \quad a, \, b \in \R; \quad 0 < x < 1 \, \}, 
\\
\cE^{+*} &:= \{\, \lambda \in \R^6\,:\,\,   
p > r, \quad 0 < q < r; \quad a, \, b \in \R; \quad 0 < x < 1 \, \},         
\end{align*}     
where the first superscript $-$, $*$ or $+$ designates the interval  
$p < 0$, $0 < p < r$ or $r < p$ respectively, while the second 
superscript is subject to the same convention with respect to $q$. 
In a similar manner we define 
$\cI := \cI^{*-} \cup \cI^{*+} \cup \cI^{-*} \cup \cI^{+*}$ by   
\begin{align*} 
\cI^{*-} &:= \{\, \lambda \in \R^6\,:\,\,  
0 < p < r, \quad q = 0; \quad a, \, b \in \R; \quad 0 < x < 1 \, \}, 
\\
\cI^{*+} &:= \{\, \lambda \in \R^6\,:\,\,  
0 < p < r, \quad q = r; \quad a, \, b \in \R; \quad 0 < x < 1 \, \}, 
\\
\cI^{-*} &:= \{\, \lambda \in \R^6\,:\,\, 
p = 0, \quad 0 < q < r; \quad a, \, b \in \R; \quad 0 < x < 1 \, \}, 
\\
\cI^{+*} &:= \{\, \lambda \in \R^6\,:\,\,   
p = r, \quad 0 < q < r; \quad a, \, b \in \R; \quad 0 < x < 1 \, \}.          
\end{align*}     
The trivial and Euler's transformations \eqref{eqn:sym0} and 
\eqref{eqn:sym1} induce the transpositions:  
\[ 
\cE^{*-} \leftrightarrow \cE^{-*}, \quad \cE^{*+} \leftrightarrow \cE^{+*} 
\quad \mbox{by \eqref{eqn:sym0}}; \qquad      
\cE^{*-} \leftrightarrow \cE^{*+}, \quad \cE^{-*} \leftrightarrow \cE^{+*} 
\quad \mbox{by \eqref{eqn:sym1}}, 
\]
with the same rules applying to the regions $\cI^{\mathbf{\star\star}}$.  
Thus we have only to deal with $\cE^{*-} \cup \cI^{*-}$. 
\subsection{Duality and Reciprocity} \label{subsec:d-r}
We introduce two symmetries or transformations, which we call duality 
and reciprocity.     
\begin{definition} \label{def:duality} 
Define a transformation 
$\lambda = (p, q, r; a, b; x) \mapsto 
\lambda' := (p', q', r'; a', b'; x')$ by   
\begin{equation} \label{eqn:duality}
p' := p, \quad q':= q, \quad r':=r; \quad 
a' := 1- \frac{2 p}{r} - a, \quad b' := 1 - \frac{2 q}{r} - b; 
\quad x':= x,       
\end{equation}
which is referred to as {\sl duality}, where the $(p,q,r;x)$-component 
of $\lambda$ is kept unchanged.  
It is a well-defined involution whenever $r$ does not vanish.  
\end{definition}
Note that duality maps each of the components $\cD^{\pm}$, $\cD^0$, 
$\cE^{* \pm}$ and $\cE^{\pm *}$ bijectively onto itself. 
\par
Given a dual pair $(\lambda, \lambda')$, 
it is convenient to think of the three-by-two matrix 
\[
\begin{pmatrix}
p  & q \\ 
a  & b \\
a' & b' 
\end{pmatrix}.  
\]
Exchange of its columns represents the trivial symmetry \eqref{eqn:sym0} 
applied to $(\lambda, \lambda')$, whereas the exchange of its middle and 
bottom rows represents duality \eqref{eqn:duality} itself. 
By the omission of its top row $(p, q)$, the matrix above 
is abbreviated to the square matrix  
\begin{equation} \label{eqn:matrix}
\begin{pmatrix} 
a  & b \\
a' & b' 
\end{pmatrix}.    
\end{equation}
Two dual pairs are said to be {\sl equivalent} if their matrices are 
transitive via the column or row exchange or the composition of them.   
This concept will be used in \S \ref{sec:dr-fc} to reduce the work 
there.     
\begin{definition} \label{def:recip} 
Define a transformation $\lambda = (p, q, r; a, b; x) \mapsto 
\check{\lambda} := 
(\check{p}, \check{q}, \check{r}; \check{a}, \check{b}; \check{x})$ by  
\begin{equation} \label{eqn:recip}
\begin{split} 
\check{p} := -p, \qquad \check{q} :=-q, \qquad 
\check{r} :=r-p-q; \qquad \check{x} :=1-x,  \\[2mm] 
\check{a} := \dfrac{(r-q)(1-a)-p b}{r-p-q}, \qquad     
\check{b} := \dfrac{(r-p)(1-b)-q a}{r-p-q},     
\end{split}  
\end{equation}
which is referred to as {\sl reciprocity}. 
This is a well-defined involution on the domain 
\begin{equation} \label{eqn:real3}
p, \, q, \, r \in \R, \quad r(r-p-q) \neq 0; \qquad a, \, b \in \R; 
\qquad 0 < x < 1.     
\end{equation}
\end{definition}  
\par
As indicated in Figure \ref{fig:D-r}, two new components 
$\cF^{\pm}$ are introduced by   
\begin{figure}[t]
\begin{center}
\unitlength 0.1in
\begin{picture}( 62.4800, 28.0800)(  6.0000,-31.2400)
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\end{picture
\end{center}
\caption{Reciprocity between $\cD^-$ and $\cF^-$.}
\label{fig:D-r} 
\end{figure}
\begin{align*}  
\cF^- &:= \{\, \lambda \in \R^6 \,:\,\, 
p < 0, \quad q < 0, \quad r > 0; 
\quad a, \, b \in \R; \quad 0 < x < 1 \, \}, 
\\
\cF^+ &:= \{\, \lambda \in \R^6 \,:\,\, 
p > r, \quad q > r, \quad r > 0; 
\quad a, \, b \in \R; \quad 0 < x < 1 \, \},         
\end{align*}
where we have only to deal with $\cF^-$, because Euler's transformation 
\eqref{eqn:sym1} swaps $\cF^-$ and $\cF^+$.    
\begin{figure}[t]
\begin{center}
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\end{center}
\caption{Reciprocity between $\cE^{*-}$ and $\cE^{-*}$.}  
\label{fig:E-r} 
\end{figure}
As is indicated in Figures \ref{fig:D-r} and \ref{fig:E-r}, 
the reciprocity induces the following transpositions:  
\[ 
\cD^{-} \leftrightarrow \cF^-, \qquad   
\cE^{*-} \leftrightarrow \cE^{-*}.   
\]
\par
A chief idea underlying duality and reciprocity is the concept of an Ebisu 
symmetry to be developed in \S \ref{sec:kummer}, based on which these 
two symmetries will be constructed in \S \ref{sec:duality} and 
\S \ref{sec:r-cf}. 
\section{Main Results} \label{sec:results} 
The main results of this article are established on the foundation 
of our previous work \cite{Iwasaki}.     
So this section begins with a review of our previous results, upon 
rearranged somewhat in a manner suitable for the purposes of the 
present as well as forthcoming papers.            
\subsection{Review of the Previous Results} \label{subsec:review}
As to Problem \ref{prob:back}, Theorem 2.3 of \cite{Iwasaki} 
contains the following statement.     
\begin{theorem} \label{thm:back} 
In region $\cD \cup \cI \cup \cE$ any solution to Problem $\rII$ 
leads back to a solution to Problem $\rI$ and hence Problems $\rI$ and 
$\rII$ are equivalent. 
\end{theorem}
\par
Thus in this region we can speak of a solution without specifying 
to which problem it is a solution. 
The next results are about solutions on the square domain $\cD$, 
which are a collection of some parts of Theorems 2.1, 2.3, 2.4, 2.7 and 
2.10, as well as of Remark 2.5 in \cite{Iwasaki}.      
\begin{theorem} \label{thm:D} 
The following hold in $\cD$. 
Let $\lambda = (p,q,r;a,b;x) \in \cD$ in what follows. 
\begin{enumerate}
\item Every solution in $\cD^0$ is elementary, while every solution in 
$\cD^{\pm}$ is non-elementary. 
All elementary solutions $\lambda$ are at the center $\bullet$ of the 
square $\cD$ in Figure $\ref{fig:D}$, or more precisely,     
\[
p = q = \frac{r}{2} > 0; \quad a = i, \quad b = j - \frac{1}{2}, 
\quad i, j \in \Z; \quad 0 < x < 1,    
\]
where $a$ and $b$ are exchangeable by symmetry while $r$ and 
$x$ are free.  
The corresponding $f(w; \lambda)$ is a degenerate hypergeometric 
functions with a dihedral monodromy group,   
\[
f(w) = S_{ij}(r w; x) \cdot \left(\ts \frac{1+\sqrt{1-x}}{2}\right)^{1-r w},   
\]
studied by Vidunas \cite[Theorem 3.1]{Vidunas2}, where $S_{ij}(w;x)$ is a 
rational function of $w$ explicitly representable in terms of terminating 
Appell's $F_3$ series (see Iwasaki \cite[Theorem 2.1]{Iwasaki}). 
\item Any solution $\lambda \in \cD^{\pm}$ falls into one of the 
following two types: 
\begin{enumerate}
\item[$(\rA)$] \, $p$, $q$, $r \in \Z$ and $0 \neq r-p-q \equiv 0 \mod 2$,   
\item[$(\rB)$] \, $p$, $q \in \frac{1}{2} + \Z$ and $r \in \Z$.  
\end{enumerate}  
\item Any solution of type $(\rA)$ in $\cD^{\pm}$ comes from contiguous relations. 
\item If $\lambda \in \cD^{\pm}$ is a solution of type $(\rB)$ then   
its duplication $2 \lambda := (2p, 2q, 2r;a,b;x) \in \cD^{\pm}$ is  
an $(\rA)$-solution, so any $(\rB)$-solution in $\cD^{\pm}$ essentially 
comes from contiguous relations. 
\end{enumerate}
\end{theorem}
\par
In \cite[Theorem 2.4]{Iwasaki} we included certain additional 
conditions involving $(a, b)$ to describe the dichotomy of types 
$(\rA)$ and $(\rB)$.  
Those conditions are omitted here because they will not be used in this 
article.      
In any case assertion (2) above implies that the $(p,q,r)$-component 
of any $(\rA)$-solution $\lambda = (p,q,r;a,b;x) \in \cD^-$ must belong 
to the integer domain  
\begin{equation} \label{eqn:D-m}
D^{-}_{\rA} := \{\, \bp = (p,q;r) \in \Z^3 \,:\, p > 0, \,\, q > 0, \,\,  
0 < r-p-q \equiv 0 \mod 2 \, \}.     
\end{equation}
In view of assertion (4) above we may concentrate our attention on 
$(\rA)$-solutions. 
Then by assertion (3) the method of contiguous relations developed in 
\cite[\S11]{Iwasaki} produces the following results about the numbers 
$x$ and the gamma product formulas (GPFs) for $(\rA)$-solutions.        
\begin{theorem} \label{thm:sol-A}
Let $\lambda = (p,q,r;a,b;x) \in \cD^-$ be any $(\rA)$-solution with 
$\bp = (p,q;r) \in D^-_{\rA}$.  
\begin{enumerate}
\item $x$ must be an algebraic number as a real root in 
$0 < z < 1$ of an algebraic equation
\begin{equation} \label{eqn:Y} 
Y(z) = Y(z; \bp) = 0, 
\end{equation}
where $Y(z) \in \Z[w]$ depends only on $\bp$ and is defined 
implicitly as follows: If we put  
\begin{align*}
\varDelta(z) &:= (p-q)^2 z^2 - 2 \{(p+q)r-2 p q\} z + r^2, \\[1mm]  
Z_{\pm}(z) &:= 
\big\{r+(p-q)z \pm \sqrt{\varDelta} \big\}^p 
\big\{r-(p-q)z \pm \sqrt{\varDelta} \big\}^q
\big\{(2 r-p-q)z-r \mp \sqrt{\varDelta} \big\}^{r-p-q},      
\end{align*}
then there exist unique polynomials $X(z)$, $Y(z) \in \Z[z]$ such that
\[ 
Z_{\pm}(z) = X(z) \pm Y(z) \, \sqrt{\varDelta}.
\]
\item There exists a positive constant $C > 0$ such that a gamma 
product formula  
\begin{equation} \label{eqn:gpf2} 
f(w; \lambda) = C \cdot d^w \cdot 
\dfrac{\prod_{i=0}^{r-1}\varGamma\left(w+\frac{i}{r} \right)}{\prod_{i=1}^r 
\varGamma\left(w+v_i \right)} 
\end{equation} 
holds true, where the number $d$ is given by 
\begin{equation} \label{eqn:d}
d = \dfrac{r^r}{\sqrt{p^p q^q (r-p)^{r-p} (r-q)^{r-q} x^r (1-x)^{p+q-r}}},    
\end{equation} 
while $v_1, \dots, v_r$ are such numbers that sum up to
\begin{equation} \label{eqn:sum} 
v_1 + \cdots + v_r = \dfrac{r-1}{2}, 
\end{equation}
and that admit a division relation in $\C[w]$,     
\begin{equation} \label{eqn:division}
\prod_{i=1}^r \left(w+v_i\right) \,\, \Big| \,\,    
\prod_{i=1}^{p-1}\left(w+{\ts\frac{i+a}{p}}\right) 
\prod_{i=1}^{q-1}\left(w+{\ts\frac{i+b}{q}}\right) 
\prod_{j=0}^{r-p-1}\left(w+{\ts\frac{j-a}{r-p}}\right) 
\prod_{j=0}^{r-q-1}\left(w+{\ts\frac{j-b}{r-q}}\right).      
\end{equation} 
\end{enumerate}
\end{theorem}
\par 
Here algebraic equation \eqref{eqn:Y} stems from 
\cite[assertion (1) of Theorem 2.10]{Iwasaki}. 
For any triple $\bp = (p, q; r) \in D^-_{\rA}$ the equation \eqref{eqn:Y} 
has at least one root in the interval $0 < z < 1$ by 
\cite[assertion (2) of Lemma 11.7]{Iwasaki}.  
Formula \eqref{eqn:d} comes from \cite[formula (20) in Theorem 2.4]{Iwasaki};  
condition \eqref{eqn:division} is from 
\cite[assertion (3) of Theorem 2.10]{Iwasaki}, whereas formula 
\eqref{eqn:gpf2} follows from a combination of 
\cite[assertion (3) of Theorem 2.10]{Iwasaki} and 
\cite[Proposition 5.4]{Iwasaki} by taking $S(w) \equiv 1$, $m=r$ and 
$u_i = (i-1)/r$ $(i = 1, \dots, r)$ in \cite[formula (57)]{Iwasaki}. 
Condition \eqref{eqn:sum} is derived by adapting the equality 
$v := v_1 + \cdots + v_r = u + s_0$ in \cite[Proposition 5.4]{Iwasaki} 
to the current situation where $u := u_1 +\cdots+ u_r = \sum_{i=1}^r 
(i-1)/r = (r-1)/2$ and $s_0 = 0$.  
In GPF \eqref{eqn:gpf2} some but not all gamma factors in the numerator 
may coincide pairwise with (and hence can be canceled by) the same 
number of gamma factors in the denominator. 
\par
Article \cite{Iwasaki} mainly concerns the domain $\cD$, 
but it also contains a few results on $\cI$ and $\cE$. 
\begin{theorem} \label{thm:I} 
The following hold in $\cI^{*-}$. 
Let $\lambda = (p, q=0, r; a, b; x) \in \cI^{*-}$ in what follows. 
\begin{enumerate}
\item $\lambda$ is an elementary solution if and only if 
$b \in \Z_{\le 0}$, in which case the hypergeometric series 
that defines $f(w; \lambda)$ is terminating, so that $f(w; \lambda)$ 
itself is a rational function of $w$. 
\item For any non-elementary solution $\lambda \in \cI^{*-}$ we have 
$p$, $r \in \Z$, $p \equiv r \mod 2$ and $b = 1/2$. 
\item Any non-elementary solution comes from contiguous relations.  
\end{enumerate}
\end{theorem}
\par
Here assertion (1) comes from \cite[Theorem 2.1]{Iwasaki}; assertion (2) 
follows from a combination of \cite[Theorem 2.3]{Iwasaki}, 
\cite[formula (80b) in Proposition 8.1]{Iwasaki} and 
\cite[Lemma 10.1]{Iwasaki}, while assertion (3) stems from 
\cite[Proposition 11.2]{Iwasaki}. 
It is remarkable that one must have $b = 1/2$ for any 
non-elementary solution in $\cI^{*-}$.  
Certainly all previously known examples enjoy this condition 
and this empirical fact has now been established logically and 
theoretically.          
\begin{proposition} \label{prop:E}
The following hold in $\cE^{*-}$. 
Let $\lambda = (p,q,r;a,b;x) \in \cE^{*-}$ in what follows.  
\begin{enumerate}
\item Every solution in $\cE^{*-}$ is non-elementary.  
\item For any solution $\lambda \in \cE^{*-}$ we must have 
$p$, $r \in \Z$ and $p \equiv r \mod 2$. 
\item Any solution $\lambda \in \cE^{*-}$ with $q \in \Z$ 
comes from contiguous relations.  
\end{enumerate} 
\end{proposition}
\par
Here assertion (1) comes from \cite[Theorem 2.1]{Iwasaki};  
assertion (2) follows from a combination of \cite[Theorem 2.3]{Iwasaki}, 
\cite[formula (80c) in Proposition 8.1]{Iwasaki} and 
\cite[Lemma 10.1]{Iwasaki}, while assertion (3) stems from 
\cite[Proposition 11.2]{Iwasaki}. 
It is not yet known whether $\cE^{*-}$ contains a solution with 
non-integral or irrational $q \in \R$; for this reason 
Proposition \ref{prop:E} is not called Theorem \ref{prop:E}. 
\subsection{Statement of the New Results} \label{subsec:new}
We proceed to stating the new results of this article. 
Most of them are results on domain $\cD^-$ and are a substantial 
sharpening of the results in the previous article \cite{Iwasaki}, 
becoming more arithmetic, while the remaining ones are on the new 
domain $\cF^-$ which is focused on for the first time in this article.    
We also include a small but important new result on domain $\cE^{*-}$.    
All these are largely indebted to the new ingredients of this article, 
that is, duality and reciprocity. 
\par
The first result is about the arithmetic properties of 
$(p,q;r)$ for $(\rA)$-solutions in $\cD^-$.  
\begin{theorem} \label{thm:pqr} 
For any $(\rA)$-solution $\lambda = (p,q,r;a,b;x) \in \cD^-$ the 
integer triple $\bp = (p,q;r)$ must belong to $D^-_{\rA}$ and 
moreover satisfy the division relations:  
\begin{equation} \label{eqn:dr} 
(\, p|r \quad \mbox{or} \quad p|(r-p-q) \,) \quad \mbox{and} \quad 
(\, q|r \quad \mbox{or} \quad q|(r-p-q) \,).
\end{equation}
\end{theorem}
\par
This theorem does not refer to which option comes true in each ``or" 
phrase of condition \eqref{eqn:dr}, while item (1) of Theorem \ref{thm:ab} 
below has something to do with this question. 
Theorem \ref{thm:pqr} is presented at the beginning because it 
concerns the principal part $\bp$ of the data $\lambda$, but its proof 
will be given rather late in \S 8 (Proposition \ref{prop:dr}), preceded 
by the proofs of Theorems \ref{thm:gpf-d}, \ref{thm:ab}, \ref{thm:ab-rf} 
and \ref{thm:D-r} below.    
The second result is regarding the duality for $(\rA)$-solutions in $\cD^-$.   
\begin{theorem}\label{thm:gpf-d} 
The duality $\lambda \mapsto \lambda'$ defined by \eqref{eqn:duality} 
yields an involution on the set of all $(\rA)$-solutions in $\cD^-$.   
For those solutions it induces a transformation of gamma product 
formulas:   
\begin{equation} \label{eqn:gpf-d}
f(w; \lambda) = C \cdot d^w \cdot 
\dfrac{\prod_{i=0}^{r-1} \varGamma\left(w+\frac{i}{r}\right)}{\prod_{i=1}^{r} \varGamma\left(w+v_i\right)} 
\quad \mapsto \quad 
f(w; \lambda') = C' \cdot d^w \cdot 
\dfrac{\prod_{i=0}^{r-1} \varGamma\left(w+\frac{i}{r}\right)}{\prod_{i=1}^{r} \varGamma\left(w+v_i'\right)},     
\end{equation}
where $C'$ is a positive constant, $d$ is the number given by 
formula \eqref{eqn:d}, while 
\begin{equation} \label{eqn:vi-p}
v_i':= 1 - \dfrac{2}{r} - v_i^* \qquad (i = 1, \dots, r), 
\end{equation}
with $v_1^*, \dots, v_r^*$ being the numbers complementary to 
$v_1, \dots, v_r$, to the effect that   
\begin{equation} \label{eqn:vi-s} 
\resizebox{0.9\hsize}{!}{$\displaystyle
\prod_{i=1}^r \left(w+v_i\right) \prod_{i=1}^r \left(w+v_i^*\right) 
= 
\prod_{i=0}^{p-1}\left(w+{\ts\frac{i+a}{p}}\right) 
\prod_{i=0}^{q-1}\left(w+{\ts\frac{i+b}{q}}\right) 
\prod_{j=0}^{r-p-1}\left(w+{\ts\frac{j-a}{r-p}}\right) 
\prod_{j=0}^{r-q-1}\left(w+{\ts\frac{j-b}{r-q}}\right). $}       
\end{equation}
\end{theorem}
\par
Notice that condition \eqref{eqn:vi-s} is well defined, since we have 
division relation \eqref{eqn:division}. 
The proof of Theorem \ref{thm:gpf-d} will be given in 
\S \ref{sec:duality} (see Proposition \ref{prop:duality}).  
Notice also that the first and second products on the right-hand side of 
division relation \eqref{eqn:division} are taken from $i = 1$, whereas 
those of equation \eqref{eqn:vi-s} are from $i = 0$. 
This subtle difference eventually grows into the following significant 
result about the arithmetic nature of $(a, b)$ as well as its dual 
$(a', b')$. 
\begin{theorem} \label{thm:ab} 
For any $(\rA)$-solution $\lambda = (p,q,r;a,b;x) \in \cD^-$ and its 
dual $(\rA)$-solution $\lambda' = (p,q,r;a',b';x) \in \cD^-$, the pair 
$(\lambda, \lambda')$ must be subject to the conditions listed in 
Table $\ref{tab:ab}$, which consist of the following four items:  
\begin{enumerate} 
\item two division relations for the integer triple $\bp = (p,q;r)$,   
\item a formula for $(a,b)$ in terms of $\bp$ and a pair  
of nonnegative integers $(i,j)$,    
\item a formula for $(a',b')$ in terms of $\bp$ and a pair 
of nonnegative integers  $(i',j')$,     
\item two $\Z$-linear equations for quadruple $(i,j;i',j') \in \Z_{\ge 0}^4$, 
\end{enumerate}
where for each $\nu \in \{1,2,3,4\}$ item $(\nu)$ is exhibited in 
column $(\nu)$ of Table $\ref{tab:ab}$. 
\begin{table}[p]
\begin{center}
\rotatebox{90}{
\begin{tabular}{|c|l|l|l|l|c|}
\hline
     &     &     &     &     &     \\[-3mm]
     & (1) & (2) & (3) & (4) & (5) \\[1mm]
\hline 
     &     &     &     &     &     \\[-3mm]
case & {\small division relations} & {\small formula for $(a, b)$} & {\small formula for $(a', b')$} & 
$\Z$-linear equations for $(i, j; i', j')$ & type \\[2mm]
\hline 
     &     &     &     &     &     \\[-3mm]
$1$ & 
$\begin{array}{l} p|r \\[1mm] q|r \end{array}$ & 
$\begin{array}{l} a = \frac{i}{r_p} \\[2mm] b = \frac{j}{r_q} \end{array}$ & 
$\begin{array}{l} a'= \frac{i'}{r_p} \\[2mm] b'= \frac{j'}{r_q} \end{array}$ & 
$\begin{array}{l} i+i'= r_p-2 \\[1mm] j+j' = r_q-2 \end{array}$ & 
$\begin{matrix} \rI & \rI \\[1mm] \rI & \rI \end{matrix}$ \\[5mm]
\hline  
     &     &     &     &     &     \\[-3mm] 
$2$ & 
$\begin{array}{l} p|(r-p-q) \\[2mm] q|(r-p-q) \end{array}$ & 
$\begin{array}{l} a  = \frac{(r-p)i-q j}{r(r-p-q)_p} \\[2mm] b  = \frac{(r-q)j-p i}{r(r-p-q)_q} \end{array}$ & 
$\begin{array}{l} a' = \frac{(r-p)i'-q j'}{r(r-p-q)_p} \\[2mm] b' = \frac{(r-q)j'-p i'}{r(r-p-q)_q} \end{array}$ & 
$\begin{array}{l} i+i' = (r-p-q)_p \\[2mm] j+j' = (r-p-q)_q \end{array}$ & 
$\begin{matrix} \rII & \rII \\[2mm] \rII & \rII \end{matrix}$ \\[6mm]
\hline 
     &     &     &     &     &     \\[-3mm]
$3$ & 
$\begin{array}{l} p|r \\[2mm] q|(r-p-q) \end{array}$ & 
$\begin{array}{l} a  = \frac{i}{r_p} \\[2mm] b  = \frac{r_p j-i}{r_p (r-p)_q} \end{array}$ &
$\begin{array}{l} a' = \frac{i'}{r_p} \\[2mm] b' = \frac{r_p j'-i'}{r_p (r-p)_q} \end{array}$ &
$\begin{array}{l} i+i' = r_p-2 \\[2mm] j+j' = (r-p-q)_q \end{array}$ & 
$\begin{matrix} \rI & \rII \\[2mm] \rI & \rII \end{matrix}$ \\[6mm]
\hline 
     &     &     &     &     &     \\[-3mm]    
$4$ & 
$\begin{array}{l} p|r \\[2mm] q|r_p(r-p-q) \end{array}$  & 
$\begin{array}{l} a = \frac{i}{r_p} \\[2mm] b  = \frac{q j}{r} \end{array}$ & 
$\begin{array}{l} a' = \frac{i'}{r_p} \\[2mm] b' = \frac{r_p j'-i'}{(r_p(r-p))_q} \end{array}$ & 
$\begin{array}{l} i+i' = r_p-2 \\[2mm] {\scriptstyle i+(r_p-1)j+r_p j' = (r_p(r-p-q))_q} 
\end{array}$ & 
$\begin{matrix} \rI & \rI \\[2mm] \rI & \rII \end{matrix}$ \\[6mm]
\hline   
     &     &     &     &     &     \\[-3mm]    
$5$ & 
$\begin{array}{l} p|(r-p-q) \\[2mm] q|r(r-p-q)_p \end{array}$  & 
$\begin{array}{l} a = \frac{(r-p)i-q j}{r(r-p-q)_p} \\[2mm] b  = \frac{(r-q)_p j-i}{(r(r-p-q)_p)_q} \end{array}$ & 
$\begin{array}{l} a' = \frac{r i'-q j'}{r(r-q)_p} \\[2mm] b' = \frac{q j'}{r} \end{array}$ & 
$\begin{array}{l} i+i' = (r-p-q)_p \\[2mm] {\scriptstyle i'+(r-q)_p j+ (r-p-q)_p j' = ((r-q)(r-p-q)_p)_q} \end{array}$ & 
$\begin{matrix} \rII & \rII \\[2mm] \rII & \rI \end{matrix}$ \\[6mm] 
\hline   
     &     &     &     &     &     \\[-3mm] 
$6$ & 
$\begin{array}{l} p|r(r-p-q) \\[2mm] q|r(r-p-q) \end{array}$ & 
$\begin{array}{l} a  = \frac{p i}{r} \\[2mm] b  = \frac{r j-p i}{(r(r-p))_q} \end{array}$ &
$\begin{array}{l} a' = \frac{r i'-q j'}{(r(r-q))_p} \\[2mm] b' = \frac{q j'}{r} \end{array}$ &
$\begin{array}{l} {\scriptstyle (r-p-q)i+(r-p)i'+q j = ((r-p)(r-p-q))_p} \\[2mm] 
{\scriptstyle (r-p-q)j'+p i' + (r-q)j = ((r-q)(r-p-q))_q} \end{array}$ & 
$\begin{matrix} \rI & \rII \\[2mm] \rII & \rI \end{matrix}$ \\[6mm]
\hline 
\end{tabular}
} 
\end{center}
\caption{Candidates for dual pairs of $(\rA)$-solutions in $\cD^-$.}
\label{tab:ab}
\end{table}
There are a total of six cases or patterns, up to equivalence, 
according to the type of the dual pair $(\lambda, \lambda')$ 
shown in column $(5)$, the notion of which will be defined 
in \S $\ref{sec:dr-fc}$ and may be skipped until there. 
Moreover,  
\begin{equation} \label{eqn:estimate}
a, b, \, a', b', \, v_1, \dots, v_r, \, v_1', \dots, v_r' \in 
\Q \cap [0, \, 1),  
\end{equation}
where $v_1, \dots, v_r$ and $v_1', \dots, v_r'$ are the numbers 
appearing in formula \eqref{eqn:gpf-d}.           
\end{theorem}
\par
Theorem \ref{thm:ab} will be proved in \S \ref{sec:dr-fc} (see 
Proposition \ref{prop:ab}) after a preliminary discussion in 
\S \ref{sec:r-cf}.   
\begin{remark} \label{rem:ab} 
A few remarks about Table \ref{tab:ab} should be in order at this stage. 
\begin{enumerate}
\item In Table \ref{tab:ab} (and also in \S\ref{sec:dr-fc}--\S\ref{sec:dr}), 
for positive integers $s$ and $t$ we write $s_t := s/t$ when and only when 
$t|s$, i.e., $t$ divides $s$.  
It is a convenient notation which indicates at once that $s_t$ stands for 
a positive integer; it also saves space when $s$ is a large expression, 
but for example $(r-p-q)_p$ should not be confused with a rising  
factorial number.    
\item The division relations in column (1) have something to do with 
those in condition \eqref{eqn:dr}.   
Indeed, in case 1 the divisions $p|r$ and $q|r$ must occur in  
the two ``or" phrases in \eqref{eqn:dr}. 
Cases 2 and 3 are similar to case 1, that is, there are unique 
choices in the two ``or"s.     
In case 4 the division $p|r$ must happen in the former ``or", 
while either option in the latter ``or" guarantees the condition 
$q|r_p(r-p-q)$.  
Case 5 is similar to case 4. 
In case 6 any combination in the two ``or"s implies $p|r(r-p-q)$ 
and $q|r(r-p-q)$. 
\item In each case of Table \ref{tab:ab} the $\Z$-linear 
equations for $(i,j;i',j') \in \Z_{\ge0}^4$ are of the form 
\begin{equation} \label{eqn:z-linear}
\mu_1 \, i + \mu_2 \, j + \mu_3 \, i' + \mu_4 \, j' = \mu_5, \qquad 
\nu_1 \, i + \nu_2 \, j + \nu_3 \, i' + \nu_4 \, j' = \nu_5, 
\end{equation}
where for each $k \in \{1,2,3,4\}$ the coefficients $\mu_k$ and $\nu_k$ 
are nonnegative integers with a positive sum $\mu_k + \nu_k \ge 1$ so that 
system \eqref{eqn:z-linear} cannot have infinitely many solutions. 
\end{enumerate}
\end{remark}
\par
A data $\lambda = (p,q,r;a,b;x) \in \cD^-$ with $\bp = (p,q;r) \in 
D_{\rA}^-$ is said to be a {\sl candidate} for an $(\rA)$-solution, 
if $\bp$ satisfies condition \eqref{eqn:dr} and $\lambda$ is subject 
to one of the six patterns in Table \ref{tab:ab}. 
By Theorems \ref{thm:pqr} and \ref{thm:ab}, a data cannot be an 
$(\rA)$-solution unless it is a candidate, but it may (perhaps quite 
frequently) happen that a candidate is not an actual $(\rA)$-solution. 
\par
Theorem \ref{thm:ab} together with assertion (1) of Theorem \ref{thm:sol-A} 
gives the following.    
\begin{theorem} \label{thm:ab-rf} 
For any $(\rA)$-solution $\lambda = (p,q,r;a,b;x) \in \cD^-$ the numbers 
$a$ and $b$ must be rational while $x$ must be algebraic. 
Given any integer triple $\bp = (p,q;r) \in D_{\rA}^-$, there are no 
or only a finite number of $(\rA)$-solutions $\lambda \in \cD^-$ with 
the prescribed principal data $\bp$.    
\end{theorem} 
\par
We turn our attention to the reciprocity $\lambda \mapsto \check{\lambda}$ 
defined by formula \eqref{eqn:recip}.  
Considering the reciprocals of $(\rA)$-solutions on $\cD^-$ not only 
produces new solutions on the target domain $\cF^-$, but also brings 
us a deeper understanding of the original solutions on the source 
domain $\cD^-$.     
\begin{theorem} \label{thm:D-r}  
For any $(\rA)$-solution $\lambda = (p,q,r;a,b;x) \in \cD^-$  
there exists a division relation
\begin{equation} \label{eqn:division2} 
\prod_{i=0}^{p-1}\left(w+{\ts\frac{i+a}{p}}\right) 
\prod_{i=0}^{q-1}\left(w+{\ts\frac{i+b}{q}}\right) 
\, \Big| \, \prod_{i=1}^r \left(w+v_i\right) 
\qquad \mbox{in \, $\Q[w]$},    
\end{equation} 
which allows us to rearrange the numbers $v_1, \dots,v_r$ in formula   
\eqref{eqn:gpf2} so that 
\begin{equation} \label{eqn:vi-pq}
\prod_{i=0}^{p-1}\left(w+{\ts\frac{i+a}{p}}\right) 
\prod_{i=0}^{q-1}\left(w+{\ts\frac{i+b}{q}}\right) 
= \prod_{i=r-p-q+1}^{r} \left(w+v_i\right).  
\end{equation}
With this convention the reciprocity $\lambda \mapsto \check{\lambda}$ 
sends any $(\rA)$-solution $\lambda \in \cD^-$ to a new solution 
$\check{\lambda}$ to Problem $\rI$ in $\cF^-$, inducing a  
transformation of gamma product formulas:   
\begin{equation} \label{eqn:gpf-c}
\resizebox{0.8\hsize}{!}{$\displaystyle
f(w; \lambda) = C \cdot d^w \cdot 
\dfrac{\prod_{i=0}^{r-1}\varGamma\left(w+\frac{i}{r} \right)}{\prod_{i=1}^r 
\varGamma\left(w+v_i \right)} \,\, \mapsto \,\,  
f(w; \check{\lambda}) = \check{C} \cdot \check{d}^w \cdot 
\dfrac{\prod_{i=0}^{r-p-q-1}\varGamma\left(w+\frac{i}{r-p-q} \right)}{
\prod_{i=1}^{r-p-q} \varGamma\left(w+ \check{v}_i \right)}, $}  
\end{equation}
where $\check{C}$ is a positive constant and 
\begin{align}
\check{d} &= (r-p-q)^{r-p-q} 
\sqrt{\frac{p^p \, q^q \, x^r}{(r-p)^{r-p} \, (r-q)^{r-q} \, (1-x)^{r-p-q}}}, 
\label{eqn:d-c} 
\\[3mm]
\check{v}_i &= v_i - \frac{1-a-b}{r-p-q} \qquad (i = 1, \dots, r-p-q). 
\label{eqn:vi-c} 
\end{align}
\end{theorem}
\par
Theorem \ref{thm:D-r} will be proved in \S\ref{sec:dr-fc} as a main part of  
Proposition \ref{prop:division}.  
This theorem concerns the reciprocity in the direction $\cD^- \to \cF^-$. 
Starting from $\cF^-$ take the reciprocity in the other way round 
$\cF^- \to \cD^-$ and then use Theorem \ref{thm:D-r} to return home in 
$\cF^-$.  
This idea leads to the following solutions to Problems  
\ref{prob:back}--\ref{prob:cfcr-ess} on the domain $\cF^-$.         
\begin{theorem} \label{thm:F} 
The following hold on $\cF^-$.  
Let $\lambda = (p, q, r; a, b; x) \in \cF^-$ in what follows.  
\begin{enumerate} 
\item Any solution $\lambda \in \cF^-$ to Problem $\rI$ or $\rII$ 
is non-elementary.    
\item If $\lambda \in \cF^-$ is an integral solution to Problem $\rII$, 
then $\lambda$ comes from contiguous relations with its reciprocal 
$\check{\lambda}$ being an $(\rA)$-solution in $\cD^-$,   
in other words, $\lambda$ is the reciprocal of an $(\rA)$-solution 
$\check{\lambda} \in \cD^-$ so that $r$ must be an even positive 
integer and $\lambda$ becomes a solution to Problem $\rI$ with  
$f(w; \lambda)$ admitting a GPF as stated in assertion $(3)$ below.  
\item If $\lambda \in \cF^-$ is a rational solution to Problem $\rII$, 
then $\lambda$ essentially comes from contiguous relations, 
$r$ must be a positive integer but not necessarily even; $a$, $b \in \Q$;   
$x$ algebraic, and $\lambda$ becomes a solution to Problem $\rI$ with 
$f(w; \lambda)$ having a GPF of the form  
\begin{equation} \label{eqn:gpf-F} 
f(w; \lambda) = C \cdot d^w \cdot 
\dfrac{\prod_{i=0}^{r-1} \varGamma \left( w+\frac{i}{r}\right) }{\prod_{i=1}^{r} 
\varGamma\left(w+v_i\right)}, 
\end{equation}
where $C$ is a positive constant, $d$ is a positive algebraic number defined 
by  
\begin{equation} \label{eqn:d-F}
d = r^r \sqrt{\dfrac{|p|^{|p|} \, |q|^{|q|} \, 
(1-x)^{r-p-q}}{(r-p)^{r-p} \, (r-q)^{r-q} \, x^r}}, 
\end{equation}
and $v_1, \dots, v_r$ are such numbers that satisfy the following conditions: 
\begin{equation} \label{eqn:vi-F} 
v_1 + \cdots + v_r = \frac{r-1}{2}, \quad  
v_1, \dots, v_r \in \left(\Q \setminus \textstyle \frac{1}{r} \, \Z\right) 
\cap \left[c, \, c+1\right), \quad c:= \frac{1-a-b}{r-p-q}.  
\end{equation}
In particular, Problems $\rI$ and $\rII$ are equivalent for 
rational data in domain $\cF^-$. 
\end{enumerate}
\end{theorem}
\begin{remark} \label{rem:F} 
We make three comments about Theorem \ref{thm:F}. 
\begin{enumerate} 
\item When $\lambda$ is integral, formula \eqref{eqn:gpf-F} is computable 
from the GPF for the $(\rA)$-solution $\check{\lambda} \in \cD^-$ through  
transformation \eqref{eqn:gpf-c}, where the roles of $\lambda$ and 
$\check{\lambda}$ are exchanged.  
Regarding duality it is possible to formulate a result similar to 
Theorem \ref{thm:gpf-d} for integral solutions in $\cF^-$.      
\item We mention how to calculate formula \eqref{eqn:gpf-F} when 
$\lambda$ is rational but not integral.    
If $k$ is the least common denominator of $p$, $q \in \Q$,  
then $k \lambda \in \cF^-$ is an integral solution to Problem $\rII$, 
so that as in item (1) the GPF for $k \lambda$ is computable from the 
GPF for the $(\rA)$-solution $(k \lambda)^{\vee} \in \cD^-$ via 
transformation \eqref{eqn:gpf-c}.      
It turns out that the result is       
\begin{equation} \label{eqn:gpf-F-k} 
f(w; k \lambda) = C_k \cdot d^{k w} \cdot 
\dfrac{\prod_{i=0}^{k r-1} 
\varGamma\left(w+\frac{i}{k r}\right)}{\prod_{i=1}^{r} \prod_{j=0}^{k-1} 
\varGamma\left(w+\frac{v_i + j}{k}\right)},   
\end{equation}
for some $C_k > 0$ and $v_1, \dots, v_r$ satisfying condition \eqref{eqn:vi-F}.    
Replacing $w$ with $w/k$ in formula \eqref{eqn:gpf-F-k} and using the 
multiplication formula \eqref{eqn:mult} for the gamma function we get the 
desired formula \eqref{eqn:gpf-F}. 
For details we refer to the proofs of Lemma \ref{lem:gpf-F} and 
Proposition \ref{prop:gpf-F}.    
\item Theorem \ref{thm:F} concerns only rational solutions in $\cF^-$.     
It is an interesting open problem to know whether $\cF^-$ contains any 
irrational solution (see Problem \ref{prob:F}).  
\end{enumerate} 
\end{remark}
\par
Assertions (1) and (2) of Theorem \ref{thm:F} will be proved in 
\S \ref{sec:SW} as Propositions \ref{prop:cfcr-F}, while assertion (3) 
will be established as Proposition \ref{prop:gpf-F} by developing the idea 
in item (2) of Remark \ref{rem:F}.  
This idea for rational solutions in $\cF^-$ can also be used to get  
information about the GPFs for $(\rB)$-solutions in $\cD^-$ because the 
latter situation is similar to the former or even simpler in the sense 
that only duplication $k = 2$ is involved for the latter class of solutions.       
\begin{theorem} \label{thm:sol-B} 
If $\lambda = (p,q,r;a,b;x) \in \cD^-$ is a $(\rB)$-solution then $a$ and 
$b$ are rational numbers with $0 \le a, \, b < 1$, $x$ is an algebraic number, 
and $\lambda$ admits a GPF of the form \eqref{eqn:gpf2} where $d$ is given by 
formula \eqref{eqn:d} and $v_1, \dots, v_r$ are rational numbers such that  
$0 \le v_1, \dots, v_r < 1$.   
\end{theorem}
\par
Only a hint for the proof of this theorem will be given in \S \ref{sec:SW} 
after the proof of Proposition \ref{prop:gpf-F}. 
In \S \ref{sec:algorithm} we shall put all the results on $\cD^- \cup \cF^-$ 
into context from an algorithmic point of view and discuss how to find all 
solutions $\lambda$ with a given bound for $r$ in a finite number of steps.   
\par
Finally we present a small but important result on the domain 
$\cE^{*-} \cup \cE^{-*}$, where Problems $\rI$ and $\rII$ are equivalent 
by Theorem \ref{thm:back}, so we can speak of a solution without 
specifying to which problem it is a solution.  
By assertion (2) of Proposition \ref{prop:E}, for any solution 
$\lambda = (p,q,r;a,b;x) \in \cE^{*-}$ we have $p$, $r \in \Z$ 
so that $\lambda$ is integral if and only if $q \in \Z$.   
Similarly, for any solution $\lambda =  (p,q,r;a,b;x) \in \cE^{-*}$ 
we have $q$, $r \in \Z$, so that $\lambda$ is integral if and only if 
$p \in \Z$.  
It is not known whether $q \in \Z$ resp. $p \in \Z$ for every solution 
$\lambda \in \cE^{*-}$ resp. $\lambda \in \cE^{-*}$.    
\begin{theorem} \label{thm:E}
Duality $\lambda \mapsto \lambda'$ in \eqref{eqn:duality} induces a 
self-bijection on the set of all integral solutions in $\cE^{*-}$. 
Similarly, reciprocity $\lambda \mapsto \check{\lambda}$ in 
\eqref{eqn:recip} induces a bijection between the set of all integral 
solutions in $\cE^{*-}$ and the set of all integral solutions in 
$\cE^{-*}$.   
\end{theorem} 
\par
The former and latter assertions of this theorem will be proved in 
\S \ref{sec:duality} and \S \ref{sec:r-cf} as Corollaries 
\ref{cor:duality1} and \ref{cor:r-cf1} respectively. 
Applications of the theorem will be discussed elsewhere.  
\section{Kummer's 24 Solutions and Ebisu Symmetries} \label{sec:kummer} 
Kummer \cite{Kummer} constructed twenty-four solutions (or more precisely 
power series representations of solutions) to the Gauss hypergeometric 
equation \eqref{eqn:hge}.  
They are known as {\sl Kummer's twenty-four solutions}, among which 
the hypergeometric series $\hgF(\ba;z)$ is the most 
representative member.   
A complete list of them can be found in Erd\'elyi 
\cite[Chap. I\!I, \S 2.9, formulas (1)--(24)]{Erdelyi}.  
Ebisu \cite[Lemma 3.2.2]{Ebisu3} showed that each of Kummer's 
solutions, say $\hgK(\ba;z)$, admits a three-term relation of the 
following form: for every integer vector $\bp = (p,q;r) \in \Z^3$,    
\begin{equation} \label{eqn:3trK}
\hgK(\ba+\bp;z) = \psi(\ba;\bp) \, r(\ba;z) \, \hgK(\ba;z) + 
\phi(\ba;\bp) \, q(\ba;z) \, \hgK(\ba+\1;z),    
\end{equation}
where $q(\ba;z)$ and $r(\ba;z)$ are the same functions as those in the 
original three-term relation \eqref{eqn:3trF} for $\hgF(\ba;z)$, 
whereas $\phi(\ba;\bp)$ and $\psi(\ba;\bp)$ are nontrivial rational 
functions of $\ba$ depending uniquely on $\hgK(\ba;z)$ and $\bp$, 
explicit formulas for which can be found in \cite{Ebisu3}.   
\par
Problem $\rII$ makes sense not only for $\hgF(\ba;z)$ but 
also for any other member $\hgK(\ba;z)$ of Kummer's solutions.    
Given an integral data $\lambda = (p,q,r;a,b;x)$, if we put 
$k(w;\lambda) := \hgK(\bal(w);x)$, $\tilde{k}(w; \lambda) 
:= \hgK(\bal(w)+\1;x)$, $\Phi(w; \lambda) := \phi(\bal(w); x)$ and 
$\Psi(w; \lambda) := \psi(\bal(w); x)$ with 
$\bal(w) := (pw+a, \, qw+b; \, rw)$, then the three-term relation 
\eqref{eqn:3trK} leads to 
\begin{equation} \label{eqn:3trk}
k(w+1; \lambda) = \Psi(w; \lambda) R(w; \lambda) \cdot k(w; \lambda) + 
\Phi(w;\lambda) Q(w; \lambda) \cdot \tilde{k}(w; \lambda),  
\end{equation}
just as the relation \eqref{eqn:3trF} leads to \eqref{eqn:3trf}. 
Notice that $\Phi(w; \lambda)$ and $\Psi(w; \lambda)$ are nontrivial 
rational functions of $w$. 
This observation gives the following important lemma.  
\begin{lemma} \label{lem:kummer} 
Let $\hgK(\ba; z)$ be any member of Kummer's twenty-four solutions.  
An integral data $\lambda$ is a solution to Problem $\rII$ for 
the function $\hgF(\ba; z)$ that comes from contiguous relations, 
if and only if the same is true for the function $\hgK(\ba; z)$.     
If this is the case, then  
\begin{equation} \label{eqn:ocf-k}
\dfrac{k(w+1; \lambda)}{k(w; \lambda)} = \Psi(w; \lambda) \cdot 
R(w; \lambda) \in \C(w),  
\end{equation}
which is corresponding to condition \eqref{eqn:ocf} for the original 
function $f(w; \lambda)$. 
\end{lemma}   
{\it Proof}. 
Recall that an integral data $\lambda$ is a solution to Problem $\rII$ 
for $\hgF(\ba; z)$ that comes from contiguous relations if and only if 
condition \eqref{eqn:cfcr} is satisfied. 
In view of formula \eqref{eqn:3trk} the corresponding condition for 
$\hgK(\ba; z)$ is $\Phi(w; \lambda) Q(w; \lambda) = 0$ in $\C(w)$.  
But this is just equivalent to condition \eqref{eqn:cfcr}, because 
$\Phi(w; \lambda)$ is nontrivial. 
Now formula \eqref{eqn:3trk} leads to condition \eqref{eqn:ocf-k}. 
\hfill $\Box$
\begin{remark} \label{rem:kummer} 
Any member $\hgK(\ba; z)$ of Kummer's $24$ solutions can be written 
$\hgK(\ba; z) = \mbox{(an elementary factor)} \times  
\hgF(\tilde{\ba}; \tilde{z})$ in terms of the original hypergeometric 
function $\hgF(\ba; z)$ and a certain transformation of variables 
$(\ba; z) \mapsto (\tilde{\ba}; \tilde{z})$. 
So Lemma \ref{lem:kummer} suggests that each $\hgK(\ba; z)$ brings a 
symmetry to Problem $\rII$ for the original function $\hgF(\ba; z)$. 
It may be referred to as an {\sl Ebisu symmetry} because it 
originates from Ebisu's observation \eqref{eqn:3trK}. 
The existence of Ebisu symmetries is an advantage of dealing with  
Problem $\rII$, whereas such a helpful structure cannot be expected 
for Problem $\rI$, although we must keep it in mind that {\sl Ebisu 
symmetries make sense only on those solutions which come from 
contiguous relations}.  
\end{remark}
\par
In this article we exhibit two special choices of Kummer's solutions 
other than the original one $\hgF(\ba;z)$. 
The resulting Ebisu symmetries will be the main players in this article, 
that is, duality and reciprocity.  
Here one choice of $\hgK(\ba; z)$ is to take 
\begin{equation} \label{eqn:hgG}
\hgG(\ba;z) := z^{1-\gamma} \hgF(\alpha-\gamma+1,\beta-\gamma+1;2-\gamma;z),    
\end{equation}
which is the solution of local exponent $1-\gamma$ at $z=0$ in Riemann 
scheme \eqref{eqn:rs}.  
Note that solutions $\hgF(\ba; z)$ and $\hgG(\ba; z)$ form a linear 
basis of local solutions to the hypergeometric equation \eqref{eqn:hge} 
at $z = 0$, unless $\gamma$ is an integer.  
The other choice is  
\begin{equation} \label{eqn:hgH}
\hgH(\ba;z) := z^{1-\gamma}(1-z)^{\gamma-\alpha-\beta} 
\hgF(1-\alpha,1-\beta;\gamma-\alpha-\beta+1;1-z), 
\end{equation}
which is an expression for the local solution of exponent  
$\gamma-\alpha-\beta$ at $z = 1$ in the scheme \eqref{eqn:rs}. 
\par
When $\hgK(\ba; z)$ is $\hgG(\ba; z)$ or $\hgH(\ba; z)$, 
we use Lemma \ref{lem:kummer} to construct duality or reciprocity, 
where we employ the following notation. 
For $\hgK(\ba; z) = \hgG(\ba; z)$ the functions $k(w;\lambda)$, $\psi(\ba;\bp)$ 
and $\Psi(w; \lambda)$ are denoted by $g(w;\lambda)$, $\psi_g(\ba;\bp)$ and 
$\Psi_g(w; \lambda)$, while for $\hgK(\ba; z) = \hgH(\ba; z)$ they are denoted 
by $h(w;\lambda)$, $\psi_h(\ba;\bp)$ and $\Psi_h(w; \lambda)$, respectively. 
Note that 
\begin{align}
g(w; \lambda) &= x^{1-r w} \hgF((p-r)w+a+1, (q-r)w+b+1; 2 -r w; x), 
\label{eqn:g} \\[1mm]  
h(w;\lambda) &= x^{1-r w}(1-x)^{(r-p-q)w-a-b} \nonumber \\
&\phantom{==}  \times \hgF(1-a-p w, \, 1-b-q w; \, (r-p-q)w+1-a-b; \, 1-x). 
\label{eqn:h} 
\end{align}
\par
From a result of Ebisu \cite[Lemma 3.2.2]{Ebisu3} we have  
\begin{align*}
\psi_g(\ba;\bp) &= (-1)^{r-p-q} 
\dfrac{(\alpha)_p(\beta)_q(\gamma-\alpha)_{r-p}(\gamma-\beta)_{r-q}}{(\gamma-1)_r (\gamma)_r},  
\\[2mm] 
\psi_h(\ba;\bp) &= (-1)^{r-p-q} 
\dfrac{(\alpha)_p(\beta)_q(\gamma-\alpha-\beta+1)_{r-p-q}}{(\gamma)_r},     
\end{align*} 
in formula \eqref{eqn:3trK}, from which we find     
\begin{align}
\Psi_g(w; \lambda) &= (-1)^{r-p-q} 
\dfrac{(p w+a)_p(q w+b)_q((r-p)w-a)_{r-p}((r-q)w-b)_{r-q}}{(r w-1)_r (r w)_r}, 
\label{eqn:Psi-g} \\[2mm]
\Psi_h(w; \lambda) &= (-1)^{r-p-q} 
\dfrac{(p w+a)_p(q w+b)_q((r-p-q)w-a-b+1)_{r-p-q}}{(r w)_r}, \label{eqn:Psi-h}  
\end{align} 
in formula \eqref{eqn:ocf-k}, where $(s)_n := 
\varGamma(s+n)/\varGamma(s)$ is Pochhammer's symbol or the rising 
factorial.     
Solution \eqref{eqn:hgG} and formula \eqref{eqn:Psi-g} will be used to 
construct duality in \S\ref{sec:duality}, while solution \eqref{eqn:hgH} 
and formula \eqref{eqn:Psi-h} will be employed to construct reciprocity 
in \S\ref{sec:r-cf}, respectively.  
\section{Duality} \label{sec:duality} 
Applying Lemma \ref{lem:kummer} to $\hgK(\ba;z) = \hgG(\ba;z)$ leads to the 
duality \eqref{eqn:duality} in Definition \ref{def:duality}. 
\begin{lemma} \label{lem:duality1} 
Let $\lambda = (p,q,r;a,b;x)$ be an integral data in domain \eqref{eqn:real2}.   
If $\lambda$ is a solution to Problem $\rII$ that comes from contiguous 
relations, with rational function $R(w; \lambda)$ in condition 
\eqref{eqn:ocf}, then its dual $\lambda' = (p',q',r';a',b';x')$ is also 
a solution to Problem $\rII$ that comes from contiguous relations,  
with the corresponding rational function  
\begin{equation} \label{eqn:R-p}
R(w;\lambda') = 
\dfrac{x^{-r}(1-x)^{r-p-q}}{\Psi_g(w'; \lambda) \, R(w';\lambda)},      
\end{equation}
where the function $\Psi_g(w; \lambda)$ is given by formula 
\eqref{eqn:Psi-g} and $w \mapsto w'$ is the reflection   
\begin{equation} \label{eqn:w-p}
w' := \frac{2}{r}-1-w. 
\end{equation}
\end{lemma} 
{\it Proof}. 
Replacing $w$ by $w+1 = \frac{2}{r} - w'$ in formula \eqref{eqn:g}, 
we observe that   
\[
\begin{split}
g(w+1;\lambda) 
&= \resizebox{0.8\hsize}{!}{$\displaystyle
x^{1-r\left(\frac{2}{r} - w'\right)} 
\hgF\left((p-r)\left(\ts\frac{2}{r} - w'\right)+a+1, \, 
(q-r)\left(\ts\frac{2}{r} - w'\right)+b+1; 
2 - r\left(\ts\frac{2}{r}-w'\right); x\right)$} \\
&= x^{r w'-1} \hgF\left((r-p) w' - a', \, (r-q) w' - b'; r w'; x\right) \\
&= x^{r w'-1} (1-x)^{(p+q-r)w'+a'+b'} f(w'; \lambda'), 
\end{split}
\]
where definition \eqref{eqn:duality} and Euler's transformation 
\eqref{eqn:transf1} are used in the second and third 
equalities respectively. 
Since the shift $w \mapsto w-1$ is equivalent to 
$w' \mapsto w'+1$, we have  
\begin{align} 
f(w'; \lambda') 
&= x^{1-r w'} (1-x)^{(r-p-q) w'-a'-b'} g(w+1;\lambda),  \nonumber \\
f(w'+1; \lambda') 
&= x^{1-r (w'+1)} (1-x)^{(r-p-q) (w'+1)-a'-b'} g(w;\lambda), 
\label{eqn:f-p} 
\end{align}   
which together with formula \eqref{eqn:ocf-k} in Lemma \ref{lem:kummer} 
yields 
\[
R(w'; \lambda') := \dfrac{f(w'+1; \lambda')}{f(w'; \lambda')} 
= x^{-r}(1-x)^{r-p-q} \cdot \dfrac{g(w;\lambda)}{g(w+1;\lambda)} = 
\dfrac{ x^{-r}(1-x)^{r-p-q}}{\Psi_g(w; \lambda) \, R(w; \lambda)}.   
\]
Replacing $w$ by $w'$ in the above and noting $w'' = w$, 
we obtain formula \eqref{eqn:R-p}. \hfill $\Box$ \par\medskip
Using Lemma \ref{lem:duality1} in domain $\cE^{*-}$ leads to an  
immediate consequence. 
\begin{corollary} \label{cor:duality1}  
Duality $\lambda \mapsto \lambda'$ in \eqref{eqn:duality} induces a 
self-bijection on the set of all solutions $\lambda = 
(p,q,r;a,b;x) \in \cE^{*-}$ with $q \in \Z$.    
\end{corollary}   
{\it Proof}. 
In view of definition \eqref{eqn:duality} the duality 
$\lambda \mapsto \lambda'$ is a bijection $\cE^{*-} \to \cE^{*-}$ in 
the data level. 
By assertion (3) of Proposition \ref{prop:E} and Lemma \ref{lem:duality1}, 
it induces a bijection in the solution level among all solutions 
$\lambda \in \cE^{*-}$ to Problem $\rII$ (and to Problem $\rI$ 
by Theorem \ref{thm:back}) with $q \in \Z$.  \hfill $\Box$ \par\medskip
The same results holds true for $(\rA)$-solutions in $\cD^-$ since 
the duality is also a bijection $\cD^- \to \cD^-$ in the data level, 
but in fact we are able to obtain more detailed results on $\cD^-$. 
\begin{lemma} \label{lem:duality2} 
If $\lambda = (p,q,r;a,b;x) \in \cD^-$ is an $(\rA)$-solution to 
Problem $\rII$, then its dual $\lambda' = (p,q,r;a',b';x) \in \cD^-$ is 
also an $(\rA)$-solution to the same problem with  
\begin{equation} \label{eqn:R-p2}
R(w;\lambda') = d \cdot  
\frac{\prod_{i=0}^{r-1} \left(w+\frac{i}{r}
\right)}{\prod_{i=1}^{r} \left(w+v_i' \right)}, 
\end{equation}
where $d$ and $v_1', \dots, v_r'$ are defined by formulas 
\eqref{eqn:d} and \eqref{eqn:vi-p} respectively.  
\end{lemma} 
{\it Proof}. 
By assertion (2) of Theorem \ref{thm:sol-A} we have the gamma product 
formula \eqref{eqn:gpf2}, so that the rational function 
$R(w; \lambda)$ in formula \eqref{eqn:R} is given by  
\begin{equation} \label{eqn:R2}
R(w; \lambda) = d \cdot \dfrac{\prod_{i=0}^{r-1} \left(w+\frac{i}{r}
\right)}{\prod_{i=1}^{r} \left(w+v_i\right)},   
\end{equation}
where $d$ is defined in formula \eqref{eqn:d}. 
On the other hand, formula \eqref{eqn:Psi-g} can be rewritten 
\[
\begin{split}
\Psi_g(w; \lambda) &= \dfrac{p^p q^q (r-p)^{r-p} (r-q)^{r-q}}{r^{2r}} \\ 
&\phantom{=} \times 
\dfrac{\prod_{i=0}^{p-1}\left(w+{\ts\frac{i+a}{p}}\right) 
\prod_{i=0}^{q-1}\left(w+{\ts\frac{i+b}{q}}\right) 
\prod_{i=0}^{r-p-1}\left(w+{\ts\frac{i-a}{r-p}}\right) 
\prod_{i=0}^{r-q-1}\left(w+{\ts\frac{i-b}{r-q}}\right)}{
\prod_{i=0}^{r-1} \left(w+\frac{i-1}{r}\right) 
\prod_{i=0}^{r-1} \left(w+\frac{i}{r}\right)},     
\end{split}  
\]
where $(-1)^{r-p-q} = 1$ is taken into account, which follows from 
assertion (2) of Theorem \ref{thm:D}. 
Thus taking the product of equations \eqref{eqn:Psi-g} and \eqref{eqn:R2} 
and using definition \eqref{eqn:vi-s}, we have 
\[
\dfrac{1}{\Psi_g(w; \lambda) \, R(w; \lambda)} = 
d \cdot x^r\cdot (1-x)^{p+q-r} \cdot 
\dfrac{\prod_{i=0}^{r-1} \left(w+\frac{i-1}{r}
\right)}{\prod_{i=1}^r \left(w+v_i^*\right)}. 
\]
Replacing $w$ by $w'$ in the above, where $w'$ is defined by 
\eqref{eqn:w-p}, and using   
$\prod_{i=0}^{r-1}\left(w'+ \frac{i-1}{r}\right) = (-1)^r 
\prod_{i=0}^{r-1}\left(w + \frac{i}{r}\right)$ and 
$\prod_{i=0}^{r-1}\left(w'+ v_i^*\right) = (-1)^r 
\prod_{i=0}^{r-1}\left(w + v_i'\right)$, we have  
\[
\dfrac{1}{\Psi_g(w'; \lambda) \, R(w'; \lambda)} = 
d \cdot x^r\cdot (1-x)^{p+q-r} \cdot 
\dfrac{\prod_{i=0}^{r-1} \left(w+\frac{i}{r}
\right)}{\prod_{i=1}^r \left(w+v_i'\right)}. 
\]
Substituting this into formula \eqref{eqn:R-p} leads to 
the desired formula \eqref{eqn:R-p2}. 
\hfill $\Box$ 
\begin{proposition} \label{prop:duality} 
If $\lambda = (p,q,r;a,b;x) \in \cD^-$ is an $(\rA)$-solution to  
Problem $\rI$ with gamma product formula \eqref{eqn:gpf2}, then 
there exists a positive constant $C' > 0$ such that 
\begin{align}
f(w; \lambda') &= C' \cdot d^w \cdot 
\dfrac{\prod_{i=0}^{r-1} \varGamma\left(w+\frac{i}{r}\right)}{\prod_{i=1}^{r} \varGamma\left(w+v_i'\right)}, 
\label{eqn:gpf3} \\[2mm]
g(w; \lambda) &= D \cdot \delta^w \cdot
\dfrac{\prod_{i=1}^{r} \sin \pi \left(w+v_i^*\right)}{\prod_{i=0}^{r-1} \sin \pi \left(w+\frac{i-1}{r}\right)}
\cdot  
\dfrac{\prod_{i=1}^{r} \varGamma\left(w+v_i^*\right)}{\prod_{i=0}^{r-1} \varGamma\left(w+\frac{i-1}{r}\right)}, 
\label{eqn:gpf-g}  
\end{align} 
where $d$, $v_i'$ and $v_i^*$ are given by formulas 
\eqref{eqn:d}, \eqref{eqn:vi-p} and \eqref{eqn:vi-s}, whereas 
$D$ and $\delta$ are defined by  
\begin{equation} \label{eqn:D-d}
D := \dfrac{x \cdot d^{2/r} \cdot C'}{(1-x)^{a+b}}, \qquad 
\delta := \dfrac{(1-x)^{r-p-q}}{d \cdot x^r}. 
\end{equation} 
\end{proposition}
{\it Proof}. 
By Lemma \ref{lem:duality2}, $\lambda'$ is a solution to Problem $\rII$ 
in $\cD^-$ with rational function $R(w;\lambda')$ in formula 
\eqref{eqn:R-p2}.   
Theorem \ref{thm:back} then implies that it leads back to a 
solution to Problem $\rI$, which is exactly the one in  
formula \eqref{eqn:gpf3}. 
Next we have 
\[
\begin{split}
g(w;\lambda) 
&= x^{r(w'+1)-1} (1-x)^{a'+b'-(r-p-q)(w'+1)} \, f(w'+1; \lambda') \\
&= \dfrac{x}{(1-x)^{a+b}} \cdot \left\{\frac{(1-x)^{r-p-q}}{x^r} \right\}^w 
\cdot f\left(\textstyle \frac{2}{r}-w; \lambda' \right) \\
&= \dfrac{x}{(1-x)^{a+b}} \cdot \left\{\frac{(1-x)^{r-p-q}}{x^r} \right\}^w 
\cdot C' \cdot d^{\frac{2}{r} -w} \cdot 
\dfrac{\prod_{i=0}^{r-1} 
\varGamma\left(\frac{2}{r}-w+\frac{i}{r} \right)}{\prod_{i=1}^r 
\varGamma\left(\frac{2}{r}-w + v_i'\right)} \\
&= \dfrac{C' \cdot x \cdot d^{\frac{2}{r}}}{(1-x)^{a+b}} \cdot 
\left\{\frac{(1-x)^{r-p-q}}{d \cdot x^r} \right\}^w  
\dfrac{\prod_{i=0}^{r-1} 
\varGamma\left(1-(w+\frac{(r-1-i)-1}{r}) \right)}{\prod_{i=1}^r 
\varGamma\left(1-(w + v_i^*)\right)} \\
&= D \cdot \delta^w \cdot \dfrac{\prod_{i=0}^{r-1} 
\varGamma\left(1-(w+\frac{i-1}{r}) \right)}{\prod_{i=1}^r 
\varGamma\left(1-(w + v_i^*)\right)}
\end{split}
\]
where the first equality follows from \eqref{eqn:f-p}, the second 
from \eqref{eqn:duality} and \eqref{eqn:w-p}, the third from 
\eqref{eqn:gpf3}, the fourth from \eqref{eqn:vi-p}, the fifth 
from \eqref{eqn:D-d} and the replacement of indices  
$i \leftrightarrow r-1-i$, respectively.    
Finally we use Euler's reflection formula for the gamma 
function \cite[Theorem 1.2.1]{AAR}:  
\begin{equation} \label{eqn:refl}
\varGamma(w) \varGamma(1-w) = \dfrac{\pi}{\sin \pi w},  
\end{equation} 
to establish the desired formula \eqref{eqn:gpf-g} 
\hfill $\Box$ \par\medskip
With the proof of formula \eqref{eqn:gpf3} in Proposition 
\ref{prop:duality}, Theorem \ref{thm:gpf-d} has been established. 
\section{Reciprocity and Connection Formula} \label{sec:r-cf} 
We shall show that applying Lemma \ref{lem:kummer} to $\hgK(\ba;z) = 
\hgH(\ba;z)$ yields the reciprocity \eqref{eqn:recip} in 
Definition \ref{def:recip}. 
First we observe that under the involution 
$\lambda \mapsto \check{\lambda}$, the transformation   
\begin{equation} \label{eqn:w-c}
w \mapsto \check{w} := w + c, \qquad 
c = c(\lambda) := \frac{1-a-b}{r-p-q}  
\end{equation}
also yields an involution, since definition \eqref{eqn:recip} implies 
$\check{c}:= c(\check{\lambda}) = - c(\lambda) = -c$.     
\begin{lemma} \label{lem:r-cf1} 
Let $\lambda = (p,q,r;a,b;x)$ be an integral data in domain \eqref{eqn:real3}. 
If $\lambda$ is a solution to Problem $\rII$ that comes from contiguous 
relations, with rational function $R(w;\lambda)$ in condition 
\eqref{eqn:ocf}, then its reciprocal $\check{\lambda} 
= (\check{p},\check{q},\check{r};\check{a},\check{b}; \check{x})$ 
is also a solution to Problem $\rII$ that comes from contiguous 
relations, with the corresponding rational function 
\begin{equation} \label{eqn:R-c}
R(w; \check{\lambda}) = x^r (1-x)^{p+q-r} \cdot 
\Psi_h(\hat{w}; \lambda) R(\hat{w}; \lambda), 
\end{equation}
where $\Psi_h(w; \lambda)$ is given by formula \eqref{eqn:Psi-h} and 
$\hat{w}$ is defined by
\begin{equation} \label{eqn:w-h}
\hat{w} := w - c, \qquad c = c(\lambda) := \frac{1-a-b}{r-p-q}.  
\end{equation}   
\end{lemma}
{\it Proof}. 
Definitions \eqref{eqn:recip} and \eqref{eqn:w-c} imply  
$(r-p-q) w+1-a-b = \check{r} \check{w}$ and   
\[
1-a - p w = 1-a - p (\check{w} - c) = \check{p} \check{w} + \check{a},    
\]
and similarly $1-b-q w = \check{q} \check{w} + \check{b}$.  
Substituting these into formula \eqref{eqn:h} we find  
$h(w; \lambda) = x^{1-r w}(1-x)^{(r-p-q)w-a-b} \,   
f(\check{w}; \check{\lambda})$, or in other words,  
\begin{equation} \label{eqn:f-h}
f(\check{w}; \check{\lambda}) = x^{r w-1}(1-x)^{(p+q-r)w+a+b} 
\, h(w; \lambda).  
\end{equation}
Since increasing $\check{w}$ by $1$ is equivalent to increasing $w$ 
by $1$, we also have  
\[
f(\check{w}+1; \check{\lambda}) = 
x^{r(w+1)-1}(1-x)^{(p+q-r)(w+1)+a+b} \, h(w+1; \lambda),    
\]
so that formula \eqref{eqn:ocf-k} in Lemma \ref{lem:kummer} and 
definition \eqref{eqn:w-c} yield  
\[
\begin{split}
R(\check{w}; \check{\lambda}) &= 
\dfrac{f(\check{w}+1; \check{\lambda})}{f(\check{w}; \check{\lambda})} 
= x^r (1-x)^{p+q-r} \dfrac{h(w+1; \lambda)}{h(w; \lambda)} = 
x^r (1-x)^{p+q-r} \cdot \Psi_h(w; \lambda) R(w;\lambda) \\
&= x^r (1-x)^{p+q-r} \cdot \Psi_h(\check{w} - c; \lambda) \, 
R(\check{w} - c; \lambda).  
\end{split} 
\]
Since $\check{w}$ is an indeterminate variable, we can replace 
$\check{w}$ by $w$ in the above to obtain 
\[
\begin{split}
R(w; \check{\lambda}) &= 
x^r (1-x)^{p+q-r} \cdot \Psi_h(w - c; \lambda) R(w - c; \lambda) \\
&= x^r (1-x)^{p+q-r} \cdot \Psi_h(\hat{w}; \lambda) \, R(\hat{w}; \lambda), 
\end{split} 
\]
where definition \eqref{eqn:w-h} is used in the last equality. 
\hfill $\Box$ \par\medskip 
Using Lemma \ref{lem:r-cf1} in domains $\cE^{*-}$ and $\cE^{-*}$ 
yields the following direct consequence. 
\begin{corollary} \label{cor:r-cf1}  
Reciprocity $\lambda \mapsto \check{\lambda}$ in \eqref{eqn:recip} 
induces a bijection between the set of all solutions $\lambda = 
(p,q,r;a,b;x) \in \cE^{*-}$ with $q \in \Z$ and the set of all solutions 
$\check{\lambda} = 
(\check{p}, \check{q}, \check{r}; \check{a}, \check{b}; \check{x}) 
\in \cE^{-*}$ with $\check{p} \in \Z$.   
\end{corollary}   
{\it Proof}.  
In view of definition \eqref{eqn:recip} the reciprocity  
$\lambda \mapsto \check{\lambda}$ is a bijection $\cE^{*-} \to \cE^{-*}$ 
in the data level. 
By Proposition \ref{prop:E} and Lemma \ref{lem:r-cf1} it induces  
a bijection in the solution level between the solutions 
$\lambda \in \cE^{*-}$ with $q \in \Z$ and the solutions 
$\check{\lambda} \in \cE^{-*}$ with $\check{p} \in \Z$.  
\hfill $\Box$ \par\medskip 
Although the reciprocity is also a bijection $\cD^- \to \cF^-$ in the 
data level, it does not immediately induce a bijection in the 
integral solution level as in Corollary \ref{cor:r-cf1}. 
This is because we have not yet known whether every integral solution 
in $\cF^-$ to Problem $\rII$ comes from contiguous relations, 
so that the backward reciprocity $\cF^- \to \cD^-$ is not established 
yet in the solution level (see Remark \ref{rem:kummer}). 
This issue is postponed until it is settled in Proposition 
\ref{prop:cfcr-F}. 
In the rest of this section we develop a detailed study of the forward 
reciprocity $\cD^- \to \cF^-$ in the integral solution level by using 
the connection formula for hypergeometric functions.    
\begin{lemma} \label{lem:r-cf2} 
If $\lambda = (p,q,r;a,b;x) \in \cD^-$ is an $(\rA)$-solution  
with GPF \eqref{eqn:gpf2}, then  
\begin{equation} \label{eqn:h2}
\resizebox{0.85\hsize}{!}{$\displaystyle 
h(w;\lambda) = \varGamma((r-p-q)w+1-a-b) \cdot \tilde{d}^w \cdot \chi(w)   
\cdot \dfrac{\prod_{i=0}^{p-1}\varGamma\left(w+\frac{i+a}{p}\right) 
\prod_{i=0}^{q-1}\varGamma\left(w+\frac{i+b}{q}\right)}{\prod_{i=1}^{r} 
\varGamma\left(w+v_i\right)}$},  
\end{equation}
where the constant $\tilde{d}$ and the function $\chi(w)$ are given by  
\begin{align}  
\tilde{d} &= \sqrt{p^p q^q (r-p)^{p-r} (r-q)^{q-r} x^{-r} (1-x)^{r-p-q}}, 
\label{eqn:d-t} \\[1mm] 
\chi(w) &= C_1 \cdot \dfrac{\sin \pi(p w+a) \sin \pi(q w+b)}{\sin(\pi r w)} 
+ C_2 \cdot \dfrac{\prod_{i=1}^r \sin \pi(w+v_i^*)}{\prod_{i=0}^{r-1} 
\sin \pi\left(w+\frac{i-1}{r}\right)}, \label{eqn:chi} \\[1mm]
C_1 &:= 2 \, (2\pi)^{(r-p-q-1)/2} \cdot C \cdot p^{a-1/2} q^{b-1/2} r^{1/2}, 
\label{eqn:C1} \\    
C_2 &:= (2\pi)^{(r-p-q-1)/2} \cdot D \cdot (r-p)^{a+1/2} (r-q)^{b+1/2} r^{-3/2}, 
\label{eqn:C2} 
\end{align} 
with $C$ and $D$ being the constants in formulas \eqref{eqn:gpf2} and 
\eqref{eqn:D-d} respectively.  
\end{lemma}
{\it Proof}. 
A connection formula in Erd\'elyi 
\cite[Chap.\! I\!I, \S 2.9, formula (43)]{Erdelyi}) reads    
\[
\hgH(\ba;z) = 
\dfrac{\varGamma(\gamma-\alpha-\beta+1) \varGamma(1-\gamma)}{\varGamma(1-\alpha) \varGamma(1-\beta)} \, \hgF(\ba;z) +
\dfrac{\varGamma(\gamma-\alpha-\beta+1) \varGamma(\gamma-1)}{\varGamma(\gamma-\alpha) \varGamma(\gamma-\beta)} \, \hgG(\ba;z).   
\]
Substituting $\ba = \bal(w) := (p w+a, \, q w+b; \, r w)$ and $z = x$ 
into the connection formula and using the definitions of 
$g(w; \lambda)$ and $h(w; \lambda)$ (just before formulas 
\eqref{eqn:g} and \eqref{eqn:h}), we have 
\begin{equation} \label{eqn:cf} 
h(w; \lambda) = C_f(w) \, f(w; \lambda) + C_g(w) \, g(w; \lambda), 
\end{equation}
where the connection coefficients $C_f(w)$ and $C_g(w)$ are given by 
\begin{align*} 
C_f(w) 
&= \frac{\varGamma((r-p-q)w+1-a-b)\, \varGamma(1-r w)}{\varGamma(1-a-p w)\, \varGamma(1-b-q w)}, 
\\[1mm] 
C_g(w)
&= \frac{\varGamma((r-p-q)w+1-a-b) \, \varGamma(r w-1)}{\varGamma((r-p)w-a) \varGamma((r-q)w-b)}.     
\end{align*}
\par
The connection coefficient $C_f(w)$ can be written  
\[
\begin{split}
C_f(w) 
&= \varGamma((r-p-q)w+1-a-b) \cdot 
\dfrac{\sin \pi(p w+a) \, \sin \pi(q w+b)}{\pi \, \sin(\pi r w)} \cdot 
\dfrac{\varGamma(p w+a) \, \varGamma(q w+b)}{\varGamma(r w)} \\
&= \varGamma((r-p-q)w+1-a-b) \cdot 
\dfrac{\sin \pi(p w+a) \, \sin \pi(q w+b)}{\pi \, \sin(\pi r w)} \\
&\phantom{=} \times (2 \pi)^{(r-p-q+1)/2} p^{a-1/2} q^{b-1/2} r^{1/2} 
\left(\frac{p^p q^q}{r^r}\right)^w 
\dfrac{\prod_{i=0}^{p-1} \varGamma\left(w+\frac{i+a}{p}\right)
\prod_{i=0}^{q-1} \varGamma\left(w+\frac{i+b}{q}\right)}{\prod_{i=0}^{r-1} 
\varGamma\left(w+\frac{i}{r}\right)}, 
\end{split}
\]
where the first equality follows from the reflection formula 
\eqref{eqn:refl}, while the second equality from Gauss's multiplication 
formula \eqref{eqn:mult} for the gamma function.  
The above expression for $C_f(w)$ is multiplied by formula 
\eqref{eqn:gpf2} to yield 
\begin{equation} \label{eqn:Cf}
\begin{split}
C_f(w) \, f(w;\lambda) 
&= \varGamma((r-p-q)w+1-a-b) \cdot 
\dfrac{\sin \pi(p w+a) \, \sin \pi(q w+b)}{\sin(\pi r w)} \\
&\phantom{=} \times C_1 \cdot 
\tilde{d}^w \cdot 
\dfrac{\prod_{i=0}^{p-1} \varGamma\left(w+\frac{i+a}{p}\right)
\prod_{i=0}^{q-1} \varGamma\left(w+\frac{i+b}{q}\right)}{\prod_{i=0}^{r-1} 
\varGamma(w+v_i)}, 
\end{split}
\end{equation}
where one uses definition \eqref{eqn:C1} and relation 
$d \cdot p^p q^q r^{-r} = \tilde{d}$ that follows from \eqref{eqn:d} 
and \eqref{eqn:d-t}. 
\par
In a similar manner the multiplication formula \eqref{eqn:mult} allows us 
to write   
\[
\begin{split}
C_g(w) &= \varGamma((r-p-q)w+1-a-b) \cdot (2\pi)^{(r-p-q-1)/2} 
(r-p)^{a+1/2} (r-q)^{b+1/2} r^{-3/2} \\
&\phantom{=} \times 
\left\{ \frac{r^r}{(r-p)^{r-p}(r-q)^{r-q}} \right\}^w \cdot 
\dfrac{\prod_{i=0}^{r-1} \varGamma\left(w+\frac{i-1}{r}\right)}{\prod_{i=0}^{r-p-1} 
\varGamma\left(w+\frac{i-a}{r-p}\right)
\prod_{i=0}^{r-q-1} \varGamma\left(w+\frac{i-b}{r-q}\right)}, 
\end{split}
\]
which is multiplied by formula \eqref{eqn:gpf-g} to yield 
\begin{align*}
C_g(w) \, g(w;\lambda) 
&= \varGamma((r-p-q)w+1-a-b) \cdot 
\dfrac{\prod_{i=1}^r \sin \pi(w+v_i)}{\prod_{i=0}^{r-1} 
\sin \pi\left(w+\frac{i-1}{r}\right)} \\
&\phantom{=} \times C_2 \cdot \tilde{d}^w \cdot 
\dfrac{\prod_{i=1}^r \varGamma(w+v_i^*)}{\prod_{i=0}^{r-p-1} 
\varGamma\left(w+\frac{i-a}{r-p}\right)
\prod_{i=0}^{r-q-1} \varGamma\left(w+\frac{i-b}{r-q}\right)}, 
\end{align*}
where one uses definition \eqref{eqn:C2} and the relation 
$\delta \cdot r^r (r-p)^{p-r} (r-q)^{q-r} = \tilde{d}$ that 
follows from \eqref{eqn:D-d} and \eqref{eqn:d-t}. 
Taking equation \eqref{eqn:vi-s} into account we have          
\begin{equation} \label{eqn:Cg}
\begin{split}
C_g(w) \, g(w;\lambda) &= \varGamma((r-p-q)w+1-a-b) \cdot 
\dfrac{\prod_{i=1}^r \sin \pi(w+v_i)}{\prod_{i=0}^{r-1} 
\sin \pi\left(w+\frac{i-1}{r}\right)} \\
&\phantom{=} \times C_2 \cdot \tilde{d}^w \cdot 
\dfrac{\prod_{i=0}^{p-1} \varGamma\left(w+\frac{i+a}{p}\right)
\prod_{i=0}^{q-1} \varGamma\left(w+\frac{i+b}{q}\right)}{\prod_{i=0}^{r-1} 
\varGamma(w+v_i)}. 
\end{split}
\end{equation}
Substituting \eqref{eqn:Cf} and \eqref{eqn:Cg} into \eqref{eqn:cf} yields 
the desired formula \eqref{eqn:h2} with \eqref{eqn:d-t} and 
\eqref{eqn:chi}. \hfill $\Box$ 
\begin{lemma} \label{lem:chi} 
In Lemma $\ref{lem:r-cf2}$ the function $\chi(w)$ in formula \eqref{eqn:chi} 
must be constant so that there exists a real constant $C_3$ such that 
formula \eqref{eqn:h2} becomes 
\begin{equation} \label{eqn:h3}
\dfrac{h(w;\lambda)}{\varGamma((r-p-q)w+1-a-b)} = C_3 \cdot \tilde{d}^w    
\cdot \dfrac{\prod_{i=0}^{p-1}\varGamma\left(w+\frac{i+a}{p}\right) 
\prod_{i=0}^{q-1}\varGamma\left(w+\frac{i+b}{q}\right)}{\prod_{i=1}^{r} 
\varGamma\left(w+v_i\right)}.   
\end{equation}
\end{lemma}
{\it Proof}. 
The left-hand side of equation \eqref{eqn:h3}, which is denoted by  
$\tilde{h}(w;\lambda)$, is an entire holomorphic function of $w$, 
since by formula \eqref{eqn:h} any pole of $h(w;\lambda)$ must be 
simple and located where $(r-p-q)w+1-a-b$ becomes a nonpositive 
integer, so that it is canceled by a zero of 
$1/\varGamma((r-p-q)w+1-a-b)$. 
It follows from formula \eqref{eqn:h2} that 
\begin{equation*}
\chi(w) = \tilde{d}^{-w} \cdot \tilde{h}(w;\lambda)      
\cdot \dfrac{\prod_{i=1}^{r} \varGamma\left(w+v_i\right)}{\prod_{i=0}^{p-1}
\varGamma\left(w+\frac{i+a}{p}\right) 
\prod_{i=0}^{q-1}\varGamma\left(w+\frac{i+b}{q}\right)},    
\end{equation*}
and so $\chi(w)$ is holomorphic in the half-plane $\rRe \, w > 
- \min\{ v_i \,:\, i = 1, \dots, n \}$. 
On the other hand formula \eqref{eqn:chi} implies that 
$\chi(w)$ is a periodic function of period one, since  
$p$, $q$, and $r$ are positive integers with $r-p-q$ even. 
Thus $\chi(w)$ must be an entire holomorphic and periodic function   
of period one. 
In particular $\chi(w)$ is bounded on the horizontal strip 
$|\rIm \, z| \le 1$.  
\par
Now notice that $\sin z$ admits a two-sided bound 
$\frac{1}{4} \, e^{|\rIm \,z|} \le |\sin z| \le e^{|\rIm \, z|}$ on 
the outer region $|\rIm \, z| \ge 1$.  
Applying it to formula \eqref{eqn:chi} yields an estimate:  
\[
|\chi(w)| \le |C_1| \, 
\dfrac{e^{\pi p |\rIm \, z|} \cdot e^{\pi q |\rIm \, z|}}{\frac{1}{4} \, 
e^{\pi r |\rIm \, z|}} + |C_2| \, \dfrac{ \prod_{i=1}^r 
e^{\pi |\rIm \, z|}}{\prod_{i=0}^{r-1} \frac{1}{4} \, e^{\pi |\rIm \, z|}} 
= 4|C_1| \, e^{-\pi(r-p-q)|\rIm \, z|} + 4^r |C_2|,  
\]
and hence $|\chi(w)| \le 4|C_1| \, e^{-\pi(r-p-q)} + 4^r |C_2|$ on 
$|\rIm \, z| \ge 1$. 
Thus the entire function $\chi(w)$ is bounded on $\C_w$. 
By Liouville's theorem $\chi(w)$ must be constant. 
\hfill $\Box$ \par\medskip
Given a positive integer $k$, we put $[k] := \{0,1,\dots, k-1\}$. 
By division relation \eqref{eqn:division} there exist subsets 
$I_p \subset [p]$, $I_q \subset [q]$, $J_p \subset [r-p]$, 
$J_q \subset [r-q]$, with $0 \not \in I_p$ and $0 \not \in I_q$,  
such that 
\begin{equation} \label{eqn:IJ}
\prod_{i=1}^r (w+v_i) = 
\prod_{i \in I_p} \left(w+{\ts\frac{i+a}{p}}\right) 
\prod_{i \in I_q} \left(w+{\ts\frac{i+b}{q}}\right) 
\prod_{j \in J_p} \left(w+{\ts\frac{j-a}{r-p}}\right) 
\prod_{j \in J_q} \left(w+{\ts\frac{j-b}{r-q}}\right).   
\end{equation}
If we put $\bar{I}_p := [p] \setminus I_p$ and $\bar{I}_q := [q] \setminus I_q$, 
then equation \eqref{eqn:h3} becomes  
\begin{equation} \label{eqn:h4}
\dfrac{h(w;\lambda)}{\varGamma((r-p-q)w+1-a-b)} = C_3 \cdot \tilde{d}^w    
\cdot \dfrac{\prod_{i\in \bar{I}_p} \varGamma\left(w+\frac{i+a}{p}\right) 
\prod_{i \in \bar{I}_q} \varGamma\left(w+\frac{i+b}{q}\right)}{
\prod_{j \in J_p} \varGamma\left(w+\frac{j-a}{r-p}\right) 
\prod_{j \in J_q} \varGamma\left(w+\frac{j-b}{r-q}\right)}.   
\end{equation}
\par
To exploit formula \eqref{eqn:h4} we need a preliminary lemma. 
A {\sl multi-set} is a set allowing repeated elements. 
For multi-sets $S = \{s_1, \dots, s_m\}$ and $T = \{t_1, \dots, t_n\}$, 
we write $S \succ T$ if $m \le n$ and there exists a re-indexing of 
$t_1, \dots, t_n$ such that $s_i - t_i \in \Z_{\ge 0}$ for every  
$i = 1, \dots, m$. 
\begin{lemma} \label{lem:nni}
Let $S = \{s_1, \dots, s_m\}$ and $T = \{t_1, \dots, t_n\}$ be 
multi-sets of real numbers. 
If  
\[ 
h(w) = \dfrac{\varGamma(w+s_1) \cdots \varGamma(w+s_m)}{\varGamma(w+t_1) \cdots \varGamma(w+t_n)}
\]
is an entire function of $w$, then $S \succ T$.  
\end{lemma} 
{\it Proof}. 
The proof is by induction on $m$. 
When $m = 0$ the assertion is obvious as the numerator of $h(w)$ is $1$. 
Let $m \ge 1$. 
We may assume $s_1 \le \cdots \le s_m$.   
An upper factor $\varGamma(w+s_1)$ of $h(w)$ has a pole at $w = -s_1$. 
In order for $h(w)$ to be holomorphic, a lower factor 
$\varGamma(w+t_j)$ must have a pole at the same point.    
After a transposition of $t_j$ and $t_1$ we may put $j = 1$. 
Then $-s_1+t_1=-r_1 \in \mathbb{Z}_{\le 0}$, that is, 
$r_1 \in \mathbb{Z}_{\ge0}$. 
Since $\varGamma(w+s_1)/\varGamma(w+t_1) = (w+t_1)_{r_1}$, we have 
\[
h(w) = (w+t_1)_{r_1} \cdot h_1(w), \qquad \mbox{where} \qquad 
h_1(w) := \dfrac{\varGamma(w+s_2) \cdots \varGamma(w+s_m)}{\varGamma(w+t_2) \cdots \varGamma(w+t_n)}. 
\]
We claim that $h_1(w)$ is entire holomorphic. 
If $r_1 = 0$ this is obvious since $h(w) = h_1(w)$. 
Let $r_1 \ge 1$. 
Any pole of $\varGamma(w+s_2) \cdots \varGamma(w+s_m)$ is separated 
from all the roots of $(w+t_1)_{r_1}$, because we have 
$-s_m \le \cdots \le -s_2 \le -s_1$ and the roots of $(w+t_1)_{r_1}$ 
are located at $-s_1+1 < -s_1+2 < \dots < -t_1$ in an increasing order.    
Thus any pole of $h_1(w)$, if it exists, cannot be 
canceled by a root of $(w+t_1)_{r_1}$.  
Accordingly, $h_1(w)$ has no poles, since $h(w)$ has no poles.  
Now we can apply the induction hypothesis to $h_1(w)$ to conclude the proof.      
\hfill $\Box$ \par\medskip
Since the left-hand side of equation \eqref{eqn:h4} is entire holomorphic 
in $w$, so must be the gamma products on the right. 
Thus Lemma \ref{lem:nni} yields   
\begin{equation} \label{eqn:succ1}
\left\{ \frac{i+a}{p}\right\}_{i\in \bar{I}_p} \bigcup \,\,  
\left\{\frac{i+b}{q}\right\}_{i \in \bar{I}_q}
\succ \, 
\left\{\frac{j-a}{r-p}\right\}_{j \in J_p} \bigcup \,\, 
\left\{\frac{j-b}{r-q}\right\}_{j \in J_q},   
\end{equation}
where the both sides above are thought of as multi-sets. 
Note that $0 \in \bar{I}_p$ and $0 \in \bar{I}_q$.  
\begin{table}[t]
\begin{center}
\begin{tabular}{|c|c|c|c|} 
\hline 
     &      &     &     \\[-4mm]
case & type & $a$ & $b$ \\[1mm]
\hline 
     &      &     &     \\[-3mm]
$1$ & $(\rI, \, \rI)$   & $\dfrac{p i}{r}$                         & $\dfrac{q j}{r}$ \\[3mm]
\hline
     &      &     &     \\[-3mm]
$2$ & $(\rII, \, \rII)$ & \hspace{2mm} $\dfrac{p \{ (r-p)i - q j \}}{r(r-p-q)}$ \hspace{2mm} 
& \hspace{2mm} $\dfrac{q \{ (r-q)j - p i \}}{r(r-p-q)}$ \hspace{2mm} \\[5mm] 
\hline 
     &      &     &     \\[-3mm]
$3$ & $(\rI, \, \rII)$  & $\dfrac{p i}{r}$                         & $\dfrac{q (r j - p i)}{r(r-p)}$ \\[4mm]
\hline
     &      &     &     \\[-3mm]
$4$ & $(\rII, \, \rI)$  & $\dfrac{p (r i -q j)}{r(r-q)}$           & $\dfrac{q j}{r}$ \\[4mm] 
\hline
\end{tabular}   
\end{center}
\caption{Formula for $(a, b)$ in terms of $(p,q,r)$ and $(i, j) \in \Z_{\ge 0}^2$.}
\label{tab:lem-ab}
\end{table}
\begin{lemma} \label{lem:ab}
For any $(\rA)$-solution $\lambda = (p,q,r;a,b;x) \in \cD^-$ there exists 
a pair $(i, j)$ of nonnegative integers such that one of the four cases 
in Table $\ref{tab:lem-ab}$ comes about.  
In particular, in either case $a$ and $b$ must be rational numbers.       
\end{lemma}
{\it Proof}.  
Since $0 \in \bar{I}_p$, condition \eqref{eqn:succ1} implies that either 
$(\rI)_p$ or $(\rII)_p$ below holds:   
\begin{enumerate}
\item[$(\rI_p)$] there exists an integer $i \in J_p + (r-p) \, \Z_{\ge0}$ such that  
$\dfrac{a}{p} = \dfrac{i-a}{r-p}$, 
\item[$(\rII_p)$] there exists an integer $i \in J_q + (r-q) \, \Z_{\ge0}$ such that 
$\dfrac{a}{p} = \dfrac{i-b}{r-q}$.   
\end{enumerate}
In a similar manner, since $0 \in \bar{I}_q$, condition \eqref{eqn:succ1} 
implies that either $(\rI)_q$ or $(\rII)_q$ below holds:    
\begin{enumerate}
\item[$(\rI_q)$] there exists an integer $j \in J_q + (r-q) \, \Z_{\ge0}$ such that  
$\dfrac{b}{q} = \dfrac{j-b}{r-q}$, 
\item[$(\rII_q)$] there exists an integer $j \in J_p + (r-p) \, \Z_{\ge0}$ such that 
$\dfrac{b}{q} = \dfrac{j-a}{r-p}$.   
\end{enumerate}
There are a total of four types $(\rI_p, \, \rI_q)$, $(\rII_p, \, \rII_q)$, 
$(\rI_p, \, \rII_q)$, $(\rII_p, \, \rI_q)$, which are exactly the Cases 
$1$--$4$ of Table \ref{tab:lem-ab} respectively, where suffixes 
$* \in \{p, \, q\}$ of $\rI_*$ and $\rII_*$ are omitted. 
In each case we have a pair of linear equations for $(a, b)$, 
which can be solved as indicated in Table \ref{tab:lem-ab}.  
\hfill $\Box$
\begin{lemma} \label{lem:ab-J} 
If $a > 0$ then $0 \not\in J_p$. 
Similarly if $b > 0$ then $0 \not\in J_q$.  
\end{lemma}
{\it Proof}. 
We use assertion (1) of Iwasaki \cite[Proposition 11.14]{Iwasaki} with  
$k = 0$.  
In the notation there it states that for each $k = 0, \dots, r-p-1$,  
one has $(w-w_k^*)|\prod_{i=1}^r(w+v_i)$ if and only if 
\begin{equation} \label{eqn:ab-J1}
(\gamma_k^*+k)_{p+1} \cdot \cF_k(\beta_k^*; \gamma_k^*;x) = 0, 
\end{equation}
unless \cite[condition (122a)]{Iwasaki} is satisfied, that is, unless  
\begin{equation} \label{eqn:ab-J2}
\tilde{\beta}_k^*, \,\, \tilde{\gamma}_k^* \in \Z, \qquad 
0 \le - \tilde{\beta}_k^* \le - \tilde{\gamma}_k^* \le r-p-k-2.  
\end{equation}
When $k = 0$, definitions in 
\cite[formula (117) and Proposition 11.12]{Iwasaki} read 
\[
w_0^* = \dfrac{a}{r-p}, \qquad  
\gamma_0^* = r w_0^* =  \dfrac{r a}{r-p}, \qquad  
\tilde{\gamma}_0^* = 2-r(w_0^* +1) = 2-r-\dfrac{r a}{r-p},  
\]
so condition $- \tilde{\gamma}_k^* \le r-p-k-2$ in \eqref{eqn:ab-J2} 
is equivalent to $a \le -p(r-p)/r \,\, (< 0)$. 
Thus if $a > 0$ then condition \eqref{eqn:ab-J2} with $k = 0$ is not 
fulfilled.  
When $k = 0$, condition \eqref{eqn:ab-J1} becomes 
$(\gamma_0^*)_{p+1} = 0$, since $\cF_0(\beta_0^*; \gamma_0^*;x) = 1$ 
by \cite[definition (116)]{Iwasaki}. 
But if $a > 0$ then $\gamma_0^* > 0$ and so $(\gamma_0^*)_{p+1} > 0$, 
which means that $w-w_0^* = w + \frac{0-a}{r-p}$ is {\sl not} a factor 
of $\prod_{i=1}^r(w+v_i)$. 
This in turn implies $0 \not\in J_p$ by formula $\eqref{eqn:IJ}$. 
The implication $b > 0 \Rightarrow 0 \not\in J_q$ is proved in 
the same way. \hfill $\Box$ 
\section{Duality Revisited and Finite Cardinality} \label{sec:dr-fc}  
Consider a solution $\lambda = (p,q,r;a,b;x) \in \cD^-$ of type $(\rA)$ 
and its dual $\lambda = (p,q,r;a',b';x) \in \cD^-$. 
Then the associated matrix \eqref{eqn:matrix} is said to be of type 
\begin{equation} \label{eqn:type}
\begin{pmatrix} 
J_1 & J_2 \\
J_3 & J_4
\end{pmatrix}, \qquad J_1, J_2, J_3, J_4 \in \{\rI, \, \rII\}, 
\end{equation}
if the top row $(a, b)$ of matrix \eqref{eqn:matrix} is of type 
$(J_1, J_2)$, while its bottom row $(a', b')$ is of type $(J_3, J_4)$ 
in the sense of Table \ref{tab:lem-ab}. 
\begin{table}[t]
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline  
     &   &   &   &   &   &   &   \\[-3mm]
case & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\[1mm]
\hline 
     &   &   &   &   &   &   &   \\[-3mm] 
type & 
$\begin{matrix} \rI & \rI \\[1mm] \rI & \rI \end{matrix}$ &  
$\begin{matrix} \rII & \rII \\[2mm] \rII & \rII \end{matrix}$ &
$\begin{matrix} \rI & \rII \\[2mm] \rI & \rII \end{matrix}$ &   
$\begin{matrix} \rI & \rI \\[2mm] \rI & \rII \end{matrix}$ & 
$\begin{matrix} \rII & \rII \\[2mm] \rII & \rI \end{matrix}$ & 
$\begin{matrix} \rI & \rII \\[2mm] \rII & \rI \end{matrix}$ &
$\begin{matrix} \rI & \rI \\[2mm] \rII & \rII \end{matrix}$ \\[4mm]
\hline 
\end{tabular}
\end{center}
\caption{Types of matrix \eqref{eqn:matrix} up to its column and/or row exchanges.}
\label{tab:type}
\end{table}
There are a total of seven types up to column and/or row exchanges of 
the matrix, which are tabulated in Table \ref{tab:type}, where 
parentheses in \eqref{eqn:type} are omitted. 
\par
The essential parts of definition \eqref{eqn:duality} for duality can be 
rewritten as  
\begin{equation} \label{eqn:duality2}
a + a' = 1- \dfrac{2 p}{r}, \qquad b + b' = 1- \dfrac{2 q}{r}. 
\end{equation}
In each case of Table \ref{tab:type} we shall see what kinds of 
consequences are derived from equations \eqref{eqn:duality2}. 
In this section we employ the notation introduced in item (1) of 
Remark \ref{rem:ab}.  
\begin{lemma} \label{lem:case1-6} 
In cases $1$--$6$ of Table $\ref{tab:type}$, equations 
\eqref{eqn:duality2} give rise to the conditions in Table $\ref{tab:ab}$, 
regarding items $(1)$--$(4)$ mentioned in Theorem $\ref{thm:ab}$.  
\end{lemma}
{\it Proof}. This lemma is proved by case-by-case treatments presented 
below. \\[1mm] 
{\bf Case 1}. By Lemma \ref{lem:ab} there exists a quadruple of 
nonnegative integers $(i,j;i',j')$ such that 
\begin{equation} \label{eqn:c1-1} 
a = \dfrac{p i}{r}, \qquad b = \dfrac{q j}{r}, \qquad 
a' = \dfrac{p i'}{r}, \qquad b' =  \dfrac{q j'}{r}. 
\end{equation}
Substituting formula \eqref{eqn:c1-1} into equations \eqref{eqn:duality2} 
yields $i + i' = \frac{r}{p} -2 \in \Z$ and $j + j' = \frac{r}{q} -2 \in \Z$, 
which implies $p|r$ and $q|r$, so that we have 
\begin{equation} \label{eqn:ls-c1}
i + i' = r_p -2, \qquad  j + j' = r_q -2, \qquad 
i, j, i', j' \in \Z_{\ge 0}. 
\end{equation}
Formula \eqref{eqn:c1-1} then becomes 
\begin{equation} \label{eqn:c1-2} 
a = \dfrac{i}{r_p}, \qquad b = \dfrac{j}{r_q}, \qquad 
a' = \dfrac{i'}{r_p}, \qquad b' =  \dfrac{j'}{r_q}. 
\end{equation}
Thus all the conditions in case 1 of Table \ref{tab:ab} have been 
obtained. \\[1mm]
{\bf Case 2}. By Lemma \ref{lem:ab} there exists a quadruple of 
nonnegative integers $(i,j;i',j')$ such that 
\begin{equation} \label{eqn:c2-1}
\begin{array}{rlrl}
a &= \dfrac{p \{(r-p)i-q j\}}{r(r-p-q)}, \qquad &  
b &= \dfrac{q \{(r-q)j-p i\}}{r(r-p-q)}, \\[4mm] 
a' &= \dfrac{p \{(r-p)i'-q j'\}}{r(r-p-q)}, \qquad &  
b' &= \dfrac{q \{(r-q)j'-p i'\}}{r(r-p-q)}. 
\end{array}
\end{equation}  
Substituting formula \eqref{eqn:c2-1} into equations \eqref{eqn:duality2} 
yields 
\[
\resizebox{0.9\hsize}{!}{$\displaystyle
(r-p) (i+i') - q(j+j') = (r-p-q) \left(\textstyle \frac{r}{p} - 2\right), 
\qquad 
-p (i+i') +(r-q)(j+j') = (r-p-q) \left(\textstyle \frac{r}{q} - 2\right). 
$} 
\]
They are solved with respect to $i+i'$ and $j+j'$ to obtain 
$i+i' = (r-p-q)/p \in \Z$ and $j+j' = (r-p-q)/q \in \Z$, which imply 
$p|(r-p-q)$ and $q|(r-p-q)$, so that we have 
\begin{equation} \label{eqn:ls-c2}
i+i' = (r-p-q)_p, \qquad j+j' = (r-p-q)_q, \qquad 
i, j, i', j' \in \Z_{\ge 0}. 
\end{equation}
Formula \eqref{eqn:c2-1} then becomes 
\begin{equation} \label{eqn:c2-2}  
a = \dfrac{(r-p)i-q j}{r(r-p-q)_p}, \quad   
b = \dfrac{(r-q)j-p i}{r(r-p-q)_q}, \quad 
a' = \dfrac{(r-p)i'-q j'}{r(r-p-q)_p}, \quad   
b' = \dfrac{(r-q)j'-p i'}{r(r-p-q)_q}. 
\end{equation}
Thus all the conditions in case 2 of Table \ref{tab:ab} have been 
obtained. \\[1mm]
{\bf Case 3}. By Lemma \ref{lem:ab} there exists a quadruple of 
nonnegative integers $(i,j;i',j')$ such that 
\begin{equation} \label{eqn:c3-1}
a = \dfrac{p i}{r}, \qquad  
b = \dfrac{q (r j- p i)}{r(r-p)}, \qquad  
a'= \dfrac{p i'}{r}, \qquad  
b'= \dfrac{q(r j'- p i')}{r(r-p)}. 
\end{equation}
Substituting formula \eqref{eqn:c3-1} into equations \eqref{eqn:duality2} 
yields $i+i' = \frac{r}{p}-2 \in \Z$ and $r(j+j')-p(i+i') = (r-p) 
\left(\frac{r}{q}-2\right)$, the former of which is put into the 
latter to yields $j+j' = (r-p-q)/q \in \Z$. 
Thus we have $p|r$ and $q|(r-p-q)$, and hence  
\begin{equation} \label{eqn:ls-c3}
i+i' = r_p -2, \qquad j+j' = (r-p-q)_q, \qquad 
i, j, i', j' \in \Z_{\ge 0}. 
\end{equation}
Formula \eqref{eqn:c3-1} then becomes 
\begin{equation} \label{eqn:c3-2} 
a = \dfrac{i}{r_p}, \qquad  
b = \dfrac{r_p \, j - i}{r_p (r-p)_q}, \qquad  
a'= \dfrac{i'}{r_p}, \qquad  
b'= \dfrac{r_p \, j' - i'}{r_p (r-p)_q}.  
\end{equation}
Thus all the conditions in case 3 of Table \ref{tab:ab} have been 
obtained. \\[1mm]
{\bf Case 4}. By Lemma \ref{lem:ab} there exists a quadruple of 
nonnegative integers $(i,j;i',j')$ such that 
\begin{equation} \label{eqn:c4-1}
a = \dfrac{p i}{r}, \qquad b = \dfrac{q j}{r}, \qquad 
a' = \dfrac{p i'}{r}, \qquad b' = \dfrac{q (r j'-p i')}{r(r-p)}. 
\end{equation}
Substituting formula \eqref{eqn:c4-1} into equations \eqref{eqn:duality2} 
yields 
\[
i+i' = \textstyle \frac{r}{p} - 2 \in \Z, \qquad 
(r-p)j - p i' + r j' = (r-p)\left(\textstyle \frac{r}{q}-2\right).  
\]
The former equation implies $p|r$ and $i+i' = r_p - 2$, while the 
division of the latter by $p$ makes $(r_p -1)j -i' + r_p \, j' = (r_p -1) 
\left(\frac{r}{q} - 2\right)$, to which $i+i'=r_p - 2$ is added to give 
$i + (r_p-1)j + r_p \, j' = r_p(r-p-q)/q \in \Z$. 
Summing up we have $q|r_p(r-p-q)$ and 
\begin{equation} \label{eqn:ls-c4}
i+i' = r_p - 2, \qquad i + (r_p-1)j + r_p \, j' = (r_p(r-p-q))_q, 
\qquad i, j, i', j' \in \Z_{\ge 0}. 
\end{equation}
Formula \eqref{eqn:c4-1} then becomes 
\begin{equation} \label{eqn:c4-2}  
a = \dfrac{i}{r_p}, \qquad b = \dfrac{q j}{r}, \qquad 
a' = \dfrac{i'}{r_p}, \qquad b' = 
\dfrac{r_p \, j'- i'}{(r_p(r -p))_q}. 
\end{equation}
Thus all the conditions in case 4 of Table \ref{tab:ab} have been 
obtained. \\[1mm]
{\bf Case 5}. By Lemma \ref{lem:ab} there exists a quadruple of 
nonnegative integers $(i,j;i',j')$ such that 
\begin{equation} \label{eqn:c5-1}
a = \dfrac{p \{(r-p)i-q j\}}{r(r-p-q)}, \qquad  
b = \dfrac{q \{(r-q) j- p i\}}{r(r-p-q)}, \qquad  
a'= \dfrac{p(r i'-q j')}{r(r-q)}, \qquad  
b'= \dfrac{q j'}{r}. 
\end{equation}
Substituting formula \eqref{eqn:c5-1} into equations \eqref{eqn:duality2} 
yields 
\begin{subequations} \label{eqn:c5-2}
\begin{align}
\resizebox{0.6\hsize}{!}{$\displaystyle (r-p)(r-q) i -q(r-q) j + r(r-p-q)i' - q(r-p-q) j'$} 
&= \resizebox{0.2\hsize}{!}{$(r-p-q)(r-q)\left(\frac{r}{p} - 2\right)$}, 
\label{eqn:c5-2a} \\ 
-p i + (r-q)j + (r-p-q) j' &= 
(r-p-q)\left(\textstyle \frac{r}{q}-2\right). \label{eqn:c5-2b}
\end{align} 
\end{subequations}
Calculating $\eqref{eqn:c5-2a} + q \times \eqref{eqn:c5-2b}$ we have  
$i+i' = (r-p-q)/p \in \Z$, which implies $p|(r-p-q)$ and so  
$i+i' = (r-p-q)_p$.  
Division of \eqref{eqn:c5-2b} by $p$ makes $-i + (r-q)_p \, j + 
(r-p-q)_p \, j' = (r-p-q)_p \left(\frac{r}{q}-2\right)$, to which 
$i+i' = (r-p-q)_p$ is added to yield $i' + (r-q)_p \, j + 
(r-p-q)_p \, j' = (r-q)(r-p-q)_p/q \in \Z$. 
Summing up we have $q|r(r-p-q)_p$ and  
\begin{subequations} \label{eqn:ls-c5}
\begin{align}
i+i' &= (r-p-q)_p,  & i, i' &\in \Z_{\ge 0}, \label{eqn:ls-c5a} \\
i' + (r-q)_p \, j + (r-p-q)_p \, j' &= ((r-q)(r-p-q)_p)_q, & 
i', j, j' &\in \Z_{\ge 0}. \label{eqn:ls-c5b}
\end{align} 
\end{subequations}
Formula \eqref{eqn:c5-1} then becomes 
\begin{equation} \label{eqn:c5-3} 
a = \dfrac{(r-p)i-q j}{r(r-p-q)_p}, \qquad  
b = \dfrac{(r-q)_p \, j- i}{(r(r-p-q)_p)_q}, \qquad  
a'= \dfrac{r i'-q j'}{r(r-q)_p}, \qquad  
b'= \dfrac{q j'}{r}. 
\end{equation}
Thus all the conditions in case 5 of Table \ref{tab:ab} have been 
obtained. \\[1mm]
{\bf Case 6}. By Lemma \ref{lem:ab} there exists a quadruple of 
nonnegative integers $(i,j;i',j')$ such that 
\begin{equation} \label{eqn:c6-1}
a = \dfrac{p i}{r}, \qquad  
b = \dfrac{q (r j- p i)}{r(r-p)}, \qquad 
a'= \dfrac{p(r i'- q j')}{r(r-q)}, \qquad  
b'= \dfrac{q j'}{r}. 
\end{equation}
Substituting formula \eqref{eqn:c6-1} into equations \eqref{eqn:duality2} 
yields
\[ 
(r-q)i - q j' + r i' = (r-q)\left(\textstyle \frac{r}{p}-2\right), \qquad 
-p i + (r-p)j'+ r j  = (r-p)\left(\textstyle \frac{r}{q}-2\right).   
\]
They are recast to $(r-p-q)i +(r-p) i' + q j = (r-p)(r-p-q)/p \in \Z$ 
and $(r-p-q)j'+ p i' + (r-q)j = (r-q)(r-p-q)/q \in \Z$. 
Thus $p|r(r-p-q)$ and $q|r(r-p-q)$, so that 
\begin{subequations} \label{eqn:ls-c6}
\begin{align}
(r-p-q)i +(r-p) i' + q j &= ((r-p)(r-p-q))_p, & i, i', j &\in \Z_{\ge0},  
\label{eqn:ls-c6a} \\ 
(r-p-q)j'+ p i' + (r-q)j &= ((r-q)(r-p-q))_q, & j', i', j &\in \Z_{\ge0}.  
\label{eqn:ls-c6b}  
\end{align}   
\end{subequations}
Formula \eqref{eqn:c6-1} then becomes 
\begin{equation} \label{eqn:c6-2}
a = \dfrac{p i}{r}, \qquad  
b = \dfrac{r j- p i}{(r(r-p))_q}, \qquad 
a'= \dfrac{r i'- q j'}{(r(r-q))_p}, \qquad  
b'= \dfrac{q j'}{r}. 
\end{equation}
Thus all the conditions in case 6 of Table \ref{tab:ab} have been 
obtained. \hfill $\Box$
\begin{lemma} \label{lem:case7} 
Case $7$ of Table $\ref{tab:type}$ cannot occur. 
\end{lemma}
{\it Proof}. 
By Lemma \ref{lem:ab} there exists a quadruple of nonnegative 
integers $(i,j;i',j')$ such that 
\[
a = \dfrac{p i}{r}, \qquad 
b = \dfrac{q j}{r}, \qquad 
a' = \dfrac{p \{(r-p) i'-q j'\}}{r(r-p-q)}, \qquad 
b' =  \dfrac{q \{(r-q) j'-p i'\}}{r(r-p-q)}. 
\]
Substituting these into relations \eqref{eqn:duality2} yields 
\begin{align*}
(r-p-q) i + (r-p) i' -q j' &= (r-p-q)\left(\textstyle \frac{r}{p} -2\right), \\
(r-p-q) j + (r-q) j' -p j' &= (r-p-q)\left(\textstyle \frac{r}{q} -2\right), 
\end{align*}
which can readily be converted into
\begin{subequations} \label{eqn:c7-1}
\begin{align}
(r-q) i + q j + r i' &= \textstyle \frac{r}{p} -2 \in \Z, \label{eqn:c7-1a} \\
p i + (r-p) j + r j' &= \textstyle \frac{r}{q} -2 \in \Z, \label{eqn:c7-1b},  
\end{align}
\end{subequations}
which imply $p|r$ and $q|r$. 
Now $r$ must be even, for otherwise $r$ is odd and hence so are 
$p$ and $q$ by $p|r$ and $q|r$, but then $r-p-q \equiv 1-1-1 \equiv 
1 \mod 2$, that is, $r-p-q$ is odd, which is absurd since it must be 
even by assertion (2) of Theorem \ref{thm:D}.     
If $i' \ge 1$ then \eqref{eqn:c7-1a} gives an absurd estimate  
$r = (r-q) \cdot 0 + q \cdot 0 + r \cdot 1 \le (r-q) i + q j + r i' = 
r_p -2 \le r-2$. 
Thus we must have $i' = 0$. 
In a similar manner equation \eqref{eqn:c7-1b} forces $j' = 0$. 
Therefore \eqref{eqn:c7-1} reduces to 
\begin{subequations} \label{eqn:c7-2}
\begin{align}
(r-q) i + q j &= r_p -2, \label{eqn:c7-2a} \\
p i + (r-p) j &= r_q -2, \label{eqn:c7-2b},  
\end{align}
\end{subequations}
If $i \ge 1$ and $j \ge 1$ then \eqref{eqn:c7-2a} gives an absurd 
estimate $r = (r-q) \cdot 1 + q \cdot 1 \le (r-q) i + q j = r_p -2 
\le r-2$. 
Thus we must have either $i = 0$ or $j = 0$. 
Here we may and shall assume $j = 0$ due to the symmetry of conditions 
\eqref{eqn:c7-2a} and \eqref{eqn:c7-2b}. 
Then \eqref{eqn:c7-2a} and \eqref{eqn:c7-2b} yield 
\begin{equation} \label{eqn:c7-3}
i = \frac{r_p-2}{r-q} = \frac{r_q-2}{p} \in \Z,  
\end{equation}
which in particular implies $p|(r_q-2)$. 
Now $r_q$ must be even, for otherwise $r_q$ is odd and hence so is 
$p$ by $p|(r_q-2)$, while $q$ is even since $r = q r_q$ is even with  
$r_q$ odd, but then $r-p-q \equiv 0-1-0 \equiv 1 \mod 2$, that is, 
$r-p-q$ is odd, which is again absurd. 
By the second equality in \eqref{eqn:c7-3} we have 
$2 q = 4 r - 2 p - r \cdot r_q = 4 r - 2 p - 2 r \cdot r_{2q}$, that is, 
$q/p = 2 r_p - 1 - r_p \cdot r_{2q} \in \Z$, where $r_{2 q}$ makes 
sense as $r_q$ is even. 
Thus $p|q$ and so we have integer equations 
$1 = 2 r_p - q_p - r_p \cdot r_{2 q} = 4 r_{2 q} 
\cdot q_p - q_p - 2 (r_{2 q})^2 q_p = 
q_p \cdot \{4 r_{2 q} - 2 (r_{2 q})^2 - 1\}$, where   
$r_p = 2 r_{2 q} \cdot q_p$ is used in the second equality. 
Thus $q_p = 1$ and $4 r_{2 q}- 2 (r_{2 q})^2 -1 = 1$, 
the former of which means $p = q$ while the latter yields  
$r_{2 q} (2-r_{2 q}) = 1$, that is, $r_{2 q} = 1$ and so 
$r = 2 q$. 
Then $r-p-q = r-2 q = 0$, which is absurd since we have 
$r-p-q > 0$ in $\cD^-$. 
This last contradiction shows that the occurrence of case $7$ in 
Table \ref{tab:type} is impossible. \hfill $\Box$
\begin{lemma} \label{lem:ab-bound} 
We must have $-1 < a, \, b, \, a', \, b' < 1$. 
\end{lemma} 
{\it Proof}. The lemma is proved by a case-by-case check. \\[1mm]
{\bf Case 1}. From condition \eqref{eqn:ls-c1} we have    
$0 \le i, \, i' \le r_p-2$ and $0 \le j, \, j' \le r_q-2$, 
so that estimate $0 \le a, \, a', \, b, \, b' < 1$ follows from 
representation \eqref{eqn:c1-2}. \\[1mm]  
{\bf Case 2}. From condition \eqref{eqn:ls-c2} we have $i \ge 0$ 
and $j \le (r-p-q)_q$ as well as $i \le (r-p-q)_p$ and $j \ge 0$.  
Thus it follows from representation \eqref{eqn:c2-2} that      
\begin{align*}
1+ a &= \dfrac{r(r-p-q)_p+(r-p)i-q j}{r(r-p-q)_p} \ge 
\dfrac{r(r-p-q)_p - q(r-p-q)_q}{r(r-p-q)_p} \\ 
&= \dfrac{r(r-p-q)_p - p (r-p-q)_p}{r(r-p-q)_p} = \dfrac{r-p}{r} > 0, \\       
1- a &= \dfrac{r(r-p-q)_p-(r-p)i+q j}{r(r-p-q)_p} \ge 
\dfrac{r(r-p-q)_p - (r-p)(r-p-q)_p}{r(r-p-q)_p} 
= \dfrac{p}{r} > 0,  
\end{align*}
which shows $-1 < a < 1$. 
In similar manners we have $-1 < b, \, a', \, b' < 1$. \\[1mm]
{\bf Case 3}. From condition \eqref{eqn:ls-c3} we have  
$0 \le i, \, i' \le r_p-2$, so that estimate $0 \le a, \, a' < 1$ 
follows from representation \eqref{eqn:c3-2}. 
Similarly, from condition \eqref{eqn:ls-c3} we have $j \ge 0$ and 
$i \le r_p-2$ as well as $j \le (r-p-q)_q$ and $i \ge 0$.  
Thus it follows from representation \eqref{eqn:c3-2} that 
\begin{align*}
1+b &= \dfrac{r_p(r-p)_q + r_p j-i}{r_p(r-p)_q} 
\ge \dfrac{r_p(r-p)_q -(r_p -2)}{r_p(r-p)_q} = 
\dfrac{r_p(r-p-q)_q + 2}{r_p(r-p)_q} > 0, \\  
1-b &= \dfrac{r_p(r-p)_q - r_p j+i}{r_p(r-p)_q} 
\ge \dfrac{r_p(r-p)_q -r_p (r-p-q)_q}{r_p(r-p)_q} = 
\dfrac{1}{(r-p)_q} > 0,   
\end{align*}
which shows $-1 < b < 1$. 
In a similar manner we have $-1 < b' < 1$.  
\\[1mm]
{\bf Case 4}. From condition \eqref{eqn:ls-c4} we have 
$0 \le i, \, i' \le r_p-2$, so that estimate $0 \le a, \, a' < 1$ 
follows from representation \eqref{eqn:c4-2}. 
Similarly, from condition \eqref{eqn:ls-c4} we have 
$0 \le (r_p -1) j \le (r_p(r-p-q))_q$, that is, 
$0 \le q j \le r(r-p-q)/(r-p)$. 
Thus representation \eqref{eqn:c4-2} yields   
\[
0 \le b = \dfrac{q j}{r} \le \dfrac{r-p-q}{r-p} < 1. 
\]
From condition \eqref{eqn:ls-c4} we have $j' \ge 0$ and $i' \le r_p-2$ 
as well as $r_p j' \le (r_p(r-p-q))_q$ and $i' \ge 0$.  
Thus it follows from representation \eqref{eqn:c4-2} that  
\begin{align*}
1+b' &= \dfrac{(r_p(r-p))_q + r_p j' - i'}{(r_p(r-p))_q} 
\ge \dfrac{(r_p(r-p))_q -(r_p -2)}{(r_p(r-p))_q} = 
 \dfrac{(r_p(r-p-q))_q + 2}{(r_p(r-p))_q} > 0, \\
1-b' &= \dfrac{(r_p(r-p))_q - r_p j' + i'}{(r_p(r-p))_q} 
\ge \dfrac{(r_p(r-p))_q -(r_p(r-p-q))_q}{(r_p(r-p))_q} = 
 \dfrac{q}{r-p} > 0.  
\end{align*}
Therefore we have $0 \le a, \, a', \, b < 1$ and $-1 < b' < 1$. \\[1mm]
{\bf Case 5}. From condition \eqref{eqn:ls-c5b} we have  
$(r-q)_p j \le ((r-q)(r-p-q)_p)_q$, that is, $q j \le r-p-q = 
p(r-p-q)_p$. 
Thus it follows from representation \eqref{eqn:c5-3} and $i \ge 0$ that 
\[
1+a = \dfrac{r(r-p-q)_p + (r-p)i-q j}{r(r-p-q)_p} 
\ge \dfrac{r(r-p-q)_p - p(r-p-q)_p}{r(r-p-q)_p} = \frac{r-p}{r} > 0.  
\]
From condition \eqref{eqn:ls-c5a} we have $i \le (r-p-q)_p$.  
It then follows from \eqref{eqn:c5-3} and $j \ge 0$ that 
\begin{align*}
1-a &= \dfrac{r(r-p-q)_p - (r-p)i+ q j}{r(r-p-q)_p} 
\ge \dfrac{r(r-p-q)_p - (r-p)(r-p-q)_p}{r(r-p-q)_p} = \frac{p}{r} > 0, \\  
1+b &= \dfrac{(r(r-p-q)_p)_q + (r-q)_p j -i}{(r(r-p-q)_p)_q} 
\ge \dfrac{(r(r-p-q)_p)_q -(r-p-q)_p}{(r(r-p-q)_p)_q} = 
\dfrac{r-q}{r} > 0.  
\end{align*}
Since $(r-q)_p j \le ((r-q)(r-p-q)_p)_q$ and $i \ge 0$, 
representation \eqref{eqn:c5-3} yields   
\[
1-b = \dfrac{(r(r-p-q)_p)_q -(r-q)_p j + i}{(r(r-p-q)_p)_q -(r-q)_p j + i} 
\ge \dfrac{(r(r-p-q)_p)_q - ((r-q)(r-p-q)_p)_q}{(r(r-p-q)_p)_q -(r-q)_p j + i} 
= \dfrac{q}{r} > 0. 
\]
From condition \eqref{eqn:ls-c5b} we have $(r-p-q)_p j' \le ((r-q)(r-p-q)_p)_q$, 
that is, $q j' \le r-q = p(r-q)_p$. 
Thus it follows from formula \eqref{eqn:c5-3} and $i' \ge 0$ that  
\[
1+a' = \dfrac{r(r-q)_p + r i'-q j'}{r(r-q)_p} 
\ge  \dfrac{r(r-q)_p - p(r-q)_p}{r(r-q)_p} = \dfrac{r-p}{r} > 0.  
\]
From condition \eqref{eqn:ls-c5a} we have $i' \le (r-p-q)_p$.  
Then it follows from \eqref{eqn:c5-3} and $j' \ge 0$ that  
\[
1-a' = \dfrac{r(r-q)_p - r i'+ q j'}{r(r-q)_p} 
\ge  \dfrac{r(r-q)_p - r(r-p-q)_p}{r(r-q)_p} = \dfrac{1}{(r-q)_p} > 0.  
\]
Finally, since $0 \le q j' \le r-q$, 
we have $0 \le b' = q j'/r \le (r-q)/r < 1$.  
\\[1mm]
{\bf Case 6}. From condition \eqref{eqn:ls-c6a} we have $(r-p-q) i \le 
((r-p)(r-p-q))_p$, that is, $p i \le r-p$. 
Thus it follows from representation \eqref{eqn:c6-2} and $i \ge 0$ that 
$0 \le a = p i/r \le (r-p)/r < 1$. 
Similarly, we have $0 \le b' < 1$. 
From condition \eqref{eqn:ls-c6b} we have $(r-p-q)j' \le ((r-q)(r-p-q))_q$, 
that is, $q j' \le r-q$. 
Thus it follows from representation \eqref{eqn:c6-2} and $i' \ge 0$ that
\[
1+ a' = \dfrac{(r(r-q))_p + r i'-q j'}{(r(r-q))_p} 
\ge \dfrac{(r(r-q))_p - (r-q)}{(r(r-q))_p} = \dfrac{r-p}{r} > 0. 
\]
On the other hand, from condition \eqref{eqn:ls-c6a} we have 
$(r-p)i' \le ((r-p)(r-p-q))_p$, that is, $p i' \le r-p-q$.  
Thus it follows from representation \eqref{eqn:c6-2} and $j' \ge 0$ that
\[
1-a' = \dfrac{(r(r-q))_p - r i' + q j'}{(r(r-q))_p} 
\ge \dfrac{(r(r-q))_p - r i'}{(r(r-q))_p}
= \dfrac{(r-q) - p i'}{r-q} \ge \dfrac{p}{r-q} > 0. 
\]
In a similar manner we have $-1 < b < 1$.  \hfill $\Box$
\begin{lemma} \label{lem:incl} 
We have $0 \le a, \, b < 1$ and there exist inclusions of multi-sets:   
\begin{subequations} \label{eqn:incl}
\begin{align}
\left\{ \frac{i+a}{p}\right\}_{i\in \bar{I}_p} \bigcup \,\,  
\left\{\frac{i+b}{q}\right\}_{i \in \bar{I}_q}
& \subset \, 
\left\{\frac{j-a}{r-p}\right\}_{j \in J_p} \bigcup \,\, 
\left\{\frac{j-b}{r-q}\right\}_{j \in J_q}, 
\label{eqn:incl-a} \\
\left\{ \frac{i+a}{p}\right\}_{i\in I_p} \bigcup \,\,  
\left\{\frac{i+b}{q}\right\}_{i \in I_q}
& \subset \, 
\left\{\frac{j-a}{r-p}\right\}_{j \in \bar{J}_p} \bigcup \,\, 
\left\{\frac{j-b}{r-q}\right\}_{j \in \bar{J}_q},   
\label{eqn:incl-b}
\end{align}    
\end{subequations}
where $\bar{I}_p := [p] \setminus I_p$, $\bar{I}_q := [q] \setminus I_q$, 
$\bar{J}_p := [r-p] \setminus J_p$ and $\bar{J}_q := [r-q] \setminus J_q$.  
\end{lemma}
{\it Proof}. Recall that $J_p \subset [r-p] = \{0, \dots, r-p-1\}$.  
By Lemma \ref{lem:ab-bound} we have $-a < 1$ so that 
$(j-a)/(r-p) < (r-p-1+1)/(r-p) = 1$ for any $j \in J_p$.  
If $a \le 0$ then obviously we have $(j-a)/(r-p) \ge 0$ for any 
$j \in J_p$.  
If $a > 0$ then we have $J_p \subset \{1, \dots, r-p-1\}$ by 
Lemma \ref{lem:ab-J} and $-a > -1$ by Lemma \ref{lem:ab-bound}, so that  
$(j-a)/(r-p) > (1-1)/(r-p) = 0$ for any $j \in J_p$. 
In either case we have $0 \le (j-a)/(r-p) < 1$ for any $j \in J_p$. 
In a similar manner we have $0 \le (j-b)/(r-q) < 1$ for any $j \in J_q$. 
So the multi-set on the right-hand side of \eqref{eqn:succ1} lies  
in the interval $[0, \, 1)$. 
Since $0 \in \bar{I}_p$ and $0 \in \bar{I}_q$, the numbers 
$a/p$ and $b/q$ belong to the multi-set on the left so that they 
must be nonnegative by the binary relation \eqref{eqn:succ1}. 
Hence we have $a, \, b \ge 0$, which together with 
Lemma \ref{lem:ab-bound} yields the estimate $0 \le a, \, b < 1$. 
\par
Since $0 \le a < 1$ and $\bar{I}_p \subset \{0, \dots, p-1\}$ we 
have $0 = (0+0)/p \le (i+a)/p < (p-1+1)/p = 1$ for any $i \in \bar{I}_p$. 
In a similar manner we have $0 \le (i+b)/q < 1$ for any $i \in \bar{I}_q$. 
Thus the multi-set on the left-hand side of \eqref{eqn:succ1} also lies 
in the interval $[0, \, 1)$, as does the multi-set on the right. 
Therefore binary relation \eqref{eqn:succ1} mus be the inclusion 
\eqref{eqn:incl-a}.  
\par
To prove inclusion \eqref{eqn:incl-b}, we use the dual version of 
inclusion \eqref{eqn:incl-a}: 
\begin{equation} \label{eqn:incl-a-p}
\left\{ \frac{i'+a'}{p}\right\}_{i' \in \bar{I}'_p} \bigcup \,\,  
\left\{\frac{i'+b'}{q}\right\}_{i' \in \bar{I}'_q}
\subset \, 
\left\{\frac{j'-a'}{r-p}\right\}_{j' \in J'_p} \bigcup \,\, 
\left\{\frac{j'-b'}{r-q}\right\}_{j' \in J'_q},  
\end{equation}
where $I'_p$, $I'_q$, $J'_p$, $J'_q$ are the dual counterparts 
of $I_p$, $I_q$, $J_p$, $J_q$, with $\bar{I}'_p := [p] \setminus 
I'_p$ and $\bar{I}'_q := [r-q] \setminus I'_q$. 
Dividing equation \eqref{eqn:vi-s} by \eqref{eqn:IJ}, we have     
\[
\prod_{i=1}^r (w+v_i^*) = 
\prod_{i \in \bar{I}_p} \left(w+{\ts\frac{i+a}{p}}\right) 
\prod_{i \in \bar{I}_q} \left(w+{\ts\frac{i+b}{q}}\right) 
\prod_{j \in \bar{J}_p} \left(w+{\ts\frac{j-a}{r-p}}\right) 
\prod_{j \in \bar{J}_q} \left(w+{\ts\frac{j-b}{r-q}}\right).     
\]
Taking the reflection $w \mapsto w'$ as in \eqref{eqn:w-p} and 
using definitions \eqref{eqn:duality} and \eqref{eqn:vi-p} there 
yield 
\begin{align*}
\prod_{i=1}^r (w+v_i') &= 
\prod_{i \in \bar{I}_p} \left(w+{\ts\frac{(p-1-i)+a'}{p}}\right) 
\prod_{i \in \bar{I}_q} \left(w+{\ts\frac{(q-1-i)+b'}{q}}\right) \\
&\phantom{=} \times 
\prod_{j \in \bar{J}_p} \left(w+{\ts\frac{(r-p-1-j)-a'}{r-p}}\right) 
\prod_{j \in \bar{J}_q} \left(w+{\ts\frac{(r-q-1-j)-b'}{r-q}}\right),    
\end{align*}
which together with the definitions of $I'_s$, $\bar{I}'_s$ and 
$J'_s$, $s \in \{p, \, q\}$, implies 
\begin{equation} \label{eqn:IJ-p}
\begin{split}
I'_s &= \{ i' = s-1-i \,:\, i \in \bar{I}_s \}, \qquad 
\bar{I}'_s = \{ i' = s -1-i \,:\, i \in I_s \}, \\   
J'_s &= \{ j' = r-s-1-j \,:\, j \in \bar{J}_s \}  
\qquad \qquad (s \in \{p, \, q\}).   
\end{split}
\end{equation}
Under the correspondences $i' \leftrightarrow i$ and 
$j' \leftrightarrow j$ in \eqref{eqn:IJ-p}, it follows from 
definition \eqref{eqn:duality} that  
\[
\dfrac{i'+c'}{s} = 1 - \frac{2}{r} - \dfrac{i+c}{s}, \qquad 
\dfrac{j'-c'}{r-s} = 1 - \frac{2}{r} - \dfrac{j-c}{r-s}, 
\qquad ((s, c) = (p, a), \, (q, b)).
\]
Thus inclusion \eqref{eqn:incl-a-p} and relation \eqref{eqn:IJ-p} 
lead to inclusion \eqref{eqn:incl-b}. \hfill $\Box$   
\begin{proposition} \label{prop:ab} 
For any $(\rA)$-solution $\lambda = (p,q,r;a,b;x) \in \cD^-$ and its 
dual $(\rA)$-solution $\lambda' = (p,q,r;a',b';x) \in \cD^-$ the pair 
$(\lambda, \lambda')$ must be in one of the six cases in 
Table $\ref{tab:ab}$ and furthermore we must have               
\begin{equation} \label{eqn:estimate2} 
a, b, \, v_1, \dots, v_r \in \Q; \qquad 
0 \le a, \, b < 1, \qquad 0 \le v_1, \dots, v_r < 1, 
\end{equation} 
along with the same condition for $a', b'$ and $v_1', \dots, v_r'$. 
\end{proposition}
{\it Proof}. 
The first assertion concerning Table \ref{tab:ab} is an immediate 
consequence of Lemmas \ref{lem:case1-6} and \ref{lem:case7}. 
To prove the second assertion \eqref{eqn:estimate2}, notice that the 
rationality of $a$ and $b$ is already proved in Lemma \ref{lem:ab}, 
while $v_1, \dots, v_r \in \Q$ follows from $a, b \in \Q$ and 
formula \eqref{eqn:IJ}.  
The estimate $0 \le a, \, b < 1$ is already proved in 
Lemma \ref{lem:incl}, while $0 \le v_1, \dots, v_r < 1$ follows from 
a combination of $0 \le a, \, b < 1$, formula \eqref{eqn:IJ} and 
Lemma \ref{lem:ab-J}.  
Thus we have condition \eqref{eqn:estimate2}. 
\hfill $\Box$ \par\medskip 
Theorem \ref{thm:ab} is the same as Proposition \ref{prop:ab},   
while Theorem \ref{thm:D-r} is contained in the following.   
\begin{proposition} \label{prop:division} 
For any $(\rA)$-solution $\lambda = (p,q,r;a,b;x) \in \cD^-$ the 
division relation \eqref{eqn:division2} holds true along with 
yet another division relation            
\begin{equation} \label{eqn:division3} 
\prod_{i=0}^{p-1} \left(w+{\ts\frac{i+a}{p}}\right) 
\prod_{i=0}^{q-1} \left(w+{\ts\frac{i+b}{q}}\right) 
\,\, \Big| \,\, 
\prod_{j=0}^{r-p-1} \left(w+{\ts\frac{j-a}{r-p}}\right)
\prod_{j=0}^{r-q-1} \left(w+{\ts\frac{j-b}{r-q}}\right) 
\qquad \mbox{in \,\, $\Q[w]$},  
\end{equation} 
and hence one can arrange $v_1, \dots, v_r$ so that equation 
\eqref{eqn:vi-pq} holds. 
With this convention the reciprocal $\check{\lambda}$ of $\lambda$ 
becomes a solution to Problem $\rI$ in $\cF^-$ with gamma product 
formula    
\begin{equation} \label{eqn:gpf-c2} 
f(w; \check{\lambda}) = \check{C} \cdot \check{d}^w \cdot 
\dfrac{\prod_{i=0}^{r-p-q-1}\varGamma\left(w+\frac{i}{r-p-q} \right)}{
\prod_{i=1}^{r-p-q} \varGamma\left(w+ \check{v}_i \right)}, 
\end{equation}
where $\check{C}$ is a positive constant, $\check{d}$ and 
$\check{v}_1, \dots, \check{v}_{r-p-q}$ are given by formulas 
\eqref{eqn:d-c} and \eqref{eqn:vi-c}. 
\end{proposition}
{\it Proof}. 
The inclusions of multi-sets \eqref{eqn:incl-a} and \eqref{eqn:incl-b} 
are equivalent to the division relations  
\begin{subequations} \label{eqn:division4}
\begin{align}
\prod_{i\in \bar{I}_p} \left(w+ \ts \frac{i+a}{p} \right)   
\prod_{i \in \bar{I}_q} \left(w+ \ts \frac{i+b}{q} \right) \,\,
& \Big | \, 
\prod_{j \in J_p} \left(w+ \ts \frac{j-a}{r-p} \right)   
\prod_{j \in J_q} \left(w+ \ts \frac{j-b}{r-q} \right), 
\label{eqn:division4a} \\
\prod_{i\in I_p} \left(w+ \ts \frac{i+a}{p} \right)    
\prod_{i \in I_q} \left(w+ \ts \frac{i+b}{q} \right) \,\,
& \Big | \, 
\prod_{j \in \bar{J}_p} \left(w+ \ts \frac{j-a}{r-p} \right)  
\prod_{j \in \bar{J}_q} \left(w+ \ts \frac{j-b}{r-q} \right),    
\label{eqn:division4b}
\end{align}    
\end{subequations}
in $\C[w]$, respectively. 
Then division relation \eqref{eqn:division2} follows from formula 
\eqref{eqn:IJ} and relation \eqref{eqn:division4a}, while division 
relation \eqref{eqn:division3} is derived by multiplying relations 
\eqref{eqn:division4a} and \eqref{eqn:division4b} together. 
\par
Now that division relation \eqref{eqn:division2} is proved, the 
convention \eqref{eqn:vi-pq} recasts formula \eqref{eqn:h3} to   
\begin{equation} \label{eqn:h5} 
h(w;\lambda) = C_3 \cdot \tilde{d}^w \cdot 
\dfrac{\varGamma((r-p-q)w+1-a-b)}{\prod_{i=1}^{r-p-q} 
\varGamma\left(w+v_i\right)}.  
\end{equation}
Replacing $w$ by $\hat{w}$ in formula \eqref{eqn:f-h}, where $\hat{w}$ 
is defined by formula \eqref{eqn:w-h}, we have 
\[
\begin{split}
f(w; \check{\lambda}) 
&= x^{r \check{w}-1} (1-x)^{(p+q-r)\check{w}+a+b} \, 
h(\hat{w}; \lambda) \\[2mm]
&= C_3 \cdot x^{r \check{w}-1} (1-x)^{(p+q-r)\check{w}+a+b} \,  
\tilde{d}^{\hat{w}} \cdot 
\dfrac{\varGamma((r-p-q) \hat{w} +1-a-b)}{\prod_{i=1}^{r-p-q} 
\varGamma\left(\hat{w} + v_i\right)} \\[2mm]
&= C_4 \cdot x^{r w} (1-x)^{(p+q-r) w} \, \tilde{d}^w \cdot 
\dfrac{\varGamma\left( (r-p-q)w \right)}{\prod_{i=1}^{r-p-q} 
\varGamma\left(w + \check{v}_i\right)},   
\end{split}
\]
where \eqref{eqn:h5} is used in the second equality, 
definitions \eqref{eqn:vi-c} and \eqref{eqn:w-h} are used in 
the third equality and $C_4$ is a real constant. 
Applying the multiplication formula \eqref{eqn:mult} to   
$\varGamma\left((r-p-q)w \right)$ yields 
\[
f(w; \check{\lambda}) 
= \check{C} \cdot \{ x^r (1-x)^{p+q-r} \, \tilde{d} \, (r-p-q)^{r-p-q} \}^w 
\cdot \dfrac{\prod_{i=0}^{r-p-q-1}
\varGamma\left(w + \frac{i}{r-p-q}\right)}{\prod_{i=1}^{r-p-q} 
\varGamma\left(w + \check{v}_i\right)}, 
\]
where $\check{C}$ is a real constant. 
This gives GPF \eqref{eqn:gpf-c2}, since 
$x^r (1-x)^{p+q-r} \, \tilde{d} \, (r-p-q)^{r-p-q} = \check{d}$ 
by definition \eqref{eqn:d-c}.  
To see $\check{C} > 0$, we look at formula \eqref{eqn:gpf-c2} 
for a large positive value of $w$. 
The right-hand side of it without constant factor $\check{C}$ is positive, 
while the left-hand side $f(w;\check{\lambda})$ is also positive since 
$\check{\lambda} \in \cF^-$.  
Here we used the fact that if $\mu \in \cF^-$ then $f(w; \mu) > 0$ 
for every large $w > 0$, which will be shown in the proof of 
Lemma \ref{lem:zeros}; see claim \eqref{eqn:f>0}. 
Thus $\check{C} > 0$. \hfill $\Box$
\begin{remark} \label{rem:division} 
In the situation of Proposition \ref{prop:division} we have 
$\check{a}$, $\check{b} \in \Q$, $\check{x}$ algebraic, and 
\begin{gather}
\check{v}_1 + \cdots + \check{v}_{r-p-q} = \frac{r-p-q-1}{2} = 
\frac{\check{r}-1}{2}, \label{eqn:sum2} \\
\check{v}_1, \dots, \check{v}_{r-p-q} \in \Q, \qquad 
-c \le \check{v}_1, \dots, \check{v}_{r-p-q} < 1-c, \qquad 
c := c(\lambda) = \frac{1-a-b}{r-p-q}. \label{eqn:estimate3}   
\end{gather}
Indeed, since $a$ and $b$ are rational by Lemma \ref{lem:ab} while 
$x$ is algebraic by assertion (1) of Theorem \ref{thm:sol-A}, so are 
$\check{a}$, $\check{b}$ and $\check{x}$ via the definition of 
reciprocity \eqref{eqn:recip}.    
Moreover summation \eqref{eqn:sum2} comes from \eqref{eqn:sum} together 
with convention \eqref{eqn:vi-pq} and definition \eqref{eqn:vi-c}.   
while condition \eqref{eqn:estimate3} follows from \eqref{eqn:estimate2} 
via definition \eqref{eqn:vi-c}.   
These remarks will play an important part in \S \ref{subsec:gpf-F}. 
\end{remark}
\section{Division Relations} \label{sec:dr} 
In Lemma \ref{lem:case1-6} we derived some arithmetical constraints 
on $\bp = (p, q; r)$ for $(\rA)$-solutions in $\cD^-$. 
In this section we exploit division relation \eqref{eqn:division3} 
to amplify this kind of study.    
In what follows let $\lambda = (p,q,r;a,b;x) \in \cD^-$ be an 
arbitrary solution of type $(\rA)$.   
\begin{lemma} \label{lem:dr1} 
If $p \not| r$ and $p \not| (r-p-q)$, then $p | 2r$ and $p | 2(r-p-q)$ 
with $p \ge 4$.  
\end{lemma} 
{\it Proof}. 
If $p = 1$ we obviously have $p|r$. 
If $p = 2$ we have $p|(r-p-q)$ since $r-p-q$ must be even. 
Hereafter we suppose $p \ge 3$.  
Since $\prod_{i=0}^{p-1}\left( w+(i+a)/p \right)$ divides the 
right-hand side of \eqref{eqn:division3}, there exists a mapping 
$\phi : [p] \to [r-p] \sqcup [r-q]$ such that for each $i \in [p]$,  
\begin{equation} \label{eqn:hm-ht}
\dfrac{i+a}{p} = 
\begin{cases}
\dfrac{\phi(i) - a}{r-p} \quad & \mbox{if} \quad \phi(i) \in [r-p] 
\quad (\mbox{$i$ is homogeneous, or of type $\rI$}), \\[4mm]
\dfrac{\phi(i) - b}{r-q} \quad & \mbox{if} \quad \phi(i) \in [r-q] 
\quad (\mbox{$i$ is heterogeneous, or of type $\rII$}). 
\end{cases}
\end{equation}
Then a sequence $\tau = (\tau_0, \tau_1, \dots, \tau_{p-1})$ of 
symbols $\rI$ and $\rII$ is defined by assigning $\tau_i = \rI$ or 
$\tau_i = \rII$ if $i$ is homogeneous or heterogeneous respectively. 
There is a dichotomy: 
\begin{enumerate}
\item There exists at least one index $i \in [p-1]$ such that 
$\tau_i = \tau_{i+1}$. 
\item The sequence $\tau$ is interlacing, that is, 
$\tau_i \neq \tau_{i+1}$ for any index $i \in [p-1]$. 
\end{enumerate}
\par
We begin with case (1). 
In this case we have either $\tau_i = \tau_{i+1} = \rI$ or 
$\tau_i = \tau_{i+1} = \rII$. 
In the former subcase, it follows from formula \eqref{eqn:hm-ht} that  
\[
\dfrac{i+a}{p} = \dfrac{\phi(i) - a}{r-p} \quad \mbox{and} \quad  
\dfrac{(i+1)+a}{p} = \dfrac{\phi(i+1) - a}{r-p}, \quad \mbox{so that} \quad 
\dfrac{1}{p} = \dfrac{\phi(i+1) - \phi(i)}{r-p}, 
\]
that is, $r-p = p \{\phi(i+1)-\phi(i)\}$, which implies $p|r$. 
In the latter subcase, similarly we have    
\[
\dfrac{i+a}{p} = \dfrac{\phi(i) - b}{r-q} \quad \mbox{and} \quad  
\dfrac{(i+1)+a}{p} = \dfrac{\phi(i+1) - b}{r-q}, \quad \mbox{so that} \quad 
\dfrac{1}{p} = \dfrac{\phi(i+1) - \phi(i)}{r-q}, 
\]
that is, $r-q = p \{\phi(i+1)-\phi(i)\}$, which implies $p|(r-p-q)$.    
\par
We proceed to case (2) with $p = 3$. 
In this case we have either $\tau = (\rI, \rII, \rI)$ or 
$\tau = (\rII, \rI, \rII)$. 
In the former subcase, it follows from \eqref{eqn:hm-ht} 
and $\tau_0 = \tau_2 = \rI$ that 
\[
\dfrac{a}{p} = \dfrac{\phi(0) - a}{r-p} \quad \mbox{and} \quad  
\dfrac{2+a}{p} = \dfrac{\phi(2) - a}{r-p}, \quad \mbox{so that} \quad 
\dfrac{2}{p} = \dfrac{\phi(2) - \phi(0)}{r-p}, 
\]
that is, $2(r-p) = p \{\phi(2)-\phi(0)\}$, which implies $p|2 r$, 
but since $p = 3$ is odd, we must have $p|r$. 
In the latter subcase, a similar reasoning with $\tau_0 = \tau_2 = 
\rII$ leads to $2(r-q) = p \{\phi(2)-\phi(0)\}$, which implies 
$p|2(r-p-q)$, but since $p = 3$ is odd, we must have $p|(r-p-q)$.   
\par
Finally we consider case (2) with $p \ge 4$. 
In this case we have either $(\tau_0, \tau_1, \tau_2, \tau_3) = 
(\rI, \rII, \rI, \rII)$ or $(\tau_0, \tau_1, \tau_2, \tau_3) = 
(\rII, \rI, \rII, \rI)$.  
In the former subcase, it follows from \eqref{eqn:hm-ht} that 
\begin{align*}
\dfrac{a}{p} &= \dfrac{\phi(0) - a}{r-p} \quad \mbox{and} \quad  
\dfrac{2+a}{p} = \dfrac{\phi(2) - a}{r-p}, \quad \mbox{so that} \quad 
\dfrac{2}{p} = \dfrac{\phi(2) - \phi(0)}{r-p}, \\[4mm]
\dfrac{1+a}{p} &= \dfrac{\phi(1)-b}{r-q} \quad \mbox{and} \quad  
\dfrac{3+a}{p} = \dfrac{\phi(3)-b}{r-q}, \quad \mbox{so that} \quad 
\dfrac{2}{p} = \dfrac{\phi(3) - \phi(1)}{r-q},
\end{align*}
that is, $2(r-p) = p \{\phi(2)-\phi(0)\}$ and 
$2(r-q) = p \{\phi(3)-\phi(1)\}$, which imply 
$p|2 r$ and $p|2(r-p-q)$. 
In the latter subcase, a similar reasoning leads to 
$2(r-p) = p \{\phi(3)-\phi(1)\}$ and $2(r-q) = p \{\phi(2)-\phi(0)\}$, 
which again imply $p|2 r$ and $p|2(r-p-q)$. 
Thus if $p \not| r$ and $p \not| (r-p-q)$, then we must be in case (2) 
with $p \ge 4$, which forces $p | 2r$ and $p | 2(r-p-q)$. \hfill $\Box$ 
\begin{lemma} \label{lem:dr2} 
If $p \not| r$ and $p \not|(r-p-q)$, then 
\begin{enumerate} 
\item there exist positive integers $s$, $t$ and $k$ such that 
\begin{equation} \label{eqn:stk}
p = 2^{k+1} \, s; \qquad p|q; \qquad r = 2^k \, t; 
\qquad s, \, t : \mbox{odd}; \qquad s|t,   
\end{equation}
\item we must also have $q \not| r$ and $q \not|(r-p-q)$. 
\end{enumerate}
\end{lemma}
{\it Proof}. 
We use Lemma \ref{lem:dr1}, upon writing $p = 2^i \, s$, $r = 2^k \, t$, 
$r-q = 2^j \, u$ with $i,j,k \in \Z_{\ge0}$ and odd $s, t, u \in \N$.  
Division relation $p| 2 r$ implies $i \le k+1$ and $s|t$, while 
$p \not| r$ yields $i \ge k+1$ and hence $i = k+1$. 
In a similar manner the division relation $p| 2 (r-q)$ implies 
$i \le j+1$ and $s|u$, while $p \not| (r-q)$ yields $i \ge j+1$ 
and hence $i = j+1$. 
In summary we have 
\[
p = 2^{k+1} \, s; \quad r = 2^k \, t; \quad 
r-q = 2^k \, u; \quad k \in \Z_{\ge0}; \quad 
s, \,t, \,u : \mbox{odd}; \quad s|t, \quad s|u.  
\]
Since $t$ and $u$ are odd, we can write $t = 2 t'+1$ and $u = 2 u'+1$ 
with $t', u' \in \Z_{\ge0}$, so that $s|t$ and $s|u$ imply 
$s|(t-u) = s|2(t'-u')$, which in turn yields $s|(t'-u')$ since 
$s$ is also odd.  
It then implies $p|q$ because $p = 2^{k+1} \, s$ and $q = r-(r-q) = 
2^k (t-u) = 2^{k+1} (t'-u')$. 
Note that $p$ is even by $p = 2^{k+1} \, s$ with $k \ge 0$, 
so $q$ is even too by $p|q$ and $r$ is also even, since $r-p-q$ is even.  
Thus we have $k \ge 1$ and all the conditions in \eqref{eqn:stk} 
have been proved. 
\par
Next we show that $q \not| r$ and $q \not|(r-p-q)$ by contradiction. 
Indeed, if $q|r$ then the division relation $p|q$ in condition 
\eqref{eqn:stk} yields $p|r$ contrary to the assumption $p \not|r$, 
while if $q|(r-p-q)$ then $p|q$ gives $p|(r-p-q)$ contrary to the 
assumption $p \not|(r-p-q)$. \hfill $\Box$ \par\medskip 
Note that Lemmas \ref{lem:dr1} and \ref{lem:dr2} remain true 
if the roles of $p$ and $q$ are exchanged. 
\begin{proposition} \label{prop:dr} 
For any $(\rA)$-solution $\lambda = (p,q,r;a,b;x) \in \cD^-$ the integer 
vector $\bp = (p,q;r) \in D^-_{\rA}$ must satisfy division relations 
\eqref{eqn:dr}.   
\end{proposition}
{\it Proof}. 
To prove the lemma by contradiction, suppose the contrary that 
\[
(\, p \not| r \quad \mbox{and} \quad p \not| (r-p-q) \,) 
\quad \mbox{or} \quad 
(\, q \not| r \quad \mbox{and} \quad q \not| (r-p-q) \,).
\]
By symmetry we may take the former condition in the ``or" sentence 
above, but the latter condition also follows from assertion (2) of 
Lemma \ref{lem:dr2}, 
so that we are led to the ``and" sentence:   
\[
(\, p \not| r \quad \mbox{and} \quad p \not| (r-p-q) \,) 
\quad \mbox{and} \quad 
(\, q \not| r \quad \mbox{and} \quad q \not| (r-p-q) \,).
\]
By a part of condition \eqref{eqn:stk} we have $p|q$ and likewise  
$q|p$ upon exchanging the role of $p$ and $q$. 
Hence $p=q$ and condition \eqref{eqn:stk} becomes 
\begin{equation} \label{eqn:stk2}
p = q = 2^{k+1} \, s; \qquad r = 2^k \, t; 
\qquad k \in \N, \quad s, \, t : \mbox{odd}; \qquad s|t,   
\end{equation}
\par
A look at Table \ref{tab:ab} shows that if $p = q$ then we must have 
$p|r$ in cases (1)--(5), while currently we have the contrary 
$p \not| r$. 
Thus we must be in case 6 of Table \ref{tab:ab}, so that $\Z$-linear 
equation \eqref{eqn:ls-c6a} must be satisfied. 
It follows from formula \eqref{eqn:stk2} that $r-p-q = 2^k s(t_s-4)$, 
$r-p = 2^k s(t_s-2)$, $q = 2^{k+1} s$ and 
$((r-p)(r-p-q))_p = 2^{k-1} s(t_s-2)(t_s-4)$, so equation 
\eqref{eqn:ls-c6a} is equivalent to 
\[
(t_s-4) i + (t_s-2) i' + 2 j = \frac{1}{2} (t_s-2)(t_s-4), \qquad 
i, i', j \in \Z_{\ge0},  
\]
where we used the notation introduced in item (1) of Remark \ref{rem:ab}.  
Notice that the left-hand side above is an integer, while the 
right-hand side is a half-integer (not an integer), because $s$ and 
$t$ are odd integers with $s|t$ so $t_s = t/s$ is also an odd integer. 
This contradiction shows that our starting assumption is false 
and condition \eqref{eqn:dr} must be true. 
\hfill $\Box$ \par\medskip 
Theorem \ref{thm:pqr} is now established because it is the same as 
Proposition \ref{prop:dr}. 
By definition \eqref{eqn:D-m}, if $\bp = (p,q;r) \in D^-_{\rA}$ then 
$\check{r} := r-p-q \in 2 \N$.  
Conversely we have the following.  
\begin{lemma} \label{lem:dr-bound}  
Given any $\check{r} \in 2 \N$, there are only a finite number of triples 
$\bp = (p,q;r) \in D^-_{\rA}$ that satisfy $r-p-q = \check{r}$ and the 
division relations \eqref{eqn:dr}; any such $\bp$ must be bounded by 
\begin{equation} \label{eqn:dr-bound}
1 \le p, \, q \le 3 \check{r}, 
\qquad 2 \le p+q \le 5 \check{r}, 
\qquad 4 \le r \le 6 \check{r}.  
\end{equation}
\end{lemma}
{\it Proof}. 
Since $p, \, q \in \N$ and $r \in 2 \N$, it is evident that 
$p, \, q \ge 1$ and $r = \check{r}+p+q \ge 4$. 
Division relation \eqref{eqn:dr} is equivalent to the condition that 
$( p|\check{r} \,\, \mbox{or} \,\, p|(\check{r}+q) )$ and 
$( q|\check{r} \,\, \mbox{or} \,\, q|(\check{r}+p) )$, which is divided 
into four cases: (i) $p|\check{r}$ and $q|\check{r}$; 
(ii) $p|\check{r}$ and $q|(\check{r}+p)$;   
(iii) $p|(\check{r}+q)$ and $q|\check{r}$; 
(iv) $p|(\check{r}+q)$ and $q|(\check{r}+p)$. 
In case (i) we have $p, \, q \le \check{r}$ and $r = \check{r}+p+q \le 3 
\check{r}$. 
In case (ii) we have $p \le \check{r}$, $q \le \check{r}+p \le 2 \check{r}$ 
and $r = \check{r}+p+q \le 4 \check{r}$. 
Case (iii) is similar to case (ii). 
In case (iv) there exist $i$, $j \in \N$ such that 
$\check{r}+q = i p$ and $\check{r}+p = j q$. 
Note that $i j \ge 2$, for otherwise $i = j = 1$ would imply 
$\check{r} = p-q = - \check{r} = 0$, a contradiction to $\check{r} \ge 4$. 
Thus the two equations above for $(p, q)$ are uniquely settled as 
$p = l \, \check{r}$, $q = m \, \check{r}$ and hence  
$r = \check{r}+p+q = n \, \check{r}$, where 
\[
l = \dfrac{j+1}{i j-1}, \qquad m = \dfrac{i+1}{i j-1}, \qquad 
n = 1+l+m = \dfrac{(i+1)(j+1)}{i j-1} \qquad (i, j \in \N; \, i j \ge 2). 
\] 
To estimate these numbers we may assume $i \ge j$ and so 
$i \ge 2$ and $j \ge 1$ by symmetry.  
Then  
\[
l \le \frac{j+1}{2j-1} = \frac{1}{2} + \frac{3}{2(2j-1)} \le 
\frac{1}{2} + \frac{3}{2} = 2, \qquad  
m \le \frac{i+1}{i-1} = 1 + \frac{2}{i-1} \le 1+2=3,  
\]
and $n = 1+l+m \le 1+ 2+3 =6$. 
This establishes the bound \eqref{eqn:dr-bound}. 
\hfill $\Box$ \par\medskip
For example, those $\bp = (p,q;r) \in D_{\rA}^-$ with 
$\check{r} = 2$ and $p \ge q$ are exactly $\bp = (1,1;4)$, $(2,1;5)$, 
$(2,2;6)$, $(3,1;6)$, $(4,2;8)$, among which only $(2,1;5)$ leads to 
no solutions (see \cite[Table 2]{Iwasaki}). 
\section{The South-West Domain} \label{sec:SW}
We discuss Problem \ref{prob:back} (regarding the relationship between 
Problems $\rI$ and $\rII$) and Problem \ref{prob:cfcr} (regarding the 
method of contiguous relations) on the south-west domain $\cF^-$. 
The corresponding work on the cross-shaped domain $\cD \cup \cI \cup \cE$ 
was carried out in \cite{Iwasaki}, where Problem \ref{prob:back} was 
treated  first (\S 5) and Problem \ref{prob:cfcr} was discussed 
later (\S 11.1). 
This logical order will be reversed on the domain $\cF^-$, 
that is, we first tackle Problem \ref{prob:cfcr} (as well as 
Problem \ref{prob:cfcr-ess}) and then proceed to Problem \ref{prob:back}. 
The main tool for this passage is the use of reciprocity.    
\subsection{Coming from Contiguous Relations} \label{subsec:cfcr-F}
In solving Problem \ref{prob:cfcr} it is important to think of the 
linear independence of $f(w; \lambda)$ and $\tilde{f}(w; \lambda)$ 
over the rational function field $\C(w)$, where $\tilde{f}(w; \lambda)$ 
is defined right after formula \eqref{eqn:3trf}. 
This issue was already discussed in \cite[\S 11.1]{Iwasaki} on the 
domain $\cD \cup \cI \cup \cE$, where the equivalence of Problems 
$\rI$ and $\rII$ (Theorem \ref{thm:back}) made it more tractable.   
Without such an advantage, our discussion on $\cF^-$ should be more 
elaborate and require some function-theoretic preliminaries.    
\begin{lemma} \label{lem:sin}
If $\alpha_1, \dots, \alpha_k$ and $\beta_1, \dots, \beta_k$ are 
real numbers with $\alpha_1 \ge \cdots \ge \alpha_k > 0$, then   
for any integer $m \ge \beta_1$ there exists a positive number 
$\rho_m \in I_m$ such that 
\begin{equation} \label{eqn:sin1}
\prod_{j=1}^k \dfrac{1}{|\sin \pi(\alpha_j w + \beta_j)|} \le C 
\qquad \mbox{on the circle} \quad |w| = \rho_m, 
\end{equation}
where $C$ is a positive constant independent of $m$ and $I_m$ is 
an open interval defined by  
\[
I_m := \left(\frac{m-\beta_1}{\alpha_1}, \,\, 
\frac{m+1-\beta_1}{\alpha_1}\right) \qquad (m \in \Z).  
\]
\end{lemma}
{\it Proof}. 
Note that $I_m$ is an open interval of width $1/\alpha_1$, whose 
endpoints are a pair of consecutive zeros of $\sin \pi(\alpha_1 w+ \beta_1)$. 
For each $j = 2, \dots,k$, all the zeros of $\sin \pi(\alpha_j w + \beta_j)$ 
form an arithmetic progression 
$w_{j,m}:=\frac{m - \beta_j}{\alpha_j}$ $(m \in \Z)$ of common difference 
$1/\alpha_j \ge 1/\alpha_1$, so that $I_m$ can contain at most one zero of 
$\sin \pi(\alpha_j w + \beta_j)$. 
Thus $I_m$ can contain at most $k-1$ zeros of $\prod_{j=2}^k 
\sin \pi(\alpha_j w+\beta_j)$, which partition $I_m$ into at most 
$k$ subintervals. 
Among them let $J_m$ be a subinterval of the largest width.   
By pigeon hole principle the width of $J_m$ cannot be smaller than 
$\frac{1}{k \alpha_1}$, so that the midpoint $\rho_m$ of $J_m$ is 
at least $\delta := \frac{1}{2 k \alpha_1}$ distant from all the 
zeros of $\prod_{j=1}^k \sin \pi(\alpha_j w + \beta_j)$. 
For $j = 1, \dots,k$ and $m \in \Z$ let $D_{j,m}$ be the open disk 
of radius $\delta$ with center at $w_{j,m}$. 
Then it is not hard to see that there exists a positive constant 
$C_j$ such that 
\[
\dfrac{1}{|\sin \pi(\alpha_j w + \beta_j)|} \le C_j \qquad 
\mbox{for all} \quad w \in \C \setminus D_j, 
\quad \mbox{with} \quad D_j := \bigcup_{m \in \Z} D_{j,m}. 
\]
Accordingly, if we set $C := C_1 \cdots C_k$ then we have  
\begin{equation} \label{eqn:sin2}
\prod_{j=1}^k \dfrac{1}{|\sin \pi(\alpha_j w + \beta_j)|} \le C 
\qquad \mbox{for all} \quad w \in K := \C \setminus \bigcup_{j=1}^k D_j. 
\end{equation}
Notice that $\rho_m \in K$ for any $m \in \Z$. 
If $m \ge \beta_1$ then $\rho_m > 0$ and the circle $|w| = \rho_m$ 
is contained in $K$, hence estimate \eqref{eqn:sin2} leads to 
estimate \eqref{eqn:sin1}. \hfill $\Box$
\begin{lemma} \label{lem:hg-estimate}
If $\lambda = (p,q,r;a,b;x)$ lies in domain \eqref{eqn:real2} 
then there exists an infinite sequence of positive numbers 
$\{ \rho_m \}_{m \ge m_0}$ such that for any integer $m \ge m_0$,  
\begin{equation} \label{eqn:hg-estimate}
\sigma m + \tau < \rho_m < \sigma(m+1) + \tau, \qquad 
|f(w; \lambda)| \le M_1 e^{c_1 \rho_m} \quad \mbox{on the circle} 
\quad |w| = \rho_m,   
\end{equation}
where $M_1$, $c_1$, $\sigma > 0$ and $\tau \in \R$ are independent of 
$m$ and $m_0$ is an integer with $\sigma m_0 + \tau \ge 0$. 
\end{lemma}
{\it Proof}.  
The hypergeometric function admits Euler's contour integral representation 
\[
\hgF(\alpha,\beta;\gamma;z) = - \dfrac{e^{-\pi i \gamma}}{4 \sin \pi \beta 
\cdot \sin \pi(\gamma-\beta) \cdot B(\beta, \, \gamma-\beta)} 
\int_{\wp} t^{\beta-1} (1-t)^{\gamma-\beta-1} (1-z t)^{-\alpha} d t,     
\]
along a Pochhammer loop $\wp$ around $t=0$ and $t=1$. 
Notice that  
\[
\begin{split}
\dfrac{1}{B(\beta, \gamma-\beta)} 
&= (1-\gamma) \cdot 
\dfrac{\sin \pi \beta \cdot \sin \pi(\gamma-\beta)}{\pi \sin \pi \gamma} \cdot 
B(1-b, \, 1-(\gamma-\beta)) \\
&= - \dfrac{(1-\gamma) e^{\pi i \gamma}}{4 \pi \cdot \sin \pi \gamma} \int_{\wp} 
t^{-\beta} (1-t)^{-(\gamma-\beta)} d t,  
\end{split}
\]
where the first equality follows from the reflection formula 
for the beta function: 
\[
B(\alpha, \beta) B(1-\alpha, 1-\beta) = 
\frac{\pi \sin \pi(\alpha+\beta)}{(1-\alpha-\beta) \cdot 
\sin \pi \alpha \cdot \sin \pi \beta}, 
\]
while the second equality stems from Euler's integral representation 
of the beta function along the Pochhammer loop $\wp$.  
Putting these together we have  
\[
\begin{split}
\hgF(\alpha, \beta; \gamma; z) &= \dfrac{1-\gamma}{16 \pi \cdot \sin \pi \beta 
\cdot \sin \pi \gamma \cdot \sin \pi(\gamma-\beta)} \\
&\phantom{=} \times 
\left(\int_{\wp} t^{-\beta}(1-t)^{-(\gamma-\beta)} d t\right) 
\left(\int_{\wp} t^{\beta-1} (1-t)^{\gamma-\beta-1} (1-z t)^{-\alpha} d t \right),  
\end{split}
\]
which evaluated at $(\alpha, \beta; \gamma; z) = (p w + a, q w + b; r w; x)$ 
yields $f(w; \lambda) = \psi_1(w) \psi_2(w)$ with 
\[
\begin{split}
\psi_1(w) &= \dfrac{1- r w}{16 \pi \cdot \sin \pi (q w+b) 
\cdot \sin(\pi r w) \cdot \sin \pi((r-q)w-b)}, \\[1mm]
\psi_2(w) &= \left(\int_{\wp} t^{- q w-b}(1-t)^{-(r-q)w+b} d t\right) 
\left(\int_{\wp} t^{q w+b-1} (1-t)^{(r-q)w-b-1} (1-x t)^{-p w-a} d t \right). 
\end{split}
\]
\par
We can apply Lemma \ref{lem:sin} to the first factor $\psi_1(w)$. 
For some constants $\sigma > 0$ and $\tau \in \R$ there exists an infinite 
sequence $\rho_m \in (\sigma m + \tau, \, \sigma(m+1) + \tau)$ such that 
$|\psi_1(w)| = O(\rho_m)$ on the circle $|w| = \rho_m$ as $m \to + \infty$. 
For the second factor $\psi_2(w)$ it is easy to see that there exists 
a constant $c_2$ such that $|\psi_2(w)| = O(e^{c_2|w|})$ as $|w| \to +\infty$, 
because the integrands in $\psi_2(w)$ admit similar exponential estimates,  
uniform in $t \in \wp$, as $|w| \to +\infty$.  
Now \eqref{eqn:hg-estimate} follows readily. \hfill $\Box$ 
\begin{lemma} \label{lem:zeros}
Let $\lambda = (p,q,r;a,b;x) \in \cF^-$ be a solution to Problem $\rII$.     
\begin{enumerate}
\item Any pole of $f(w;\lambda)$ is simple and lies in the arithmetic 
progression $\{w_j := -j/r\}_{j=0}^{\infty}$. 
Conversely, $f(w; \lambda)$ actually has a pole at $w = w_j$ for every 
sufficiently large integer $j$, in particular it has infinitely many poles.        
\item $f(w;\lambda)$ has infinitely many zeros.   
\item $f(w;\lambda)$ and $\tilde{f}(w;\lambda)$ are linearly independent 
over the rational function field $\C(w)$. 
\end{enumerate} 
\end{lemma}
{\it Proof}. Assertion (1). 
It is evident from definition \eqref{eqn:f} that every pole of 
$f(w;\lambda)$ is simple and lies in the sequence $\{w_j\}_{j=0}^{\infty}$. 
By Lemma 4.1 of Iwasaki \cite{Iwasaki} we have   
\[ 
\underset{\scriptstyle w = w_j}{\mathrm{Res}} \, f(w) = C_j \cdot 
\hgF(a_j, \, b_j; \, j+2; \, x) \qquad (j = 0,1,2,\dots),  
\]
where $a_j := pw_j+j+a+1$, $b_j := qw_j+j+b+1$ and 
\[
C_j := \frac{(-1)^j}{r} \cdot
\frac{(p w_j+a)_{j+1} \, (q w_j+b)_{j+1}}{j! \, (j+1)!} \, x^{j+1}. 
\]
Since $\lambda \in \cF^-$, we have $p < 0$, $q < 0$, $r > 0$ and 
$0 < x < 1$, so there exists an integer $j_0$ such that 
$p w_j + a > 0$, $q w_j + b > 0$ and hence $(-1)^j C_j > 0$ 
for every $j \ge j_0$. 
Notice also that if $j \ge j_0$ then $a_j > p w_j + a > 0$, 
$b_j > q w_j + b > 0$ and thus $\hgF(a_j, b_j; j+2; x) > 0$.  
Therefore $f(w)$ actually has a simple pole with a non-vanishing 
residue at $w = w_j$ for every $j \ge j_0$.  
\par
Assertion (2). 
Suppose the contrary that $f(w;\lambda)$ has at most finitely many zeros. 
Then it follows from assertion (1) that $u(w) := f(w;\lambda)/\varGamma(r w)$ 
is an entire holomorphic function with at most finitely many zeros. 
Here we have a uniform estimate $|1/\varGamma(r w)| = 
O\left(e^{r |w| (\log|w| + c_3)}\right)$ as $|w| \to \infty$, 
which follows from Stirling's formula: for any $\varepsilon \in 
(0, \, \pi)$ one has uniformly 
\begin{equation} \label{eqn:stirling}
\dfrac{1}{\varGamma(w)} \sim 
\dfrac{1}{\sqrt{2\pi}} \, w^{1/2} \, e^{w(1-\log w)} \qquad 
\mbox{as} \quad w \to \infty \quad \mbox{in} \quad |\arg w| < 
\pi - \varepsilon, 
\end{equation}
combined with the reflection formula \eqref{eqn:refl} for 
the gamma function.  
By this estimate and Lemma \ref{lem:hg-estimate} we have  
$|u(w)| \le M_2 \, e^{r \rho_m (\log \rho_m + c_4)}$ on the circle 
$|w| = \rho_m$ for every $m \ge m_0$. 
Given any $z \in \C$ with sufficiently large $|z|$, take an $m \ge m_0$ 
so that $\rho_{m-1} < |z| \le \rho_m$. 
Since $u(w)$ is entire and $\rho_m < \rho_{m-1} + 2 \sigma < |z| + 2 \sigma$ 
by estimate \eqref{eqn:hg-estimate}, the maximum principle yields 
\[
|u(w)| \le M_2 \, e^{r \rho_m (\log \rho_m + c_4)} 
\le M_2 \, e^{r (|z|+ 2 \sigma) \{\log (|z| + 2 \sigma) + c_4\}} 
\le M_3 \, e^{r |z| (\log |z| + c_5)},  
\]
which means that $u(w)$ is an entire function of order at most $1$.  
Since $u(w)$ has at most finitely many zeros, Hadamard's factorization 
theorem allows us to write $u(w) = u_0(w) \, e^{c_6 w + c_7 w^2}$, 
where $u_0(w)$ is a nonzero polynomial (or constant) and 
$c_6$, $c_7 \in \C$, so that   
\[
\dfrac{u(w+1)}{u(w)} = \dfrac{u_0(w+1)}{u_0(w)} 
\cdot e^{c_6 + c_7 + 2 c_7 w} \sim e^{c_6 + c_7 + 2 c_7 w} 
\qquad \mbox{as} \quad w \to \infty.  
\]
On the other hand, since $\lambda$ is a solution to Problem $\rII$, 
that is, satisfies condition \eqref{eqn:ocf}, we have  
\[
\dfrac{u(w+1)}{u(w)} = \dfrac{f(w+1; \lambda)}{f(w; \lambda)} 
\cdot \dfrac{\varGamma(r w)}{\varGamma(r(w+1))} = 
R(w;\lambda) \cdot \dfrac{\varGamma(r w)}{\varGamma(r(w+1))}.  
\]
As a rational function, $R(w;\lambda) \sim M_4 \, w^{k_1}$ 
as $w \to \infty$ for some $M_4 \in \C^{\times}$ and $k_1 \in \Z$, 
while by Stirling's formula \eqref{eqn:stirling} we have   
$\varGamma(r w)/\varGamma(r(w+1)) \sim (r w)^{-r}$ as 
$\R \ni w \to +\infty$. 
Thus 
\[
\dfrac{u(w+1)}{u(w)} \sim M_4 \, r^{-r} \cdot w^{k_1-r} 
\quad \mbox{and so} \quad w^{r-k_1} e^{2 c_7 w} 
\sim M_4 \, r^{-r} e^{-c_6-c_7} \quad  
\mbox{as} \quad \R \ni w \to + \infty, 
\]
which forces $c_7 = 0$ and hence $u(w) = u_0(w) \, e^{c_6 w}$. 
As a polynomial, $u_0(w) \sim M_5 \, w^{k_2}$ as 
$w \to \infty$ for some $M_5 \in \C^{\times}$ and 
$k_2 \in \Z_{\ge0}$. 
It follows from estimates \eqref{eqn:hg-estimate} and 
\eqref{eqn:stirling} that 
\[
\begin{split}
1 &= \left|\dfrac{u(\rho_m)}{u_0(\rho_m) \, e^{c_6 \rho_m}} \right| 
= \left|\dfrac{f(\rho_m; \lambda)}{u_0(\rho_m) \, e^{c_6 \rho_m}} 
\cdot \frac{1}{\varGamma(r \rho_m)} \right| \underset{\sim}{<} 
\dfrac{M_1 \, e^{c_1 \rho_m} \cdot (r \rho_m)^{1/2} 
e^{r \rho_m \{1-\log(r \rho_m) \}}}{|M_5| \, \rho_m^{k_2} \cdot e^{c_6 \rho_m} \cdot 
\sqrt{2\pi}} \\
&= M_6 \cdot \rho_m^{\frac{1}{2}-{k_2}} \cdot e^{\rho_m (c_8 - r \log \rho_m)} 
\to 0 \qquad \mbox{as} \quad m \to \infty, 
\end{split}
\]
where $M_6 := \frac{M_1}{|M_5|}\sqrt{\frac{r}{2\pi}}$ and 
$c_8 := c_1 - c_6 + r - r \log r$. 
This contradiction shows that $f(w;\lambda)$ must have infinitely many 
zeros. 
\par
Assertion (3) can be established by modifying the proof of Lemma 11.1 in 
Iwasaki \cite{Iwasaki}. 
Suppose the contrary that $f(w; \lambda)$ and $\tilde{f}(w; \lambda)$ are 
linearly dependent over $\C(w)$, so that there exists a rational function 
$T(w) \in \C(w)$ such that $\tilde{f}(w; \lambda) = T(w) f(w;\lambda)$, 
because $f(w; \lambda)$ does not vanish identically. 
Then as in the proof of \cite[Lemma 11.1]{Iwasaki} we have 
\begin{equation} \label{eqn:f-f1}
f(w; \lambda) f_1(w; \lambda) = (1-x)^{(r-p-q)w-a-b-1},  
\end{equation}
where $f_1(w; \lambda)$ is defined by    
\[
\begin{split}
f_1(w; \lambda) &:= \hgF( (p-r)w+a+1, (q-r)w+b+1; 1-r w; x) \\
&\phantom{:=} + \dfrac{(p w+a)(q w+b) x}{r w(r w-1)} \cdot T(w) \cdot 
\hgF( (p-r)w+a+1, (q-r)w+b+1; 2-r w; x).    
\end{split}
\]
Observe that any pole of $f_1(w; \lambda)$ is either a pole of 
$T(w)$ or in the discrete set $\frac{1}{r} \Z_{\ge0}$. 
Thus $f_1(w; \lambda)$ cannot have infinitely many 
poles off the positive real axis $\R_+$.  
On the other hand, applying Euler's transformation \eqref{eqn:transf1} 
to definition \eqref{eqn:f}, we have   
\[
f(w;\lambda) = (1-x)^{(r-p-q)w-a-b} \hgF( (r-p)w-a, (r-q)w-b; r w; x), 
\]
where $r-p > 0$, $r-q > 0$ and $r > 0$ while $0 < x < 1$ 
by the assumption $\lambda \in \cF^-$.  
Thus 
\begin{equation} \label{eqn:f>0}
f(w;\lambda) > 0 \qquad \mbox{for every real} \quad  
w > \max \{a/(r-p), \, b/(r-q), \, 0\}, 
\end{equation}
so $f(w;\lambda)$ cannot have infinitely many zeros on $\R_+$.   
Assertion (2) then implies that $f(w; \lambda)$ must have  
infinitely many zeros off $\R_+$.  
Therefore $f(w; \lambda)$ admits a zero $w_0 \in \C \setminus \R_+$ 
that is not a pole of $f_1(w; \lambda)$. 
Substituting $w = w_0$ into equation \eqref{eqn:f-f1} yields a 
contradiction $0 = f(w_0; \lambda) f_1(w_0; \lambda) = 
(1-x)^{(r-p-q)w_0-a-b} \neq 0$, which shows that 
$f(w; \lambda)$ and $\tilde{f}(w; \lambda)$ are linearly independent 
over the rational function field $\C(w)$. \hfill $\Box$ 
\begin{proposition} \label{prop:cfcr-F}
Let $\lambda = (p,q,r;a,b;x) \in \cF^-$ in what follows.   
\begin{enumerate}
\item Any solution to Problem $\rI$ or $\rII$ in $\cF^-$ is non-elementary.  
\item Any integral solution in $\cF^-$ to Problem $\rII$ comes from 
contiguous relations. 
\item For any integral solution $\lambda \in \cF^-$ to Problem $\rII$, 
its reciprocal $\check{\lambda}$ is an $(\rA)$-solution in $\cD^-$ 
so that $r \equiv 0 \mod 2$ and $\lambda = \lambda^{\vee\vee}$ 
is the reciprocal of an $(\rA)$-solution $\check{\lambda} \in \cD^-$.     
\item Any rational solution in $\cF^-$ to Problem $\rII$ 
essentially comes from contiguous relations. 
\end{enumerate} 
\end{proposition}
{\it Proof}. 
Assertion (1) follows directly from assertion (1) of Lemma \ref{lem:zeros}. 
With assertion (3) of Lemma \ref{lem:zeros} the proof of assertion (2) 
is exactly the same as that of \cite[Proposition 11.2]{Iwasaki}.  
The proof of assertion (3) proceeds as follows.  
Since the $\lambda$ in assertion (3) comes from contiguous relations by 
assertion (2), Lemma \ref{lem:r-cf1} can be used to infer that its 
reciprocal $\check{\lambda}$ is a solution to Problem $\rII$ and hence 
becomes an $(\rA)$-solution in $\cD^-$ by Theorems \ref{thm:back} 
and \ref{thm:D}. 
In other words, $\lambda = \lambda^{\vee\vee}$ is the reciprocal of an 
$(\rA)$-solution $\check{\lambda} \in \cD^-$. 
Applying assertion (2) of Theorem \ref{thm:D} to $\check{\lambda}$ 
we have $r = \check{r} - \check{p} - \check{q} \equiv 0 \mod 2$. 
Assertion (4) is an immediate consequence of assertion (2) applied to 
a multiplication $k \lambda$, where $k \in \N$ is chosen so that 
$k \lambda$ becomes integral. \hfill $\Box$
\subsection{Gamma Product Formulas} \label{subsec:gpf-F} 
Now that Proposition \ref{prop:cfcr-F} is established, the reciprocity 
can be used to tackle Problem \ref{prob:back} for {\sl rational} data 
in $\cF^-$ through their suitable multiplications.   
\begin{lemma} \label{lem:gpf-F} 
If $\lambda = (p,q,r;a,b;x) \in \cF^-$ is a rational solution to 
Problem $\rII$ with 
\begin{equation} \label{eqn:ocf-F}
\dfrac{f(w+1; \lambda)}{f(w; \lambda)} = d 
\dfrac{(w+u_1)\cdots(w+u_m)}{(w+v_1)\cdots(w+v_n)},   
\end{equation}
for some $d \in \C^{\times}$, $u_1, \dots, u_m \in \C$ and 
$v_1, \dots, v_m \in \C$, then $m = n$, $d$ is a positive number given 
by formula \eqref{eqn:d-F}, and $\lambda$ becomes a solution to Problem 
$\rI$ with gamma product formula     
\begin{equation} \label{eqn:gpf-F2}
f(w; \lambda) = C \cdot d^w \cdot 
\dfrac{\varGamma(w+u_1) \cdots \varGamma(w+u_m)}{\varGamma(w+v_1) 
\cdots \varGamma(w+v_m) },
\end{equation}
where $C$ is a positive constant. 
\end{lemma}
{\it Proof}.  
Let $k \in \N$ be the least common denominator of $p$, $q$, $r \in \Q$. 
Then the multiplication $k \lambda = (k p, k q, kr; a, b; x) \in \cF^-$ 
is an integral solution to Problem $\rII$.   
It follows from assertion (3) of Proposition \ref{prop:cfcr-F} that 
$(k \lambda)^{\vee}$ is an $(\rA)$-solution in $\cD^-$ and $k \lambda 
= (k \lambda)^{\vee\vee}$ is its reciprocal.   
Accordingly we can apply Proposition \ref{prop:division} to the 
$(\rA)$-solution $(k \lambda)^{\vee} = (k|p|, k|q|, k(r-p-q); \, \check{a}, 
\check{b}; \, \check{x}) \in \cD^-$. 
Adapting GPF \eqref{eqn:gpf-c2} to the current situation, we have 
\begin{equation} \label{eqn:gpf-F3}
f(w; k \lambda) = f\left(w; (k \lambda)^{\vee\vee} \right) 
= \check{C} \cdot \delta^{k w} \cdot \dfrac{\prod_{i=0}^{k r-1} 
\varGamma\left(w + \frac{i}{k r} \right)}{\prod_{i=1}^{k r} 
\varGamma(w+ \xi_i)}, 
\end{equation}
where $r-p-q$ in the general formula \eqref{eqn:gpf-c2} is now $k r$, 
the constant $\check{d}$ there is now $\delta^k$ with 
\begin{equation} \label{eqn:d-F2}
\delta := r^r \sqrt{\dfrac{|p|^{|p|} \, |q|^{|q|} \, 
(1-x)^{r-p-q}}{(r-p)^{r-p} \, (r-q)^{r-q} \, x^r}},   
\end{equation}
which is different from $\delta$ in \eqref{eqn:D-d}, while the numbers 
$\check{v}_i$ there are now written $\xi_i$.  
Under these circumstances the conditions \eqref{eqn:sum2} and 
\eqref{eqn:estimate3} with $\lambda$ replaced by $(k \lambda)^{\vee}$ 
are represented as  
\begin{gather} 
\xi_1 + \cdots + \xi_{k r} = \frac{k r -1}{2}, \label{eqn:sum3} \\
\xi_1, \dots, \xi_{k r} \in \Q; \qquad 
\frac{c}{k} \le \xi_1, \dots, \xi_{k r} < \frac{c}{k} + 1, \qquad 
c = c(\lambda) := \dfrac{1-a-b}{r-p-q},  \label{eqn:estimate4} 
\end{gather}
where we used $c\left((k \lambda)^{\vee}\right) = - c(k \lambda) = 
- c(\lambda)/k = -c/k$ to derive condition \eqref{eqn:estimate4}.    
\par
On the one hand formula \eqref{eqn:gpf-F3} leads to a closed-form 
expression 
\begin{equation} \label{eqn:ocf-F2}
\frac{f(w+1; k \lambda)}{f(w; k \lambda)} 
= \delta^k \cdot \dfrac{\prod_{i=0}^{k r-1} 
\left( w + \frac{i}{k r} \right)}{\prod_{i=1}^{k r} (w+ \xi_i) },  
\end{equation}
whereas on the other hand applying formula \eqref{eqn:R-mult} to 
assumption \eqref{eqn:ocf-F} yields 
\begin{equation} \label{eqn:ocf-F3}
\frac{f(w+1; k \lambda)}{f(w; k \lambda)} 
= d^k \cdot k^{k(m-n)} \cdot 
\prod_{i=0}^{k-1} 
\dfrac{\left(w+ \frac{u_1 + i}{k}\right) \cdots 
\left(w+ \frac{u_m +i}{k}\right)}{\left(w+ \frac{v_1 + i}{k}\right) 
\cdots \left(w+ \frac{v_n +i}{k}\right)}. 
\end{equation}
Since \eqref{eqn:ocf-F2} and \eqref{eqn:ocf-F3} must be exactly the same,   
they must be so asymptotically, that is, 
$\delta^k \sim d^k \cdot k^{k(m-n)} \cdot w^{k(m-n)}$ as $w \to \infty$,  
which forces $m = n$ and $\delta^k = d^k$. 
Formula \eqref{eqn:ocf-F} together with \eqref{eqn:f>0} and $m = n$ 
yields $0 < f(w+1; \lambda)/f(w; \lambda) \to d$ as $\R \ni w \to +\infty$ 
and so $d \ge 0$.   
Since $\delta$ is positive by definition \eqref{eqn:d-F2}, 
equation $\delta^k = d^k$ gives $d = \delta$.  
Thus in view of definition \eqref{eqn:d-F2}, $d$ must be given 
by formula \eqref{eqn:d-F}. 
The coincidence of \eqref{eqn:ocf-F2} and \eqref{eqn:ocf-F3} yields  
\[
\dfrac{\prod_{i=0}^{k r-1} 
\left( w + \frac{i}{k r} \right)}{\prod_{i=1}^{k r} (w+ \xi_i) }
= 
\prod_{i=0}^{k-1} 
\dfrac{\left(w+ \frac{u_1 + i}{k}\right) \cdots 
\left(w+ \frac{u_m +i}{k}\right)}{\left(w+ \frac{v_1 + i}{k}\right) 
\cdots \left(w+ \frac{v_m +i}{k}\right)},  
\]
so that gamma product formula \eqref{eqn:gpf-F3} can be recast to  
\begin{equation} \label{eqn:gpf-F4}
f(w; k \lambda) = \check{C} \cdot d^{k w} \cdot \prod_{i=0}^{k-1} 
\dfrac{\varGamma\left(w+ \frac{u_1 + i}{k}\right) \cdots 
\varGamma\left(w+ \frac{u_m +i}{k}\right)}{\varGamma 
\left(w+ \frac{v_1 + i}{k}\right) 
\cdots \varGamma \left(w+ \frac{v_m +i}{k}\right)}, 
\end{equation}
where $m = n$ and $d = \delta$ are also incorporated. 
Replacing $w$ by $w/k$ in \eqref{eqn:gpf-F4} and using the multiplication 
formula \eqref{eqn:mult} for the gamma function, we obtain formula 
\eqref{eqn:gpf-F2} with $C = \check{C} \cdot k^{v-u}$, where 
$u := u_1+\cdots+u_m$ and $v := v_1+\cdots+v_m$ are real numbers 
as $\lambda$ is a real data.   
Since $\check{C}$ is positive, so is the constant $C$.  \hfill $\Box$
\begin{proposition} \label{prop:gpf-F} 
If $\lambda = (p,q,r;a,b;x) \in \cF^-$ is a rational solution to Problem 
$\rII$, then $r \in \N$, $a, b \in \Q$, $x$ algebraic, and $\lambda$ is 
a solution to Problem $\rI$ with GPF 
\eqref{eqn:gpf-F} where $C$ is a positive constant, $d$ is given by 
formula \eqref{eqn:d-F} and $v_1, \dots, v_r$ satisfy condition 
\eqref{eqn:vi-F}.  
\end{proposition}
{\it Proof}. 
By assertion (1) of Lemma \ref{lem:zeros} the function $f(w; \lambda)$ 
has infinitely many poles so we must have $m \ge 1$ in formula 
\eqref{eqn:gpf-F2}. 
There exists an integer $s$ with $0 \le s \le m$ such that 
\begin{center}
(i) \,\, $u_i - v_j \not\in \Z$ \, for any \, $i, j = 1, \dots, s$, 
\qquad   
(ii) \,\, $u_i - v_i \in \Z$ \, for any \, $i = s+1, \dots, m$, 
\end{center}
after suitable rearrangements of $\{u_1, \dots, u_m\}$ and 
$\{v_1, \dots, v_m \}$, where condition (i) resp. (ii) should be 
ignored if $s = 0$ resp. $s = m$. 
In view of property (ii) a repeated use of the recursion formula 
$\varGamma(w+1) = w \varGamma(w)$ allows us to rewrite formula 
\eqref{eqn:gpf-F2} as 
\begin{equation} \label{eqn:gpf-F5}
f(w; \lambda) = S(w) \cdot d^w \cdot 
\dfrac{\varGamma(w+u_1) \cdots \varGamma(w+u_s)}{\varGamma(w+v_1) 
\cdots \varGamma(w+v_s) },
\end{equation}
where $S(w)$ is a nontrivial rational function of $w$. 
We must have $s \ge 1$, for otherwise $f(w; \lambda) = S(w) \cdot 
d^w$ could not have infinitely many poles, contradicting assertion (1) 
of Lemma \ref{lem:zeros}. 
\par
Take a nonnegative integer $i$ sufficiently large so that neither 
$\omega_0 := -u_1-i$ nor $\omega_1 := -u_1-i-1$ are zeros of $S(w)$. 
Then by property (i), $w = \omega_0$ and $w = \omega_1$ are actually 
poles of $f(w; \lambda)$, so by assertion (1) of Lemma \ref{lem:zeros} 
there exist nonnegative integers $j_0$ and $j_1$ with $j_0 < j_1$ 
such that $\omega_0 = -j_0/r$ and $\omega_1 = -j_1/r$. 
Thus we have $1 = \omega_0 - \omega_1 = (j_1 - j_0)/r$, namely,  
$r = j_1 - j_0 \in \N$ and hence the number $k$ in the proof of 
Lemma \ref{lem:gpf-F} is the least common denominator of $p$, $q \in \Q$.   
The assertion that $a$, $b \in \Q$ and $x$ is rational follows 
from the first part of Remark \ref{rem:division} applied to 
$(k \lambda)^{\vee}$ in place of $\lambda$. 
\par
Again by assertion (1) of Lemma \ref{lem:zeros}, for a sufficiently 
large $j_2$, the set  
\[ 
\left\{-\frac{j}{r} \right\}_{j \ge j_2 \, r} = 
\coprod_{i=1}^r \left\{ -j- \frac{i-1}{r} \right\}_{j \ge j_2}  
\]
constitutes all but a finite number of poles of $f(w; \lambda)$. 
The same is true with the multi-set
\[ 
\bigcup_{i=1}^s \left\{ -j- u_i \right\}_{j \ge j_3} \qquad 
(\mbox{union as multi-sets}),    
\]
due to formula \eqref{eqn:gpf-F5} and property (i). 
Since all poles are simple, we have $u_i - u_j \not\in \Z$ for every 
distinct $i,j = 1, \dots, s$, so the union of multi-sets above is 
just a disjoint union of ordinary sets. 
As the two sets above can differ only by a finite number of elements, 
we must have 
\[
r = s, \qquad u_i - (i-1)/r \in \Z \qquad (i = 1, \dots,r), 
\]
after taking a suitable rearrangement of $u_1, \dots, u_r$. 
Thus property (i) is equivalent to 
\begin{equation} \label{eqn:vi-F2}
v_1, \dots, v_r \not\in \frac{1}{r} \, \Z,  
\end{equation}
and a further repeated use of the recursion formula 
$\varGamma(w+1) = w \varGamma(w)$ converts \eqref{eqn:gpf-F5} to 
\begin{equation} \label{eqn:gpf-F6}
f(w; \lambda) = S(w) \cdot d^w \cdot \dfrac{ \prod_{i=0}^{r-1} 
\varGamma\left(w+\frac{i}{r} \right)}{ \prod_{i=1}^r 
\varGamma(w+v_i)},
\end{equation}
with a possibly different rational function $S(w)$ of $w$, where we may 
assume   
\begin{equation} \label{eqn:vi-F3}
c \le \mathrm{Re} \, v_1, \dots, \mathrm{Re}\, v_r < c + 1, 
\qquad c := \frac{1-a-b}{r-p-q},  
\end{equation}
after translating $v_1, \dots, v_r$ by suitable integers and making yet 
another use of the recursion formula $\varGamma(w+1) = w \varGamma(w)$ 
with an ensuing modification of the rational function $S(w)$. 
\par
Replacing $w$ by $k w$ in formula \eqref{eqn:gpf-F6} and using 
formulas \eqref{eqn:f-mult} and \eqref{eqn:mult}, we have 
\begin{equation} \label{eqn:gpf-F7}
f(w; k \lambda) = k^{(r-1)/2-v} \cdot S(k w) \cdot d^{k w} \cdot 
\dfrac{ \prod_{i=0}^{k r-1} \varGamma\left(w+\frac{i}{k r} \right)}{ 
\prod_{i=1}^{k r} \varGamma(w+\eta_i)}, 
\end{equation}
where $v := v_1 + \cdots + v_r$ and $\eta_1, \dots, \eta_{k r}$ are the 
numbers defined by the multi-set 
\begin{equation} \label{eqn:eta} 
\{\eta_1, \dots, \eta_{k r}\} := 
\{v_{i,j} \,:\, i = 1, \dots, r, \, j = 0, \dots, k-1\}, \qquad  
v_{i,j} := \frac{v_i + j}{k}. 
\end{equation} 
In view of this definition the estimate \eqref{eqn:vi-F3} leads to 
\begin{equation} \label{eqn:vi-F4}
\frac{c}{k} \le \mathrm{Re} \, \eta_1, \dots, 
\mathrm{Re} \, \eta_{k r} < \frac{c}{k} + 1.       
\end{equation}
Comparing formulas \eqref{eqn:gpf-F3} and \eqref{eqn:gpf-F7} with 
$\delta = d$ taken into account, we find 
that  
\[ 
\dfrac{ \prod_{i=1}^{k r} \varGamma(w+ \eta_i)}{\prod_{i=1}^{k r} 
\varGamma(w+\xi_i)} = \left(\check{C} \right)^{-1} k^{(r-1)/2-v} 
\cdot S(k w),    
\]
which must be a rational function of $w$, having only at most finitely 
many poles and zeros. 
This forces $\eta_i - \xi_i\in \Z$ for every $i = 1, \dots, k r$, 
after a suitable rearrangement of $\xi_1, \dots, \xi_{k r}$. 
But in view of conditions \eqref{eqn:estimate4} and \eqref{eqn:vi-F4},  
this coincidence modulo $\Z$ must be an exact coincidence 
\begin{equation} \label{eqn:eta-xi} 
\eta_i = \xi_i \qquad (i = 1, \dots, k r).  
\end{equation} 
So $\left(\check{C} \right)^{-1} k^{(r-1)/2-v} \cdot S(k w) = 1$  
and thus $S(w) = \check{C} k^{v-(r-1)/2} =: C$ must be a positive 
constant and formula \eqref{eqn:gpf-F6} reduces to GPF 
\eqref{eqn:gpf-F}. 
We can arrange $\eta_1, \dots, \eta_{k r}$ so that 
$\eta_i = v_{i,0} = v_i/k$ for each $i = 1, \dots, r$. 
Then coincidence \eqref{eqn:eta-xi} and the rationality in 
\eqref{eqn:estimate4} imply $v_i = k \eta_i = k \xi_i \in \Q$ for 
every $i = 1, \dots, r$. 
This together with \eqref{eqn:vi-F2} and \eqref{eqn:vi-F3} gives 
$v_1, \dots, v_r \in \left(\Q \setminus \frac{1}{r} \, \Z\right)  
\cap \left[c, \, c+1\right)$ in conditions \eqref{eqn:vi-F}. 
To prove the remaining condition in \eqref{eqn:vi-F} we observe that 
\begin{equation} \label{eqn:xi-eta}
\sum_{i=1}^{k r} \xi_i = \sum_{i=1}^{k r} \eta_i = 
\sum_{i=1}^r \sum_{j=0}^{k-1} v_{i,j} \\
= \sum_{i=1}^r \sum_{j=0}^{k-1} \frac{v_i + j}{k} = 
\sum_{i=1}^r v_i + \frac{(k-1) r}{2}, 
\end{equation}
where we used \eqref{eqn:eta-xi} in the first equality and 
\eqref{eqn:eta} in the second and third equalities. 
Comparing \eqref{eqn:xi-eta} with \eqref{eqn:sum3} 
we have $v_1 + \cdots + v_r = (r-1)/2$ and hence proves conditions 
\eqref{eqn:vi-F}. \hfill $\Box$ \par\medskip 
It is evident that Theorem \ref{thm:F} follows from Propositions 
\ref{prop:cfcr-F} and \ref{prop:gpf-F}. 
The proof of Theorem \ref{thm:sol-B} is virtually contained in the proof  
of Proposition \ref{prop:gpf-F} as a special case $k = 2$; the only 
necessary modification is to deduce the condition 
$0 \le v_1, \dots, v_r < 1$ for a $(\rB)$-solution from a similar 
condition for an $(\rA)$-solution.  
\section{From an Algorithmic Point of View} \label{sec:algorithm}
We put all the results on $\cD^- \cup \cF^-$ (in \S \ref{sec:results})  
into context from an algorithmic point of view and discuss how to 
enumerate all (rational) solutions $\lambda = (p,q,r;a,b,x)$ with a 
prescribed value of $\bp = (p,q;r)$, or perhaps how to prove the 
nonexistence of such solutions, both in finite steps.      
\subsection{Integral Solutions} \label{subsec:integral} 
If $\lambda = (p,q,r;a,b;x)$ is an $(\rA)$-solution in $\cD^-$ then 
$\bp = (p,q;r) \in D^-_{\rA}$ by assertion (2) of Theorem \ref{thm:D}, 
where $D^-_{\rA}$ is defined by formula \eqref{eqn:D-m}. 
Putting Theorems \ref{thm:back}, \ref{thm:sol-A}, \ref{thm:pqr} and 
\ref{thm:ab} together along with Theorem \ref{thm:truncate} below, 
we are able to develop an algorithm to find all $(\rA)$-solutions 
$\lambda \in \cD^-$ with a prescribed value of $\bp \in D^-_{\rA}$. 
To describe it we introduce some notation.  
Let $\langle \varphi(z) \rangle_k := \sum_{j=0}^k c_j \, z^j$ 
be the {\sl truncation} at degree $k$ of a power series $\varphi(z) = 
\sum_{j=0}^{\infty} c_j \, z^j$. 
For a data $\lambda \in \cD^-$ with $\bp \in D^-_{\rA}$ we consider 
two ``truncated hypergeometric products":       
\begin{align*} 
V(w;\lambda) 
&:= (r w)_{r-1} \, \langle \, \hgF(\bal^*(w); z) \cdot 
\hgF(\bv-\bal^*(w+1);z) \, \rangle_{k} \big|_{z = x},  
\\[2mm]
P(w;\lambda) 
&:= (r w)_r \, \langle \, \hgF(\bal^*(w); z) \cdot 
\hgF(\1-\bal^*(w+1);z) \, \rangle_{k}\big|_{z = x},          
\end{align*}
where $(s)_n := \varGamma(s+n)/\varGamma(s)$, $\bal^*(w) := 
((r-p)w-a, \, (r-q)w-b; \, r w)$, $\bv := (1,1;2)$, 
$\1 := (1,1;1)$ and $k := \max\{r-p-1, \, r-q-1\}$. 
Note that $V(w; \lambda)$ is written $\Phi(w; \lambda)$ in 
the previous paper \cite{Iwasaki} but the latter symbol is 
already used in a different context in \S \ref{sec:kummer} 
of the present article and hence altered.    
It is shown in \cite[Lemma 11.10]{Iwasaki} that $V(w; \lambda)$ is a 
polynomial of degree at most $r-1$ in $w$ and so admits an expansion 
\[
V(w; \lambda) = \sum_{\nu=0}^{r-1} V_{\nu}(\lambda) \, w^{\nu} 
= \sum_{\nu=0}^{r-1} V_{\nu}(a,b;x) \, w^{\nu} ,  
\]
where $V_{\nu}(\lambda)$ is written $V_{\nu}(a,b; x)$ when 
$\bp$ is understood to be a priori given.     
It is also known that $P(w; \lambda)$ is a polynomial of degree 
at most $r$. 
Our algorithm is based on the following.   
\begin{theorem} \label{thm:truncate} 
$\lambda \in \cD^-$ is a solution of type $(\rA)$ if and only if 
$\bp \in D^-_{\rA}$ and $V(w; \lambda)$ vanishes identically as a 
polynomial of $w$, that is, $(a,b;x)$ is a simultaneous root of 
algebraic equations    
\begin{equation} \label{eqn:cfcr2}
V_{\nu}(a, b; x) = 0 \qquad (\nu = 0, \dots, r-1),  
\end{equation} 
in which case $P(w; \lambda)$ is exactly of degree $r$ in $w$  
and $R(w; \lambda)$ in formula \eqref{eqn:ocf} is given by 
\begin{equation} \label{eqn:R-l}
R(w;\lambda) = (1-x)^{r-p-q-1} \cdot \dfrac{(r w)_r}{P(w;\lambda)}. 
\end{equation}
\end{theorem} 
\par
This theorem follows from \cite[Theorems 2.9 and 2.10]{Iwasaki}. 
Now we have the following.   
\begin{algorithm} \label{algorithm}
To enumerate all $(\rA)$-solutions $\lambda \in \cD^-$ with any  
prescribed $\bp \in D^-_{\rA}$:     
\begin{enumerate}
\item Check if $\bp = (p,q;r)$ satisfies division relation \eqref{eqn:dr} 
in Theorem \ref{thm:pqr}.  
\item If it is alright, then following Theorem \ref{thm:ab} find a 
candidate for $\ba =(a,b)$ explicitly in terms of $\bp$, which must be 
in one of Cases 1--6 of Table \ref{tab:ab}.      
\item Substitute the ensuing candidate $\ba$ into \eqref{eqn:cfcr2} 
to derive algebraic equations for $x$ over $\Q$:       
\begin{equation} \label{eqn:V}
V_{\nu}(x) = 0 \qquad (\nu = 0, \dots, r-1),   
\end{equation}
where $V_{r-1}(x) = 0$ is equivalent to equation \eqref{eqn:Y} in 
Theorem \ref{thm:sol-A} (see \cite[Remark 2.11]{Iwasaki}). 
\item Check if equations \eqref{eqn:V} admits a simultaneous root $x$ 
with $0 < x < 1$.  
If so, we certainly get an $(\rA)$-solution $\lambda = (\bp; \ba; x) \in 
\cD^-$; otherwise the current candidate gives no solution.    
\item If we actually have a solution $\lambda$ in step (4), put it into 
formula \eqref{eqn:R-l} to find $R(w;\lambda)$ explicitly. 
If $R(w;\lambda)$ is of the form \eqref{eqn:R} (inevitably with $m = n$),  
then we have a GPF \eqref{eqn:gpf} by Theorem \ref{thm:back}, where the 
constant $C$ can be determined explicitly by evaluating $f(w; \lambda)$ 
at such a point $w = w_0$ that the hypergeometric series $f(w_0; \lambda)$ 
is terminating or more specifically it is vanishing except for its 
leading term.      
\item Repeat the procedures (2)--(5) until all candidates are exhausted. 
By Theorem \ref{thm:ab-rf} there are only a finite number of candidates   
so that the algorithm terminates in finite steps.   
\end{enumerate}
\end{algorithm}
\par
Empirically, it almost surely occurs that algebraic equations 
\eqref{eqn:V} have no roots in common for all candidates, 
in which case Algorithm \ref{algorithm} serves as a rigorous proof of 
the nonexistence of any solution in $\cD^-$ with a given value of 
$\bp \in D^-_{\rA}$.     
Observe also that for any $\bp = (p,q;r) \in D_{\rA}^-$ with  
division relation \eqref{eqn:dr} there is an estimate 
$1 \le p, \, q \le r/2$.  
Thus by the use of Algorithm \ref{algorithm} the enumeration of all 
$(\rA)$-solutions in $\cD^-$ with any  prescribed bound for $r$ 
terminates in finite steps after producing at most finitely many solutions.    
In view of estimate \eqref{eqn:dr-bound} in Lemma \ref{lem:dr-bound},  
this is true even with any prescribed bound for $\check{r} := r-p-q$. 
For example, there are exactly seven $(\rA)$-solutions in $\cD^-$ with 
$\check{r} = 2$, all of which are contained in \cite[Table 2]{Iwasaki}.   
\par
We proceed to the treatment of integral solutions in $\cF^-$, which 
is based on the following.  
\begin{theorem} \label{thm:D-F} 
Reciprocity \eqref{eqn:recip} induces a bijection between the set of all 
$(\rA)$-solutions in $\cD^-$ and the set of all integral solutions in 
$\cF^-$. 
\end{theorem}
\par
This theorem is an immediate consequence of Theorems \ref{thm:D-r} and 
\ref{thm:F}. 
It implies that all integral solutions $\lambda = (p,q,r;a,b;x) \in \cF^-$ 
with a given $\bp = (p,q;r)$ is in one-to-one correspondence with all 
$(\rA)$-solutions $\check{\lambda} = 
(\check{p}, \check{q}, \check{r}; \check{a}, \check{b}; \check{x}) \in \cD^-$ 
with a given $\check{\bp} = (\check{p}, \check{q}; \check{r}) = 
(-p,-q;r-p-q) = (|p|,|q|;r+|p|+|q|)$, so that the enumeration of former 
solutions is accomplished through the enumeration of latter solutions 
by the use of Algorithm \ref{algorithm}.  
\begin{corollary} \label{cor:D-F} 
There are at most finitely many integral solutions $\lambda = 
(p,q,r;a,b;x) \in \cF^-$ with any prescribed bound for $r$. 
Moreover any integral solution $\lambda \in \cF^-$ must satisfy  
\[
-3 r \le p, \, q \le -1, \qquad -5 r \le p + q \le -2.  
\]
\end{corollary} 
\par
These inequalities follow from estimates \eqref{eqn:dr-bound} in 
Lemma \ref{lem:dr-bound} by replacing $\lambda$ with $\check{\lambda}$.      
\subsection{Rational Solutions} \label{subsec:rational}
We turn our attention to dealing with rational solutions in 
$\cD^- \cup \cF^-$ that are non-integral. 
In $\cD^-$ this amounts to considering solutions of type 
$(\rB)$, since all solutions in $\cD^-$ are rational and the 
dichotomy of types $(\rA)$ and $(\rB)$ is exactly that of 
`integral' and `non-integral'.     
\par
A solution $\lambda = (p, q, r; a, b; x)$ to Problem $\rI$ or $\rII$ 
is said to be {\sl divisible} by an integer $k \ge 2$, if the data 
$\lambda/k := (p/k, q/k, r/k; a, b; x)$ is also a solution to the 
same problem, in which case $\lambda/k$ is referred to as the 
{\sl division} of $\lambda$ by $k$, in particular, $\lambda/2$ 
is a {\sl half} of $\lambda$. 
\begin{lemma} \label{lem:divisible}
For an integer $k \ge 2$, an integral solution $\lambda = (p,q,r;a,b;x) 
\in \cD^- \cup \cF^-$ is divisible by $k$ if and only if $k|r$ and there 
exist $d > 0$ and $v_1, \dots, v_s \in \Q$ with $s := r/k$ such that  
\begin{equation} \label{eqn:divisible}
R(w; \lambda) = d \cdot 
\dfrac{\prod_{i=0}^{r-1} \left(w+ \frac{i}{r} \right)}{\prod_{i=1}^{s} 
\prod_{j=0}^{k-1} \left(w+ \frac{v_i + j}{k} \right)}.  
\end{equation}
\end{lemma}
{\it Proof}. 
Essential is the ``if" part. 
Since Problems $\rI$ and $\rII$ are equivalent for integral (or more 
generally rational) data in $\cD^- \cup \cF^-$ by Theorem \ref{thm:back} 
and assertion (3) of Theorem \ref{thm:F}, the closed-form condition 
\eqref{eqn:divisible} lifts to a ``multiplied" gamma product formula  
\[ 
f(w; \lambda) = C \cdot d^w \cdot 
\dfrac{\prod_{i=0}^{r-1} 
\varGamma \left(w+ \frac{i}{r} \right)}{\prod_{i=1}^{s} 
\prod_{j=0}^{k-1} \varGamma \left(w+ \frac{v_i + j}{k} \right)} 
= C \cdot d^w \cdot \prod_{j=0}^{k-1} 
\dfrac{\prod_{i=0}^{s-1}  
\varGamma \left(w+ \frac{(i/s) + j}{k} \right)}{\prod_{i=1}^{s} 
\varGamma \left(w+ \frac{v_i + j}{k} \right)},  
\]
which in turn descends through the multiplication formula 
\eqref{eqn:mult} to a GPF  
\[ 
f(w; \lambda/k) = f(w/k; \lambda) = C \cdot d^{w/k} \cdot 
\dfrac{ \prod_{i=0}^{s-1} 
\varGamma \left(w+ \frac{i}{s} \right)}{\prod_{i=1}^{s} 
\varGamma \left(w+ v_i \right)},   
\]
with a possibly different $C$. 
Thus $\lambda/k$ is a solution and hence $\lambda$ is divisible by $k$. 
The proof of ``only if" part is a simple reversing of the 
argument so far, with the positivity $d > 0$ coming from 
formulas \eqref{eqn:d} and \eqref{eqn:d-F}, while the 
rationality $v_1, \dots, v_s \in \Q$ from properties 
\eqref{eqn:estimate} and \eqref{eqn:vi-F}. 
\hfill $\Box$ \par\medskip 
By assertions (2) and (4) of Theorems \ref{thm:D} and 
assertion (3) of Theorem \ref{thm:F},  
\begin{enumerate}
\item any $(\rB)$-solution in $\cD^-$ is a {\sl half} of an 
$(\rA)$-solution $\lambda_* = (p_*, q_*, r_*; a_*, b_*; x_*) 
\in \cD^-$ such that $p_*$ and $q_*$ are odd positive integers 
and $r_*$ is an even positive integer,   
\item any rational solution in $\cF^-$ is the division 
$\lambda_*/k$ of an integral solution $\lambda_* \in \cF^-$ 
by a {\sl divisor} $k$ of $r_*$, where $r_*$ is necessarily 
an even positive integer.     
\end{enumerate}
Thus finding a rational solution consists of finding an integral   
solution $\lambda_*$ by using the recipe in \S \ref{subsec:integral} 
and verifying if $\lambda_*$ is divisible by a divisor of $r_*$ 
based on Lemma \ref{lem:divisible}. 
All rational solutions are obtained in this manner. 
Note that any seed solution $\lambda_*$ produces only 
a finite number of division solutions. 
All $(\rB)$-solutions $\lambda = (p,q,r;a,b;x) \in \cD^-$ with any 
prescribed bound for $r$ are only of a finite cardinality, since  
only division by $2$ is involved in this case.  
It is not known whether this is also the case for rational solutions 
in $\cF^-$, because $r_*$ and $k$ may be arbitrary large while 
$r = r_*/k$ is kept bounded (see Problem \ref{prob:F}). 
\subsection{Examples of Solutions in the South-West Domain} \label{subsec:sol-F}
\begin{table}[t]
\[
\begin{array}{c|c|c|c|c|c|c|cccc|l}
\hline
  &    &    &             &                  &      &     &      &       &       &      &               \\[-4mm]
r & p  & q  & x           & d                & a    & b   & v_1  & v_2   & v_3   & v_4  & \mbox{remark} \\[1mm]
\cline{1-12}
  &    &    &             &                 &              &             &              &               &  &  &                            \\[-3mm]  
2 & -1 & -1 & \frac{1}{9} & \frac{2^8}{3^5} & \frac{11}{8} & \frac{9}{8} & \frac{5}{24} & \frac{7}{24}  &  &  & \mbox{GS \cite[(6.6)]{GS}} \\[2mm]
\cline{6-12}
  &    &    &             &                 &              &             &              &               &  &  &                            \\[-3mm]   
  &    &    &             &                 & \frac{5}{8}  & \frac{7}{8} & \frac{1}{24} & \frac{11}{24} &  &  & \mbox{GS \cite[(6.5)]{GS}} \\[2mm]
\cline{6-12}
  &    &    &             &                 &              &             &              &               &  &  &                  \\[-3mm]   
  &    &    &             &                 & \frac{5}{4}  & \frac{3}{4} & \frac{1}{12} & \frac{5}{12}  &  &  & \mbox{self-dual} \\[2mm]  
\cline{1-12}
  &    &    &             &                    &             &             &              &              &               &               &                         \\[-3mm]   
4 & -1 & -1 & \frac{1}{5} & \frac{2^{14}}{5^6} & \frac{9}{8} & \frac{5}{8} & \frac{3}{40} & \frac{7}{40} & \frac{23}{40} & \frac{27}{40} & \mbox{divisible by $2$} \\[2mm]
\cline{6-12}
  &    &    &             &                    &             &             &              &              &               &               &                         \\[-3mm]   
  &    &    &             &                    & \frac{3}{8} & \frac{7}{8} & \frac{1}{40} & \frac{9}{40} & \frac{21}{40} & \frac{29}{40} & \mbox{divisible by $2$} \\[2mm]
\cline{1-12}
  &    &    &                           &                          &             &             &              &              &     & & \\[-3mm]   
2 & -2 & -2 & \frac{1}{4}(3 \sqrt{3}-5) & \frac{3^4}{2^7} \sqrt{3} & \frac{5}{3} & \frac{4}{3} & \frac{1}{12} & \frac{5}{12} &     & & \mbox{self-dual} \\[2mm]  
\cline{2-12}
  &    &    &                &                              &             &             &              &              &   &  &    \\[-3mm]   
  & -3 & -1 & 9 -4 \sqrt{5}  & \frac{2^8}{5^3}(5-2\sqrt{5}) & \frac{9}{4} & \frac{5}{4} & \frac{3}{20} & \frac{7}{20} &   &  &    \\[2mm]
\cline{6-12}
  &    &    &                &                              &             &             &              &              &   &  &    \\[-3mm]   
  &    &    &                &                              & \frac{7}{4} & \frac{3}{4} & \frac{1}{20} & \frac{9}{20} &   &  &    \\[2mm]
\cline{2-12}
  &    &    &                &                                   &             &             &              &              & & &                   \\[-3mm]   
  & -4 & -2 & 17-12 \sqrt{2} & \frac{2^{10}}{3^3}(17-12\sqrt{2}) & \frac{5}{2} & \frac{3}{2} & \frac{1}{12} & \frac{5}{12} & & &  \mbox{self-dual} \\[2mm]
\hline  
\end{array}
\]
\caption{Some integral solutions in $\cF^-$; reciprocals of the $(\rA)$-solutions in \cite[Table 2]{Iwasaki}.} 
\label{tab:sol-F}
\end{table} 
Some examples of $(\rA)$-solutions in $\cD^-$ were presented in the previous article 
\cite[Table 2]{Iwasaki}. 
Their reciprocal solutions in $\cF^-$ are given in the corresponding places of 
Table \ref{tab:sol-F}, which exhibits for each solution the data 
$\lambda = (p,q,r;a,b;x) \in \cF^-$ itself as well as the values of the numbers 
$d$ and $v_1, \dots, v_r$ in GPF \eqref{eqn:gpf-F}, with a brief remark in the  
last column, if any. 
Solutions in the top and second row of Table \ref{tab:sol-F} are Gosper's conjectural  
identities in Gessel and Stanton \cite[formulas (6.6) and (6.5)]{GS}, proved later 
by Koepf \cite[Table 5]{Koepf} using an extension of Wilf-Zeilberger method \cite{WZ}.    
Self-dual solutions are indicated so in the remark column, while any pair of 
consecutive solutions without this indication is a dual pair of solutions.  
Two solutions with $(p,q;r) = (-1,-1;4)$ are divisible by $2$, halves of 
which are tabulated in Table \ref{tab:sol-F-half}. 
These two solutions in $\cF^-$ are ``dual" to each other as well as 
``reciprocal" to the $(\rB)$-solutions in \cite[Table 3]{Iwasaki}, 
although we have to be careful in applying these terminologies to 
non-integral solutions (as mentioned in Remark \ref{rem:kummer}). 
\begin{table}[b]
\[
\begin{array}{c|c|c|c|c|c|c|cc}
\hline
  &              &              &             &                    &             &             &              &              \\[-4mm]
r & p            & q            & x           & d                  &  a          & b           & v_1          & v_2          \\[1mm]
\cline{1-9}
  &              &              &             &                    &             &             &              &              \\[-3mm]   
2 & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{5}  & \frac{2^{7}}{5^3}  & \frac{9}{8} & \frac{5}{8} & \frac{3}{20} & \frac{7}{20} \\[2mm]
\cline{6-9}
  &              &              &             &                    &             &             &              &              \\[-3mm]   
  &              &              &             &                    & \frac{3}{8} & \frac{7}{8} & \frac{1}{20} & \frac{9}{20} \\[2mm]
\hline  
\end{array}
\]
\caption{Two non-integral rational solutions in $\cF^-$.}  
\label{tab:sol-F-half}
\end{table} 
\section{Concluding Discussions} \label{sec:CD} 
The main orientation of the field has been toward searching for as 
many solutions as possible, but the converse orientation of confining 
solutions into as slim a region as possible or perhaps of proving 
the nonexistence of solutions other than those already known is equally 
important for the ultimate goal of a complete enumeration. 
In the latter direction there are such interesting studies as 
\cite{Maier,Wimp,Zeilberger}. 
The principal aim of our studies is also in this direction.    
With this in mind we close this article by giving a couple of  
problems that should be discussed in the future.        
\par
A solution to Problem $\rI$ or $\rII$ is said to be {\sl primitive} 
if it is a multiplication of no other solution. 
Since any solution is a multiplication of a primitive solution, the 
enumeration of all solutions boils down to that of primitive ones. 
As a milestone toward the complete enumeration we pose the following 
problem which is still very ambitious.        
\begin{problem} \label{prob:primitive} 
Are there infinitely many {\sl primitive} solutions in domain $\cD^-$  
or only a finite number of them ?  
Ask the same question for domain $\cE^{*-}$ or region $\cI^{*-}$  
or anywhere else.       
\end{problem}
\par
We have no idea as to whether the cardinality of them is finite or infinite.  
As $r$ becomes larger, it is increasingly more difficult that the $r$ 
algebraic equations \eqref{eqn:V} have a root in common, but there 
is no logical reasoning that prohibits this miracle.       
Either answer is welcome as it contributes to the topic 
in one or the other orientation.  
Solving Problem \ref{prob:primitive} seems to require an amazingly 
new idea.     
\par
For solutions in $\cD^-$ such a variety of results are at our 
disposal as Theorems \ref{thm:D}, \ref{thm:sol-A}, 
\ref{thm:pqr}--\ref{thm:ab}, \ref{thm:ab-rf}, \ref{thm:D-r}, 
\ref{thm:sol-B}, while only Proposition \ref{prop:E} and 
Theorem \ref{thm:E} are available for solutions in $\cE^{*-}$  
besides Theorem \ref{thm:back} which is common with $\cD^-$ and $\cE^{*-}$.    
\begin{problem} \label{prob:E} 
Lift our understanding of solutions in $\cE^{*-}$ to the same level  
as that of solutions in $\cD^-$ by establishing appropriate   
$\cE^{*-}$-counterparts of the above-mentioned results on $\cD^-$.      
\end{problem} 
\par
For any solution $\lambda = (p,q,r;a,b;x) \in \cD^-$ the numbers $a$ and 
$b$ must be rational by Theorems \ref{thm:ab-rf} and \ref{thm:sol-B}.  
This remains true for any rational solution in $\cF^-$ 
by Theorem \ref{thm:F}.  
However, it is not always the case in the south domain $\cE^{*-}$. 
Indeed, for each pair $(i, j)$ of integers with $j > k > 0$,  
Iwasaki \cite[formula (123)]{Iwasaki} gives a one-parameter family 
of solutions $\lambda = (p,q,r;a,b;x) = (j-k, k-j, j+k; c, 1-c; 1/2) 
\in \cE^{*-}$ with gamma product formula  
\begin{equation} \label{eqn:sol-E}
\begin{split}
& \hgF(\, (j-k)w+c, \, -(j-k)w+1-c; \, (j+k)w; \, 1/2) \\[2mm] 
& \qquad = \dfrac{\sqrt{2} \, k^{c/2}}{j^{(c-1)/2} \, (j+k)^{1/2}} 
\left\{\dfrac{(j+k)^{j+k}}{2^{j+k} \, j^j \, k^k} \right\}^w 
\dfrac{\prod_{\nu=0}^{j+k-1} \varGamma\big(w+\frac{\nu}{j+k}\big)}{\prod_{\nu=0}^{j-1} 
\varGamma\big(w+\frac{c}{2j}+\frac{\nu}{j}\big) \prod_{\nu=0}^{k-1} 
\varGamma\big(w+\frac{1-c}{2k}+\frac{\nu}{k}\big)},       
\end{split}
\end{equation}
where $c$ is a free parameter.  
If $c$ is irrational, so are $a = c$ and $b = 1-c$.  
\begin{problem} \label{prob:E-rat}    
In spite of the existence of such solutions as \eqref{eqn:sol-E}, it might 
be expected that $a$ and $b$ must be rational for ``almost all" solutions 
in $\cE^{*-}$. 
Is this true and in what sense ?  
\end{problem}
\par
Thanks to reciprocity our knowledge of integral solutions in $\cF^-$ 
is at the same level as that of $(\rA)$-solutions in $\cD^-$. 
However, we know less about (non-integral) rational solutions and 
even nothing about irrational solutions in $\cF^-$. 
Thus we pose the following.   
\begin{problem} \label{prob:F} 
For rational solutions $\lambda = (p,q,r;a,b;x) \in \cF^-$ how large 
can the least common denominator of $p$, $q \in \Q$ become ?  
Moreover, as mentioned in item (3) of Remark \ref{rem:F} we can also ask: 
Is there any irrational solution in $\cF^-$?   
\end{problem} 
\par
These are among many interesting problems in searching for, or rather 
confinement of, or perhaps proving the nonexistence of ``strange" 
hypergeometric identities (see also \cite{AZ} for a related discussion).  
\par\vspace{5mm} \noindent
{\bf Acknowledgment}. The author thanks Akihito Ebisu 
for many stimulating discussions.       
