\section{Introduction}
Let $\mathbb{Z}_{2^s}$ be the ring of integers modulo $2^s$ with $s\geq1$. The set of
$n$-tuples over $\mathbb{Z}_{2^s}$ is denoted by $\mathbb{Z}_{2^s}^n$. In this paper,
the elements of $\mathbb{Z}^n_{2^s}$ will also be called vectors. 
A code over $\mathbb{Z}_2$ of length $n$ is a nonempty subset of $\mathbb{Z}_2^n$,
and it is linear if it is a subspace of $\mathbb{Z}_{2}^n$. Similarly, a nonempty
subset of $\mathbb{Z}_{2^s}^n$ is a $\mathbb{Z}_{2^s}$-additive code if it is a subgroup of $\mathbb{Z}_{2^s}^n$. A $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-additive code is a subgroup of $\mathbb{Z}_2^{\alpha_1}  \times \mathbb{Z}_4^{\alpha_2} \times \mathbb{Z}_8^{\alpha_3}$. Note that a $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-additive code is a linear code over $\mathbb{Z}_2$ when $\alpha_2=\alpha_3=0$,  a $\mathbb{Z}_4$-additive or $\mathbb{Z}_8$-additive code when $\alpha_1=\alpha_3=0$ or $\alpha_1=\alpha_2=0$, respectively, and a $\mathbb{Z}_2\mathbb{Z}_4$-additive code when $\alpha_3=0$.
The order of a vector $u\in \mathbb{Z}_{2^s}$, denoted by $o(u)$, is the smallest positive integer $m$ such that $m u =(0,\dots,0)$. Also, the order of a vector $\mathbf u\in \mathbb{Z}_2^{\alpha_1}\times\mathbb{Z}_4^{\alpha_2} \times\mathbb{Z}_8^{\alpha_3}$, denoted by $o(\mathbf u)$, is the smallest positive integer $m$ such that $m \mathbf u =(0,\dots,0\mid 0,\dots,0\mid 0,\dots,0)$.


The Hamming weight of a vector $u\in\mathbb{Z}_{2}^n$, denoted by ${\rm wt}_H(u)$, is
the number of nonzero coordinates of $u$. The Hamming distance of two
vectors $u,v\in\mathbb{Z}_{2}^n$, denoted by $d_H(u,v)$, is the number of
coordinates in which they differ.  Note that $d_H(u,v)={\rm wt}_H(u-v)$. The minimum distance of a code $C$ over $\mathbb{Z}_2$ is $d(C)=\min \{ d_H(u,v) : u,v \in C, u \not = v  \}$. 


In \cite{Sole}, a Gray map  from $\mathbb{Z}_4$ to $\mathbb{Z}_2^2$ is defined as
$\phi(0)=(0,0)$, $\phi(1)=(0,1)$, $\phi(2)=(1,1)$ and $\phi(3)=(1,0)$. There exist different generalizations of this Gray map, which go from $\mathbb{Z}_{2^s}$ to
$\mathbb{Z}_2^{2^{s-1}}$ \cite{Carlet,Codes2k,dougherty,Nechaev,Krotov:2007}.
The one given in \cite{Nechaev} can be defined in terms of the elements of a Hadamard code \cite{Krotov:2007}, and Carlet's Gray map \cite{Carlet} is a particular case of the one given in \cite{Krotov:2007} 
satisfying $\sum \lambda_i \phi(2^i) =\phi(\sum \lambda_i 2^i)$ \cite{KernelZ2s}. 
In this paper, we focus on Carlet's Gray map \cite{Carlet}, from $\mathbb{Z}_{2^s}$ to $\mathbb{Z}_2^{2^{s-1}}$, which is also a particular case of the one given in \cite{ShiKrotov2019}. Specifically, \begin{gather}\label{eq:GrayMapCarlet}
\phi_s(u)=(u_{s-1},u_{s-1},\dots,u_{s-1})+(u_0,\dots,u_{s-2})Y_{s-1},
\end{gather}
where $u\in\mathbb{Z}_{2^s}$; $[u_0,u_1,\dots,u_{s-1}]_2$ is the binary expansion of $u$, that is, $u=\sum_{i=0}^{s-1}u_i2^i$ with $u_i\in \{0,1\}$; and $Y$ is the  matrix of size $(s-1)\times 2^{s-1}$ whose columns are all the vectors in $\mathbb{Z}_2^{s-1}$. Without loss of generality, we assume that the columns of $Y_{s-1}$ are ordered in ascending order, by considering the elements of $\mathbb{Z}_2^{s-1}$ as the binary expansions of the elements of $\mathbb{Z}_{2^{s-1}}$. Note that $\phi_1$ is the identity map, and $(u_{s-1},\dots,u_{s-1})$ and $(u_0,\dots,u_{s-2})Y_{s-1}$ are binary vectors of length $2^{s-1}$, and that the rows of $Y_{s-1}$ form a basis of a first order Reed-Muller code
after adding the all-one row. We define $\Phi_s:\mathbb{Z}_{2^s}^n \rightarrow \mathbb{Z}_2^{n2^{s-1}}$ as the component-wise extended map of $\phi_s$.
We can also define a Gray map $\Phi$ from $\mathbb{Z}_2^{\alpha_1}  \times \mathbb{Z}_4^{\alpha_2} \times \mathbb{Z}_8^{\alpha_3}$ to $\mathbb{Z}_2^n$, where $n=\alpha_1+2\alpha_2+4\alpha_3$, as follows:
$$
\Phi(u_1\mid u_2 \mid u_3)=(u_1, \Phi_2(u_2),\Phi_3(u_3)),
$$
for any $u_i \in \mathbb{Z}_{2^i}^{\alpha_i}$, where $1\leq i\leq 3$.


Let ${\cal C} \subseteq \mathbb{Z}_{2^s}^n$ be a $\mathbb{Z}_{2^s}$-additive code of length $n$. We say that its Gray map image
$C=\Phi_s({\cal C})$ is a $\mathbb{Z}_{2^s}$-linear code of length $n2^{s-1}$. 
Since ${\cal C}$ is a subgroup of
$\mathbb{Z}_{2^s}^n$, it is isomorphic to an abelian structure
$\mathbb{Z}_{2^s}^{t_1}\times\mathbb{Z}_{2^{s-1}}^{t_2}\times
\dots\times\mathbb{Z}_p^{t_s}$, and we say that ${\cal C}$, or equivalently
$C=\Phi_s({\cal C})$, is of type $(n;t_1,\dots,t_{s})$.
Note that $|{\cal C}|=p^{st_1}p^{(s-1)t_2}\cdots p^{t_s}$. Similarly, if ${\cal C} \subseteq \mathbb{Z}_2^{\alpha_1}  \times \mathbb{Z}_4^{\alpha_2} \times \mathbb{Z}_8^{\alpha_3}$
is a $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-additive code,  we say that its Gray map image $C=\Phi({\cal C})$ is a $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear code of length $\alpha_1+2\alpha_2+4\alpha_3$.
Since ${\cal C}$ can be seen as a subgroup of
$\mathbb{Z}_8^{\alpha_1+\alpha_2+\alpha_3}$, it is isomorphic to an abelian structure
$\mathbb{Z}_8^{t_1}\times\mathbb{Z}_4^{t_2} \times\mathbb{Z}_2^{t_3}$, and we say that ${\cal C}$, or equivalently
$C=\Phi({\cal C})$, is of type $(\alpha_1, \alpha_2,\alpha_3;t_1,t_2,t_3)$.
Note that $|{\cal C}|=8^{t_1}4^{t_2} 2^{t_3}$. Unlike linear codes over finite fields,
linear codes over rings do not have a basis, but there
exists a generator matrix for these codes having minimum number of rows, that is, $t_1+\dots+t_s$ rows.
If $\alpha_1=\alpha_3=0$ (respectively, $\alpha_1=\alpha_2=0$), then they coincide with $\mathbb{Z}_4$-additive codes (respectively, $\mathbb{Z}_8$-additive codes). If $\alpha_3=0$, then they are also known as $\mathbb{Z}_2\mathbb{Z}_4$-additive codes, and their Gray map images as $\mathbb{Z}_2\mathbb{Z}_4$-linear codes. In the last case, we also say that the code, or equivalently  the Gray map image of the code, is of type $(\alpha_1, \alpha_2;t_1,t_2)$. Note that there are no $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear Hadamard codes neither with only $\alpha_1=0$  nor with only $\alpha_2=0$ \cite{Krotov:2007, ShiKrotov2019}.



Two structural properties of codes over $\mathbb{Z}_2$ are the rank and
dimension of the kernel. The rank of a code $C$ over $\mathbb{Z}_2$ is simply the
dimension of the linear span, $\langle C \rangle$,  of $C$.
The kernel of a code $C$ over $\mathbb{Z}_2$ is defined as
$\mathrm{K}(C)=\{\textbf{x}\in \mathbb{Z}_2^n : \textbf{x}+C=C \}$ \cite{BGH83,pKernel}. If the all-zero vector belongs to $C$,
then $\mathrm{K}(C)$ is a linear subcode of $C$.
Note also that if $C$ is linear, then $K(C)=C=\langle C \rangle$.
We denote the rank of $C$ as $\operatorname{rank}(C)$ and the dimension of the kernel as $\operatorname{ker}(C)$.
These parameters can be used to distinguish between nonequivalent codes, since equivalent ones have the same rank and dimension of the kernel.

A binary code of length $n$, $2n$ codewords and minimum distance $n/2$ is called a Hadamard code. Hadamard codes can be constructed from Hadamard matrices \cite{Key,WMcwill}.
Note that linear Hadamard codes are in fact first order Reed-Muller codes, or
equivalently, the dual of extended Hamming codes \cite[Ch.13 \S 3]{WMcwill}. The $\mathbb{Z}_{2^s}$-additive codes such that after the Gray map $\Phi_s$ give
Hadamard codes are called $\mathbb{Z}_{2^s}$-additive Hadamard codes and the
corresponding images are called $\mathbb{Z}_{2^s}$-linear Hadamard codes. 
Similarly, the $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-additive codes such that after the Gray map $\Phi$ give
Hadamard codes are called $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-additive Hadamard codes and the
corresponding images are called $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear Hadamard codes. 

It is known that $\mathbb{Z}_4$-linear Hadamard codes (that is, $\mathbb{Z}_2\mathbb{Z}_4$-linear Hadamard code with $\alpha_1 =0$) and $\mathbb{Z}_2\mathbb{Z}_4$-linear Hadamard codes with $\alpha_1\not = 0$ can be classified by using either the rank or the dimension of the kernel \cite{Kro:2001:Z4_Had_Perf,PRV06}. Moreover, in \cite{KV2015}, it is shown that each $\mathbb{Z}_4$-linear Hadamard code is  equivalent to a  $\mathbb{Z}_2\mathbb{Z}_4$-linear Hadamard code with $\alpha_1 \not =0$.
Later, in \cite{KernelZ2s,HadamardZps,EquivZ2s,ZpsEquivalance}, an iterative construction for $\mathbb{Z}_{p^s}$-linear Hadamard codes is described, the linearity is established, and a partial classification by using the dimension of the kernel is obtained,
giving the exact amount of nonequivalent such codes for some parameters.
In \cite{fernandez2019mathbb}, a complete classification of $\mathbb{Z}_8$-linear Hadamard codes by using the rank and dimension of the kernel is provided, giving the exact amount of nonequivalent such codes.  For any $t\geq 2$, the full classification of $\mathbb{Z}_p\mathbb{Z}_{p^2}$-linear Hadamard codes of length $p^t$, with $\alpha_1\neq 0$, $\alpha_2\neq 0$, and $p\geq 3$ prime, is given in \cite{ZpZp2Construction,WCC2022,ZpZp2Classification}, by using just the dimension of the kernel.




  This paper is focused on $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear Hadamard codes with $\alpha_1\not =0$, $\alpha_2\not =0$, and $\alpha_3\not =0$, generalizing some results given for $\mathbb{Z}_2\mathbb{Z}_4$-linear Hadamard codes with $\alpha_1\not =0$  in \cite{PRV06,RSV08} related to the construction, linearity, kernel and classification of such codes. These codes are also compared with the $\mathbb{Z}_4$-linear, $\mathbb{Z}_8$-linear, and in general $\mathbb{Z}_{2^s}$-linear Hadamard codes with $s\geq 2$.
This paper is organized as follows.
In Section \ref{Sec:GrayMap}, we recall the definition of the Gray map considered in this paper and some of its properties.
In Section \ref{Sec:construction}, we describe the construction of  $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear Hadamard codes of type $(\alpha_1,\alpha_2, \alpha_3;t_1,t_2, t_3)$ with $\alpha_1\not =0$, $\alpha_2\not =0$, and $\alpha_3\not =0$.
We see that they are not included neither in the family of  $\mathbb{Z}_4$-linear Hadamard codes, nor in the family of  $\mathbb{Z}_8$-linear Hadamard codes, nor in the family of  $\mathbb{Z}_2\mathbb{Z}_4$-linear Hadamard codes with $\alpha_1 \not =0$.
Indeed, we see that all the nonlinear $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear Hadamard codes of lenght $2^t$ with $\alpha_1\not =0$, $\alpha_2\not =0$, and $\alpha_3\not =0$ are not equivalent to any $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear Hadamard code of any other type, any $\mathbb{Z}_2\mathbb{Z}_4$-linear Hadamard code, and  any $\mathbb{Z}_{2^s}$-linear Hadamard code, with $s\geq 2$, of the same length $2^t$.

\section{Preliminary results on the Gray map}\label{Sec:GrayMap}


In this section, we focus on the generalized Gray maps considered in this paper for elements of $\mathbb{Z}_4$ and $\mathbb{Z}_8$, and in general of $\mathbb{Z}_{2^s}$, $s\geq 2$. We include some of its properties used in the paper.


We consider the Carlet's Gray map from $\mathbb{Z}_{2^s}$ to $\mathbb{Z}_2^{2^{s-1}}$ \cite{Carlet} given in (\ref{eq:GrayMapCarlet}).
For $s=2$ and $s=3$, the Gray maps $\phi_2$ and $\phi_3$ considered in the paper for the elements of $\mathbb{Z}_4$ and $\mathbb{Z}_8$, respectively, are the following:
\begin{equation*}
\left.\begin{array}{cccccccccl}
\phi_2: &\mathbb{Z}_4 \longrightarrow \mathbb{Z}_2^2 &&& \phi_3: & \mathbb{Z}_8 \longrightarrow \mathbb{Z}_2^4\\
&0\mapsto(0,0) &&& & 0\mapsto (0,0,0,0) \\
&1\mapsto(0,1)&&& &1\mapsto (0,1,0,1)\\
&2\mapsto(1,1)&&& &2 \mapsto (0,0,1,1)\\
&3\mapsto(1,0)&&& &3 \mapsto (0,1,1,0)\\
 &            &&& &4 \mapsto (1,1,1,1)\\
  &           &&& &5 \mapsto (1,0,1,0)\\
   &          &&& &6 \mapsto (1,1,0,0)\\
    &         &&& &7 \mapsto (1,0,0,1).\\
\end{array}\right.
\end{equation*}


 
From \cite{HadamardZps}, we have the following results: 




\begin{corollary}\cite{HadamardZps}\label{coro4}
 Let $\lambda, \mu \in \mathbb{Z}_2$. Then,  $\phi_s(\lambda \mu 2^{s-1})= \lambda \phi_s(\mu 2^{s-1})=\lambda \mu \phi_s(2^{s-1})$.
\end{corollary}









\begin{corollary}\cite{HadamardZps} \label{lemma2}
Let $u, v$ $\in \mathbb{Z}_{2^s}$. Then, $\phi_s(2^{s-1}u+v)= \phi_s(2^{s-1}u)+\phi_s(v)$.
\end{corollary}






\begin{proposition}\cite{HadamardZps}\label{disweight}
Let $u,v\in\mathbb{Z}_{2^s}$. Then, $$d_H(\phi_s(u),\phi_s(v))={\rm wt}_H(\phi_s(u-v)).$$
\end{proposition}

By Proposition \ref{disweight}, the $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear codes obtained from the Gray map $\Phi$ are distance invariant, that is, the Hamming weight distribution is invariant under translation by a codeword. Therefore, their minimum distance coincides with the minimum weight.

\section{Construction of $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-additive Hadamard codes}
\label{Sec:construction}

The description of generator matrices having minimum number of rows for $\mathbb{Z}_4$-additive, $\mathbb{Z}_{2^s}$-additive, and in general for $\mathbb{Z}_{p^s}$-additive Hadamard codes, with $s\geq 2$ and $p$ prime,  are given in \cite{Kro:2001:Z4_Had_Perf}, \cite{KernelZ2s}, and \cite{HadamardZps}, respectively. 
Similarly, generator matrices having minimum number of rows for
$\mathbb{Z}_p\mathbb{Z}_{p^2}$-additive Hadamard codes with $\alpha_1\not =0, \alpha_2\not =0$ and $p$ prime, as long as an iterative construction of these matrices, are given in \cite{PRV06,RSV08} when $p=2$ and in \cite{ZpZp2Construction, WCC2022, ITW2022} when $p\geq 3$.
In this section, we generalize these results for $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-additive Hadamard codes with $\alpha_1\not =0$, $\alpha_2\not =0$, and $\alpha_3\not =0$. Specifically, we define an iterative construction for the generator matrices of these codes and establish that they generate $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-additive Hadamard codes.


 Let ${\mathbf{0}}, {\mathbf{1}},{\mathbf{2}},\ldots, \mathbf{7}$ be the vectors having the elements $0, 1, 2, \ldots, 7$  repeated in each coordinate, respectively. If $A$ is a generator matrix of a $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-additive code, that is, a subgroup of $\mathbb{Z}_2^{\alpha_1} \times \mathbb{Z}_4^{\alpha_2} \times \mathbb{Z}_8^{\alpha_3}$ for some integers $\alpha_1,\alpha_2,\alpha_3\geq 0$, then we denote by $A_1$ the submatrix of $A$ with the first $\alpha_1$ columns over $\mathbb{Z}_2$, $A_2$ the submatrix with the next $\alpha_2$ columns over $\mathbb{Z}_4$, and $A_3$ the submatrix with the last $\alpha_3$ columns over $\mathbb{Z}_8$. We have that $A=(A_1\mid A_2\mid A_3)$, where the number of columns of $A_i$ is $\alpha_i$ for $i\in \{1,2,3\}$.


Let $t_1\geq 1$, $t_2\geq 0$, and $t_3\geq 1$ be integers. Now, we construct recursively matrices $A^{t_1,t_2,t_3}$ having $t_1$ rows of order $8$, $t_2$ rows of order $4$, and $t_3$ rows of order $2$ as follows. First, we consider the following matrix:
\begin{equation}\label{eq:recGenMatrix0}
A^{1,0,1}=
\left(\begin{array}{cc|c|c}
1 & 1  & 2  &4 \\
0  & 1 &1  &1  \\
\end{array}\right).
\end{equation}
Then, we apply the following constructions. 
If we have a matrix  $A^{\ell-1,0,1}=(A_1 \mid A_2 \mid A_3)$, with $\ell \geq 2$, we may construct the matrix
\begin{equation}\label{eq:recGenMatrix1}
\footnotesize
A^{\ell,0,1}=
\left(\begin{array}{cc|ccccc|ccccc}
A_1 & A_1 &M_1 &A_2 &A_2 &A_2 &A_2 &M_2 &A_3 &A_3 &\cdots &A_3 \\
\mathbf{0}  & \mathbf{1} & \mathbf{1}  &\mathbf{0} &\mathbf{1}  &\mathbf{2} &\mathbf{3} &\mathbf{1} &\mathbf{0} &\mathbf{1} &\cdots &\mathbf{7} \\
\end{array}\right),
\end{equation}
where $M_1=\{\mathbf{z}^T:\mathbf{z}\in\lbrace2\rbrace\times \lbrace0,2\rbrace^{\ell-1}\}$ and $M_2=\{\mathbf{z}^T: \mathbf{z}\in\lbrace4\rbrace\times \lbrace0,2,4,6\rbrace^{\ell-1}\}$. We perform construction (\ref{eq:recGenMatrix1}) until $\ell=t_1$. If we have a matrix $A^{t_1,\ell-1,1}=(A_1\mid A_2 \mid A_3)$, with $t_1 \geq 1$ and $\ell\geq 1$, we may construct the matrix 
\begin{equation}\label{eq:recGenMatrix2}
\footnotesize
A^{t_1,\ell,1}=
\left(\begin{array}{cc|ccccc|cccc}
A_1 & A_1 &M_1 &A_2 &A_2 &A_2 &A_2 &A_3 &A_3 &A_3 &A_3 \\
\mathbf{0}  & \mathbf{1} & \mathbf{1}  &\mathbf{0} &\mathbf{1}  &\mathbf{2} &\mathbf{3}  &\mathbf{0} &\mathbf{2} &\mathbf{4} &\mathbf{6} \\
\end{array}\right),
\end{equation}
where $M_1=\{\mathbf{z}^T:\mathbf{z}\in\lbrace2\rbrace\times \lbrace0,2\rbrace^{t_1+\ell-1}\}$. We repeat construction (\ref{eq:recGenMatrix2}) until $\ell=t_2$. Finally, if we have a matrix $A^{t_1,t_2,\ell-1}=(A_1\mid A_2 \mid A_3)$, with $t_1\geq 1$, $t_2\geq 0$, and $\ell\geq 2$, we may construct the matrix
\begin{equation}\label{eq:recGenMatrix3}
A^{t_1,t_2,\ell}=
\left(\begin{array}{cc|cc|cc}
A_1 & A_1 & A_2 & A_2 &A_3 &A_3 \\
\mathbf{0}  & \mathbf{1} & \mathbf{0} &\mathbf{2}  &\mathbf{0}  &\mathbf{4}  \\
\end{array}\right).
\end{equation}
We repeat construction (\ref{eq:recGenMatrix3}) until $\ell=t_3$. Thus, in this way, we obtain $A^{t_1,t_2,t_3}$.

Summarizing, in order to achieve $A^{t_1,t_2,t_3}$ from $A^{1,0,1}$, first we add $t_1-1$ rows of order $8$  by applying construction (\ref{eq:recGenMatrix1}) $t_1-1$ times, starting from $A^{1,0,1}$ up to obtain $A^{t_1,0,1}$; then we add $t_2$ rows of order $4$ by applying construction (\ref{eq:recGenMatrix2}) $t_2$ times, up to generate $A^{t_1,t_2,1}$; and,  finally, we add $t_3-1$ rows of order $2$ by applying construction (\ref{eq:recGenMatrix3}) $t_3-1$ times to achieve $A^{t_1,t_2,t_3}$. Note that in the first row there is always the row $({\mathbf{1}} \mid {\mathbf{2}} \mid {\mathbf{4}})$. 


\begin{example} \label{ex:MatricesA}
By using the constructions described in (\ref{eq:recGenMatrix1}), (\ref{eq:recGenMatrix2}), and (\ref{eq:recGenMatrix3}), we obtain the following matrices $A^{2,0,1}$, $A^{1,1,1}$ and $A^{1,1,2}$, respectively, starting from the matrix $A^{1,0,1}$ given in (\ref{eq:recGenMatrix0}):
\begin{equation}\label{eq:A201}
A^{2,0,1}=
\left(\begin{array}{cc|cc|cc}
11 &11 &22 &2 2 2 2 &4 4 4 4 &4 4 4 4 4 4 4 4 \\
0  1 &0  1 &0 2 &1 1 1 1 &0 2 4 6 &1 1 1 1 1 1 1 1 \\
0 0 &1 1 &1 1 &0 1 2 3 &1 1 1 1 &0 1 2 3 4 5 6 7\\ 
\end{array}\right),
\end{equation}
\begin{equation}\label{eq:A111}
A^{1,1,1}=
\left(\begin{array}{cc|cc|c}
 11 &11 &22&2222 &4444\\
 01&01 &02&1111 &1111\\
 00&11 &11&0123 &0246\\ 
\end{array}\right),
\end{equation}
$$
A^{1,1,2}=
\left(\begin{array}{cc|cc|cc}
 1111 & 1111 &222222 &222222 &4444 &4444\\
 0101 & 0101 &021111 &021111 &1111 &1111\\
 0011 & 0011 &110123 &110123 &0246 &0246\\ 
 0000 &1111  &000000 &222222 &0000 &4444\\
\end{array}\right).
$$
\end{example}


In order to obtain $A^{2,1,1}$, we start with $A^{1,0,1}$,  we apply construction (\ref{eq:recGenMatrix1}) to obtain $A^{2,0,1}=(A_1\mid A_2\mid A_3)$ given in (\ref{eq:A201}), and then we apply (\ref{eq:recGenMatrix2}) to obtain 
 \begin{equation*}\label{eq:A_211}
A^{2,1,1}=\left(\begin{array}{cc|@{}cccccc|cccc}
      A_1  &A_1 &&\begin{matrix}
            22 2 2\\
            0 0 2 2\\
            0 2 0 2\\
      \end{matrix} &A_2  &A_2  &A_2  &A_2 &A_3 &A_3 &A_3 &A_3\\
      \mathbf{0} &\mathbf{1} &&\mathbf{1} &\mathbf{0} &\mathbf{1} &\mathbf{2} &\mathbf{3} &\mathbf{0} &\mathbf{2} &\mathbf{4} &\mathbf{6} \\
   \end{array}\right).
\end{equation*}


The $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-additive code generated by $A^{t_1,t_2,t_3}$ is denoted by ${\cal{H}}^{t_1,t_2,t_3}$, and the corresponding $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear code $\Phi( {\cal{H}}^{t_1,t_2,t_3})$ by $H^{t_1,t_2, t_3}$.

\begin{lemma}\label{relation:t1a1}
Let $t_1\geq 1$ and $t_2\geq 0$ be integers. Let ${\cal{H}}^{t_1,t_2,1}$ be the $\mathbb{Z}_2\mathbb{Z}_4 \mathbb{Z}_8$-additive code of type $(\alpha_1, \alpha_2, \alpha_3; t_1,t_2,1)$ generated by the matrix $A^{t_1,t_2,1}$. Then, $2^{t_1+t_2}=\alpha_1$, $4^{t_1+t_2}=\alpha_1+2\alpha_2$ and $8^{t_1}4^{t_2}=\alpha_1+2\alpha_2+ 4\alpha_3$.
\end{lemma}
\begin{proof}
First, we prove this lemma for the code ${\cal{H}}^{t_1,0,1}$ by induction on $t_1\geq 1$. Note that the lemma is true for the code ${\cal{H}}^{1,0,1}$ of type $(2,1,1;1,0,1)$. Assume that the lemma is true for the code ${\cal{H}}^{t_1,0,1}$ of type $(\alpha_1, \alpha_2, \alpha_3; t_1,0,1)$, that is, 
\begin{align}\label{t1_0_1}
    2^{t_1}=\alpha_1,
    4^{t_1}=\alpha_1+2\alpha_2~and~
    8^{t_1}=\alpha_1+2\alpha_2+4\alpha_3. 
\end{align}
By using construction (\ref{eq:recGenMatrix1}), the type of ${\cal{H}}^{t_1+1,0,1}$ is $(\alpha'_1, \alpha'_2, \alpha'_3; t_1+1,0,1)$, where 
\begin{align}\label{t1p1_0_1}
     \alpha'_1=2\alpha_1,
    \alpha'_2=2^{t_1}+4\alpha_2~and~
    \alpha'_3=4^{t_1}+8\alpha_3.
\end{align}
Thus, from (\ref{t1_0_1}) and (\ref{t1p1_0_1}), $2^{t_1+1}=2\alpha_1=\alpha'_1$, $4^{t_1+1}=4\alpha_1+8\alpha_2=2\alpha_1+2\alpha_1+8\alpha_2=\alpha'_1+2^{t_1+1}+8\alpha_2=\alpha'_1+2\alpha'_2$ and $8^{t_1+1}=8\alpha_1+16\alpha_2+32\alpha_3=2\alpha_1+(2\alpha_1+8\alpha_2)+(4\alpha_1+8\alpha_2+32\alpha_3)=2\alpha_1+(2^{t_1+1}+8\alpha_2)+(4^{t_1+1}+32\alpha_3)=\alpha'_1+2\alpha'_2+4\alpha'_3$. Therefore, the lemma is true for the code ${\cal{H}}^{t_1,0,1}$.

Next, we prove this lemma for the code ${\cal{H}}^{t_1,t_2,1}$ by induction on $t_2\geq 0$. Assume that the lemma holds for the code ${\cal{H}}^{t_1,t_2,1}$ of type $(\alpha_1, \alpha_2, \alpha_3; t_1,t_2,1)$, that is, 
\begin{align}\label{t1_t2_1}
      2^{t_1+t_2}=\alpha_1,
    4^{t_1+t_2}=\alpha_1+2\alpha_2,~and~
    8^{t_1}4^{t_2}=\alpha_1+2\alpha_2+4\alpha_3.
\end{align}
By using construction (\ref{eq:recGenMatrix2}), the type of ${\cal{H}}^{t_1,t_2+1,1}$ is $(\alpha'_1, \alpha'_2, \alpha'_3; t_1,t_2+1,1)$, where 
\begin{align}\label{t1_t2p1_1}
    \alpha'_1=2\alpha_1,
    \alpha'_2=2^{t_1+t_2}+4\alpha_2 ~and ~
    \alpha'_3=4\alpha_3.
\end{align}
Thus, from (\ref{t1_t2_1}) and (\ref{t1_t2p1_1}), $2^{t_1+(t_2+1)}=2\alpha_1=\alpha'_1$, $4^{t_1+(t_2+1)}=4\alpha_1+8\alpha_2=2\alpha_1+2\alpha_1+8\alpha_2=\alpha'_1+2^{t_1+t_2+1}+8\alpha_2=\alpha'_1+2\alpha'_2$ and $8^{t_1}4^{t_2+1}=4\alpha_1+8\alpha_2+16\alpha_3=2\alpha_1+(2\alpha_1+8\alpha_2)+16\alpha_3=\alpha'_1+(2^{t_1+t_2+1}+8\alpha_2)+4\alpha'_3=\alpha'_1+2\alpha'_2+4\alpha'_3$. Therefore, the lemma is true for the code ${\cal{H}}^{t_1,t_2+1,1}$. This completes the proof.
\end{proof}

\begin{proposition}\label{relation:t1a1a3}
Let $t_1\geq 1$, $t_2\geq 0$, and $t_3\geq 1$ be integers. Let ${\cal{H}}^{t_1,t_2,t_3}$ be the $\mathbb{Z}_2\mathbb{Z}_4 \mathbb{Z}_8$-additive code of type $(\alpha_1, \alpha_2, \alpha_3; t_1,t_2,t_3)$ generated by the matrix $A^{t_1,t_2,t_3}$. Then,  
\begin{align} \label{eq:propt1t2t3}
\begin{split}
    &\alpha_1=2^{t_1+t_2+t_3-1},\\ 
    &\alpha_1+2\alpha_2=4^{t_1+t_2}2^{t_3-1},\\ 
    &\alpha_1+2\alpha_2+ 4\alpha_3= 8^{t_1}4^{t_2}2^{t_3-1}.
\end{split}
\end{align}
\end{proposition}
\begin{proof}
We prove this result for the code ${\cal{H}}^{t_1,t_2,t_3}$ by induction on $t_3\geq 1$. By Lemma \ref{relation:t1a1}, the proposition is true for $t_3=1$, that is, for the code ${\cal{H}}^{t_1,t_2,1}$.
Assume that it holds for the code ${\cal{H}}^{t_1,t_2,t_3}$ of type $(\alpha_1, \alpha_2, \alpha_3; t_1,t_2,t_3)$, that is, (\ref{eq:propt1t2t3}) holds.
By using construction (\ref{eq:recGenMatrix3}), the type of ${\cal{H}}^{t_1,t_2,t_3+1}$ is $(\alpha'_1, \alpha'_2, \alpha'_3; t_1,t_2,t_3+1)$, where 
\begin{align}\label{t1_t2p1_t_3}
    \alpha'_1=2\alpha_1,
    \alpha'_2=2\alpha_2, ~\mbox{and} ~
    \alpha'_3=2\alpha_3.
\end{align}
Thus, from (\ref{eq:propt1t2t3}) and (\ref{t1_t2p1_t_3}), $2^{t_1+t_2+t_3}=2\alpha_1=\alpha'_1$, $4^{t_1+t_2}2^{t_3}=2\alpha_1+4\alpha_2=\alpha'_1+2\alpha'_2$ and $8^{t_1}4^{t_2}2^{t_3}=2\alpha_1+4\alpha_2+8\alpha_3=\alpha'_1+2\alpha'_2+4\alpha'_3$. Therefore, the proposition is true for the code ${\cal{H}}^{t_1,t_2,t_3+1}$. This completes the proof.
\end{proof}

\begin{corollary} \label{coro:Length}
Let $t_1\geq 1$, $t_2\geq 0$, and $t_3\geq 1$ be integers. Let ${\cal{H}}^{t_1,t_2,t_3}$ be the $\mathbb{Z}_2\mathbb{Z}_4 \mathbb{Z}_8$-additive code of type $(\alpha_1, \alpha_2, \alpha_3; t_1,t_2,t_3)$ generated by the matrix $A^{t_1,t_2,t_3}$.
 Then,
\begin{align*}
\begin{split}
    &\alpha_1=2^{t_1+t_2+t_3-1},\\ 
    &\alpha_2=4^{t_1+t_2}2^{t_3-2}-2^{t_1+t_2+t_3-2},\\ &\alpha_3=8^{t_1}4^{t_2-1}2^{t_3-1}-4^{t_1+t_2-1}2^{t_3-1}.
\end{split}
\end{align*}
\end{corollary}

\begin{remark}
By Corollary \ref{coro:Length},  we have that the $\mathbb{Z}_2\mathbb{Z}_4 \mathbb{Z}_8$-additive codes $\mathcal{H}^{t_1,t_2,t_3}$ of type $(\alpha_1, \alpha_2, \alpha_3; t_1,t_2,t_3)$ generated by the matrix $A^{t_1,t_2,t_3}$, so constructed recursively from (\ref{eq:recGenMatrix1}), (\ref{eq:recGenMatrix2}), and (\ref{eq:recGenMatrix3}), satisfy that $\alpha_1\not =0$, $\alpha_2\not =0$, and $\alpha_3\not =0$. 
\end{remark}

\begin{remark}
We can see the construction of the generator matrices $A^{t_1,t_2,t_3}$ as a generalization of the recursive construction of the generator matrices of the $\mathbb{Z}_2\mathbb{Z}_4$-additive Hadamard codes of type $(\alpha_1,\alpha_2;t_1,t_2)$ with $\alpha_1\not =0$ and $\alpha_2 \not =0$, given in \cite{RSV08}. Note that if we do not consider the coordinates over $\mathbb{Z}_8$ in constructions (\ref{eq:recGenMatrix1}), (\ref{eq:recGenMatrix2}), and (\ref{eq:recGenMatrix3}), we have that (\ref{eq:recGenMatrix1}) and (\ref{eq:recGenMatrix2}) become 
\begin{equation}\label{eq:recZ2Z4GenMatrix1}
\footnotesize
A^{\ell,1}=
\left(\begin{array}{cc|ccccc}
A_1 & A_1 &M_1 &A_2 &A_2 &A_2 &A_2  \\
\mathbf{0}  & \mathbf{1} & \mathbf{1}  &\mathbf{0} &\mathbf{1}  &\mathbf{2} &\mathbf{3} \\
\end{array}\right),
\end{equation}
where $A^{\ell-1,1} =(A_1\mid A_2)$ and  $M_1=2A_1=\{\mathbf{z}^T:\mathbf{z}\in\lbrace2\rbrace\times \lbrace0,2\rbrace^{\ell-1}\}$ (up to a column permutation); and
construction (\ref{eq:recGenMatrix3}) become 
\begin{equation}\label{eq:recZ2Z4GenMatrix3}
A^{t_1,\ell}=
\left(\begin{array}{cc|cc}
A_1 & A_1 & A_2 & A_2 \\
\mathbf{0}  & \mathbf{1} & \mathbf{0} &\mathbf{2}   \\
\end{array}\right),
\end{equation}
where $A^{t_1, \ell-1} =(A_1\mid A_2)$.  Then, starting from the following matrix:
\begin{equation}\label{eq:recZ2Z4GenMatrix0}
A^{1,1}=
\left(\begin{array}{cc|c}
1 & 1  & 2  \\
0  & 1 &1    \\
\end{array}\right),
\end{equation}
and applying (\ref{eq:recZ2Z4GenMatrix1}) and (\ref{eq:recZ2Z4GenMatrix3}) in the same way as above, we obtain the generator matrices $A^{t_1,t_2}$ of the known $\mathbb{Z}_2\mathbb{Z}_4$-additive Hadamard codes of type $(\alpha_1,\alpha_2;t_1,t_2)$ with $\alpha_1\not =0$ and $\alpha_2 \not =0$ \cite{PRV06,RSV08}. The $\mathbb{Z}_2\mathbb{Z}_4$-additive code generated by $A^{t_1,t_2}$ is denoted by ${\cal{H}}^{t_1,t_2}$, and the corresponding $\mathbb{Z}_2\mathbb{Z}_4$-linear code $\Phi( {\cal{H}}^{t_1,t_2})$ by $H^{t_1,t_2}$.
\end{remark}


When we include all the elements of  $\mathbb{Z}_{2^i}$, where $1\leq i\leq 3$,  as coordinates of a vector, we place them in increasing order.
For a set $S \subseteq \mathbb{Z}_{2^i}$ and $\lambda \in \mathbb{Z}_{2^i}$, where $i\in \{1,2,3\}$, we define $\lambda S=\{\lambda j: j\in S\}$ and $S+\lambda=\{j+\lambda: j\in S\}$. As before, when including all the elements in those sets as coordinates of a vector, we place them in increasing order. 
For example,
$2\mathbb{Z}_8=\{0,4,6,8\}$, $(\mathbb{Z}_4,\mathbb{Z}_4)=(0,1,2,3,0,1,2,3) \in \mathbb{Z}_4^8$ and $(\mathbb{Z}_2 \mid \mathbb{Z}_4\mid 2\mathbb{Z}_8, 4\mathbb{Z}_8)=(0,1 \mid 0,1,2,3\mid 0,2,4,6, 0,4) \in \mathbb{Z}_2^2 \times \mathbb{Z}_4^4 \times \mathbb{Z}_8^6$.



\begin{lemma}\label{Lemma:cases}
Let $1\leq i\leq 3$ and  $j \in \{0, 1,\dots, i-1\}$. 
\begin{enumerate} 
\item \label{2z22} If $\mu \in 2^j\mathbb{Z}_{2^i}$, then $2^j\mathbb{Z}_{2^i}+\mu = 2^j\mathbb{Z}_{2^i}$.
\item\label{item:vector} If $\mu \in 2^j\mathbb{Z}_{2^i}$, then $(2^j\mathbb{Z}_{2^i}, \stackrel{m}{\dots}, 2^j\mathbb{Z}_{2^i})+ \mu {\mathbf{1}}$, where $m \geq  1$, is a permutation of the vector $(2^j\mathbb{Z}_{2^i}, \stackrel{m}{\dots}, 2^j\mathbb{Z}_{2^i})$.
\item \label{2z2_}If $\mu \in 2\mathbb{Z}_{2^i}$, then $(\mathbb{Z}_{2^i}{\setminus}2\mathbb{Z}_{2^i})+\mu = \mathbb{Z}_{2^i}{\setminus}2\mathbb{Z}_{2^i}$.
\item\label{item:N2^-} If $\mu \in\mathbb{Z}_{2^i}$, then
$({\mathbf{0}}, \dots, \mathbf{2^i-1})+(\mu,\stackrel{\ell\cdot 2^i}{\dots}, \mu)$, where $\ell\geq 1$ and  $\mathbf{k}=(k,\stackrel{\ell}{\dots}, k)$ for $k \in \mathbb{Z}_{2^i}$, is a permutation of  $(\mathbb{Z}_{2^i},\stackrel{\ell}{\dots}, \mathbb{Z}_{2^i})$.
\end{enumerate}
\end{lemma}
\begin{proof}
Item 1 follows from the fact that $\mathbb{Z}_{2^i}$ is a ring and $2^j\mathbb{Z}_{2^i}$ is an ideal of $\mathbb{Z}_{2^i}$. Item 2 follows from Item 1. 



For Item 3, it $x\in (\mathbb{Z}_{2^i}{\setminus}2\mathbb{Z}_{2^i})+\mu$, then $x-\mu \in \mathbb{Z}_{2^i}{\setminus} 2\mathbb{Z}_{2^i}$. Assume that $x \notin \mathbb{Z}_{2^i}{\setminus} 2\mathbb{Z}_{2^i}$, so $x \in 2\mathbb{Z}_{2^i}$. Since $2\mathbb{Z}_{2^i}$ is an ideal of $\mathbb{Z}_{2^i}$, we have that $x-\mu\in 2\mathbb{Z}_{2^i}$, which is a contradiction. Thus, $x \in \mathbb{Z}_{2^i}{\setminus} 2\mathbb{Z}_{2^i}$ and hence $(\mathbb{Z}_{2^i}{\setminus} 2\mathbb{Z}_{2^i})+\mu \subseteq \mathbb{Z}_{2^i}{\setminus} 2\mathbb{Z}_{2^i}$. In the same way,  $(\mathbb{Z}_{2^i}{\setminus} 2\mathbb{Z}_{2^i})-\mu \subseteq \mathbb{Z}_{2^i}{\setminus} 2\mathbb{Z}_{2^i}$. Hence, $\mathbb{Z}_{2^i}{\setminus} 2\mathbb{Z}_{2^i} \subseteq (\mathbb{Z}_{2^i}{\setminus} 2\mathbb{Z}_{2^i})+\mu $ and therefore
$(\mathbb{Z}_{2^i}{\setminus} 2\mathbb{Z}_{2^i})+\mu = \mathbb{Z}_{2^i}{\setminus}2\mathbb{Z}_{2^i}$.


For Item 4, note that $({\mathbf{0}}, \dots, \mathbf{2^i-1})+(\mu,\stackrel{\ell\cdot 2^i}{\dots}, \mu)$ is a permutation of 
\begin{equation}\label{a}
   (\mathbb{Z}_{2^i},\stackrel{\ell}{\dots}, \mathbb{Z}_{2^i})+(\mu,\stackrel{\ell\cdot 2^i}{\dots}, \mu).
\end{equation}
Since $\mathbb{Z}_{2^i}+\mu =\mathbb{Z}_{2^i}$, (\ref{a}) is a permutation of $(\mathbb{Z}_{2^i},\stackrel{\ell}{\dots}, \mathbb{Z}_{2^i})$.
\end{proof}

\begin{lemma}\label{lemm:uu+lambda}
Let $1\leq i\leq 3$, $\lambda\in\mathbb{Z}_{2^i}{\setminus} 2\mathbb{Z}_{2^i}$, and $u\in\mathbb{Z}_{2^i}^n$. Then,
$$
(u,\stackrel{2^i}{\dots},u)+\lambda({\mathbf{0}}, \dots, \mathbf{2^i-1})
$$
is a permutation of $(\mathbb{Z}_{2^i},\stackrel{n}{\dots}, \mathbb{Z}_{2^i})$.
\end{lemma}
\begin{proof}
Since  $\lambda\in\mathbb{Z}_{2^i}{\setminus} 2\mathbb{Z}_{2^i}$, $\lambda({\mathbf{0}}, \dots, \mathbf{2^i-1})$ is a permutation of $({\mathbf{0}}, \dots, \mathbf{2^i-1})$ and we may consider $\lambda=1$. 
Then, $(u,\dots,u)+({\mathbf{0}}, \dots, \mathbf{2^i-1})$ is a permutation of $(u_1+\mathbb{Z}_{2^i},\dots,u_n+\mathbb{Z}_{2^i})=(\mathbb{Z}_{2^i},\stackrel{n}{\dots}, \mathbb{Z}_{2^i})$, where  $u=(u_1,\dots,u_n)$.
\end{proof}

\begin{lemma}\label{lemm:ord8_u2}
Let  $u=(\mu, \stackrel{m}{\dots}, \mu, 2\mathbb{Z}_{4}, \stackrel{n}{\dots}, 2\mathbb{Z}_{4}, \mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4},\stackrel{r}{\dots},\mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4})\in \mathbb{Z}_4^{m+2n+2r}$, where  $m,n,r\geq 0$ and $\mu \in \mathbb{Z}_4{\setminus} 2\mathbb{Z}_4 =\{1,3\}$. Then, 
 $$
 (u,u,u,u)+({\mathbf{0}}, {\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}})
 $$
 is a permutation of $(2\mathbb{Z}_{4}, \stackrel{4n}{\dots}, 2\mathbb{Z}_{4}, \mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4},\stackrel{4r+2m}{\dots},\mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4})$.
\end{lemma}
\begin{proof}
 By Items \ref{2z22} and \ref{2z2_} of Lemma \ref{Lemma:cases}, $u+\bf{2}$ is a permutation of $(\mu+2, \stackrel{m}{\dots}, \mu+2, 2\mathbb{Z}_{4}, \stackrel{n}{\dots}, 2\mathbb{Z}_{4}, \mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4},\stackrel{r}{\dots},\mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4})$. Let ${\bf k}=(\mu, \stackrel{m}{\dots}, \mu)$. Since $\mu\in\{1,3\}$, we have that
$
({\bf k},{\bf k},{\bf k},{\bf k})+({\mathbf{0}},{\mathbf{2}},{\mathbf{0}},{\mathbf{2}})
$
is a permutation of $(\mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4},\stackrel{2m}{\dots},\mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4})$. Therefore, $(u,u,u,u)+({\mathbf{0}}, {\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}})$ is a permutation of $(2\mathbb{Z}_{4}, \stackrel{4n}{\dots}, 2\mathbb{Z}_{4}, \mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4},\stackrel{4r+2m}{\dots},\mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4})$.
\end{proof}

\begin{lemma}\label{lemm:ord8_u3} 
Let  $u=(\mu', \stackrel{m'}{\dots}, \mu', \mu'', \stackrel{m'}{\dots}, \mu'', 2\mathbb{Z}_{8}, \stackrel{n'}{\dots}, 2\mathbb{Z}_{8}, \mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8},\stackrel{r'}{\dots},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8})\in \mathbb{Z}_8^{2m'+4n'+4r'}$, where  $m', n',r'\geq 0$ and $\mu, \mu'\in \mathbb{Z}_8{\setminus}2\mathbb{Z}_8= \{1,3,5,7\}$. Then, 
\begin{enumerate}
    \item\label{ord8u3_case1} $(u,u,u,u)+({\mathbf{0}}, {\mathbf{2}},{\mathbf{4}}, \mathbf{6})$ is a permutation of $(2\mathbb{Z}_{8}, \stackrel{4n'}{\dots}, 2\mathbb{Z}_{8}, \mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8},\stackrel{4r'+2m'}{\dots},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8})$;
    \item \label{ord8u3_case2}  $(u,u,u,u)+({\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}})$ is a permutation of $(\mu',\stackrel{4m'}{\dots}, \mu', \mu'+4,\stackrel{4m'}{\dots},\mu'+4, 2\mathbb{Z}_{8}, \stackrel{4n'}{\dots}, 2\mathbb{Z}_{8}, \mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8},\stackrel{4r'}{\dots},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8})$ if $\mu' =\mu''$ or $\mu'=\mu''+4$, or a permutation of $(2\mathbb{Z}_{8}, \stackrel{4n'}{\dots}, 2\mathbb{Z}_{8},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8},\stackrel{4r'+2m'}{\dots},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8})$ otherwise.
\end{enumerate}
\end{lemma}
\begin{proof}
For Item 1, by Items \ref{2z22} and  \ref{2z2_} of Lemma \ref{Lemma:cases}, if $j\in\{0,2,4,6\}$, then $u+\bf{j}$ is a permutation of $(\mu'+j, \stackrel{m'}{\dots}, \mu'+j, \mu''+j, \stackrel{m'}{\dots}, \mu''+j, 2\mathbb{Z}_{8}, \stackrel{n'}{\dots}, 2\mathbb{Z}_{8}, \mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8},\stackrel{r'}{\dots},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8})$. Let ${\bf k}'=(\mu', \stackrel{m'}{\dots}, \mu', \mu'' \stackrel{m'}{\dots}, \mu'')$.  Since $\mu',\mu''\in\{1,3,5,7\}$,  we have that
$
({\bf k}', \stackrel{4}{\dots}, {\bf k}')+({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})
$ 
is a permutation of $(\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8},\stackrel{2m'}{\dots},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8})$ and hence $(u,u,u,u)+({\mathbf{0}}, {\mathbf{2}},{\mathbf{4}}, \mathbf{6})$ is a permutation of $(2\mathbb{Z}_{8}, \stackrel{4n'}{\dots}, 2\mathbb{Z}_{8}, \mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8},\stackrel{4r'+2m'}{\dots},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8})$. 

For item 2, we have that
$
({\bf k}', \stackrel{4}{\dots}, {\bf k}')+({\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}})
$
is a permutation of $(\mu',\stackrel{4m'}{\dots}, \mu', \mu'+4,\stackrel{4m'}{\dots},\mu'+4)$ if $\mu' =\mu''$ or $\mu'=\mu''+4$, or a permutation of $(\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8},\stackrel{2m'}{\dots},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8})$ otherwise. Therefore,  $(u,u,u,u)+({\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}})$ is a permutation of $(\mu',\stackrel{4m'}{\dots}, \mu', \mu'+4,\stackrel{4m'}{\dots},\mu'+4, 2\mathbb{Z}_{8}, \stackrel{4n'}{\dots}, 2\mathbb{Z}_{8}, \mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8},\stackrel{4r'}{\dots},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8})$ if $\mu' =\mu''$ or $\mu'=\mu''+4$, or a permutation of $(2\mathbb{Z}_{8}, \stackrel{4n'}{\dots}, 2\mathbb{Z}_{8},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8},\stackrel{4r'+2m'}{\dots},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8})$ otherwise.
\end{proof}

\begin{lemma}\label{lemm:ord4_u3}
Let $u=(\mu,\stackrel{m}{\dots}, \mu,4\mathbb{Z}_8,\stackrel{n}{\dots},4\mathbb{Z}_8,2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8, \stackrel{r}{\dots}, 2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8)\in \mathbb{Z}_8^{m+2n+2r}$, where  $m, n,r\geq 0$ and $\mu \in 2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8=\{2,6\}$.
Then,
\begin{enumerate}
    \item\label{lemm:ord4_case1} $(u,u,u,u)+({\mathbf{0}}, {\mathbf{2}},{\mathbf{4}}, \mathbf{6})$ is a permutation of $(2\mathbb{Z}_8, \stackrel{2r+2n+m}{\dots},2\mathbb{Z}_8)$;
    \item \label{lemm:ord4_case2} $(u,u,u,u)+({\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}})$ is a permutation of $(4\mathbb{Z}_8,\stackrel{4n}{\dots},4\mathbb{Z}_8,2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8, \stackrel{4r+2m}{\dots}, 2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8)$.
\end{enumerate}
\end{lemma}
\begin{proof}
By Item \ref{2z22} of Lemma \ref{Lemma:cases}, if $j\in \{0,4\}$, then $u+\mathbf{j}$ is a permutation of $ (\mu+j,\stackrel{m}{\dots}, \mu+j,4\mathbb{Z}_8,\stackrel{n}{\dots},4\mathbb{Z}_8,2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8, \stackrel{r}{\dots}, 2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8)$. Similarly,  if $j\in \{2,6\}$, then $u+\mathbf{j}$ is a permutation of $ (\mu+j,\stackrel{m}{\dots}, \mu+j,4\mathbb{Z}_8,\stackrel{r}{\dots},4\mathbb{Z}_8,2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8, \stackrel{n}{\dots}, 2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8)$. Let $\mathbf{k}=(\mu, \stackrel{m}{\dots}, \mu)$.  

For Item 1, since $\mu \in \{2,6\}$, we have that 
    $
    (\mathbf{k}, \stackrel{4}{\dots}, \mathbf{k})+({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})
    $
    is a permutation of $(2\mathbb{Z}_8,\stackrel{m}{\dots}, 2\mathbb{Z}_8)$, and hence $(u,u,u,u)+({\mathbf{0}}, {\mathbf{2}},{\mathbf{4}}, \mathbf{6})$ is a permutation of $(2\mathbb{Z}_8, \stackrel{2r+2n+m}{\dots},2\mathbb{Z}_8)$. 
    
    For Item 2,  we have that 
    $
    (\mathbf{k}, \stackrel{4}{\dots}, \mathbf{k})+({\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}})
    $
    is a permutation of $(2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8, \stackrel{2m}{\dots}, 2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8)$. Therefore, $(u,u,u,u)+({\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}})$ is a permutation of $(4\mathbb{Z}_8,\stackrel{4n}{\dots},4\mathbb{Z}_8,2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8, \stackrel{4r+2m}{\dots}, 2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8)$.
\end{proof}

\medskip
 Let $t_1\geq1, t_2\geq 0$, and $t_3\geq1$ be integers.  Let ${\cal{G}}^{t_1,t_2,t_3}$ be the set of all codewords of the code generated by the matrix obtained from $A^{t_1,t_2,t_3}$ after removing the row $(\mathbf{1\mid 2\mid 4})$.


\begin{lemma}\label{lem:z_t1p1_0_1}
    Let $t_1\geq 1$ be an integer. Let $$\mathbf{z}=(u_1,u_1 \mid x_1,u_2, u_2, u_2, u_2\mid x_2,u_3,\stackrel{8}{\dots},u_3)\in {\cal{G}}^{t_1+1,0,1},$$  where $\mathbf{u}=(u_1\mid u_2\mid u_3) \in {\cal{G}}^{t_1,0,1}$ and $x_{i-1}\in (2\mathbb{Z}_{2^i})^{2^{(i-1)t_1}}$ for $i \in \{2,3\}$.
  
    Then,
    \begin{enumerate}
    \item if $o(\mathbf{z})=8$, then $x_{i-1}$ is a permutation  of $(2\mathbb{Z}_{2^{i}},\stackrel{2^{(i-1)(t_1-1)}}{\dots}, 2\mathbb{Z}_{2^{i}})$ for $i \in \{2,3\}$.
        \item if $o(\mathbf{z})=4$, then $x_1={\mathbf{0}}$ and $x_2$ is a permutation of $(4\mathbb{Z}_8, \stackrel{2\cdot 4^{t_1-1}}{\dots}, 4\mathbb{Z}_8)$.
        \item if $o(\mathbf{z})=2$, then $x_1={\mathbf{0}}$ and $x_2={\mathbf{0}}$.
    \end{enumerate}
\end{lemma}
\begin{proof}
    Let $\mathbf{w}_j$, where $j\in \{1,\dots,t_1+2\}$, be the $j$th row of the matrix $A^{t_1+1,0,1}$. Note that $\mathbf{w}_1=(\mathbf{1\mid 2\mid 4})$, and $\mathbf{w}_2,\dots, \mathbf{w}_{t_1+2}$ are the rows of order 8, where $\mathbf{w}_{t_1+2}=({\mathbf{0}}, {\mathbf{1}} \mid {\mathbf{1}},{\mathbf{0}}, {\mathbf{1}}, {\mathbf{2}},  \mathbf{3}\mid {\mathbf{1}}, {\mathbf{0}}, \dots, \mathbf{7})$. 
    Since any element of ${\cal{G}}^{t_1+1,0,1}$ can be written as $\mathbf{z}+\lambda \mathbf{w}_{t_1+2}$, where $\lambda \in \mathbb{Z}_8$, then $\mathbf{z}=\sum_{j=2}^{t_1+1}r_j\mathbf{w}_j=(u_1,u_1 \mid x_1,u_2, u_2, u_2, u_2\mid x_2,u_3,\stackrel{8}{\dots},u_3)$, where $r_j \in \mathbb{Z}_8$.  By construction, 
   
    $x_1$ and $x_2$ are generated by the rows of $M'_1=\{\mathbf{z}^T:\mathbf{z}\in \lbrace 0,2\rbrace^{t_1}\}$ and $M'_2=\{\mathbf{z}^T: \mathbf{z}\in \lbrace0,2,4,6\rbrace^{t_1}\}$, respectively. Thus,  $x_1={\mathbf{0}}$ or $x_1$ is a permutation of $(2\mathbb{Z}_4, \stackrel{2^{t_1-1}}{\dots}, 2\mathbb{Z}_4)$, and $x_2={\mathbf{0}}$ or $x_2$ is a permutation of $(2\mathbb{Z}_8, \stackrel{4^{t_1-1}}{\dots}, 2\mathbb{Z}_8)$ or $(4\mathbb{Z}_8, \stackrel{2\cdot 4^{t_1-1}}{\dots}, 4\mathbb{Z}_8)$.
    
    For Item 1, we have that there exists at least one $j\in \{2,\dots, t_1+1\}$ such that $r_j \in \{1,3,5,7\}$. Therefore, by  Item \ref{2z22} of Lemma \ref{Lemma:cases}, $x_{i-1}$ is a permutation  of $(2\mathbb{Z}_{2^{i}},\stackrel{2^{(i-1)(t_1-1)}}{\dots}, 2\mathbb{Z}_{2^{i}})$ for $i \in \{2,3\}$. 

    For Item 2, we have that $r_j\in 2\mathbb{Z}_8$ for all $j\in \{2,\dots, t_1+1\}$ and there exist at least one $j\in \{2,\dots, t_1+1\}$ such that $r_j\in \{2,6\}$. Therefore, $x_1={\mathbf{0}}$ and, by Item \ref{2z22} of Lemma \ref{Lemma:cases}, $x_2$ is a permutation of $(4\mathbb{Z}_8, \stackrel{2\cdot 4^{t_1-1}}{\dots}, 4\mathbb{Z}_8)$.

    For Item 3,  we have that $r_j\in 4\mathbb{Z}_8$ for all $j\in \{2,\dots, t_1+1\}$ and there exist at least one $j\in \{2,\dots, t_1+1\}$ such that  $r_j=4$. Therefore, $x_1={\mathbf{0}}$ and $x_2={\mathbf{0}}$.
\end{proof}


\begin{lemma}\label{lem:z_t1t2p1_1}
     Let $t_1\geq 1$ and $t_2\geq 0$ be integers.   Let $$\mathbf{z}=(u_1,u_1 \mid x_1,u_2, u_2, u_2, u_2\mid u_3,u_3, u_3, u_3) \in {\cal{G}}^{t_1,t_2+1,1},$$ where $\mathbf{u}=(u_1\mid u_2\mid u_3) \in {\cal{G}}^{t_1,t_2,1}$ and $x_1\in (2\mathbb{Z}_4)^{2^{t_1+t_2}}$.
    Then,
    \begin{enumerate}
    \item if $o(\mathbf{z})=8$, then $x_1$ is a permutation  of $(2\mathbb{Z}_4,\stackrel{2^{t_1+t_2-1}}{\dots}, 2\mathbb{Z}_4)$.
        \item if $o(\mathbf{z})=4$, then $x_1={\mathbf{0}}$ if $u_1={\mathbf{0}}$, and $x_1$ is a permutation of $(2\mathbb{Z}_4,\stackrel{2^{t_1+t_2-1}}{\dots}, 2\mathbb{Z}_4)$ otherwise.
        \item if $o(\mathbf{z})=2$, then $x_1={\mathbf{0}}$.
    \end{enumerate}
\end{lemma}
\begin{proof}
    Let $\mathbf{w}_i$, where $i\in \{1,\dots,t_1+t_2+2\}$, be the $i$th row of the matrix $A^{t_1, t_2+1,1}$. Note that $\mathbf{w}_1=(\mathbf{1\mid 2\mid 4})$, $\mathbf{w}_2, \dots, \mathbf{w}_{t_1+1}$ are the rows of order $8$, and $\mathbf{w}_{t_1+2}, \dots, \mathbf{w}_{t_1+t_2+2}$ are the rows of order $4$, where $\mathbf{w}_{t_1+t_2+2}=({\mathbf{0}}, {\mathbf{1}} \mid {\mathbf{1}}, {\mathbf{0}}, {\mathbf{1}}, {\mathbf{2}}, \mathbf{3}\mid  {\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$. 
    Since any element of ${\cal{G}}^{t_1,t_2+1,1}$ can be written as $\mathbf{z}+\lambda \mathbf{w}_{t_1+t_2+2}$, where $\lambda \in \{0,1,2,3\}$, then $\mathbf{z}=\sum_{i=2}^{t_1+t_2+1}r_i\mathbf{w}_i=(u_1,u_1 \mid x_1,u_2, u_2, u_2, u_2\mid u_3,u_3, u_3, u_3)$, where $r_i \in \mathbb{Z}_8$ for $i \in \{2,\dots, t_1+1\}$ and $r_i\in \{0,1,2,3\}$ for $i\in \{t_1+2,\dots, t_1+t_2+1\}$. By construction, 
   
    $x_1$ is generated by the rows of $M'_1=\{\mathbf{z}^T:\mathbf{z}\in \lbrace 0,2\rbrace^{t_1+t_2}\}$. Thus,  $x_1={\mathbf{0}}$ or $x_1$ is a permutation of $(2\mathbb{Z}_4, \stackrel{2^{t_1+t_2-1}}{\dots}, 2\mathbb{Z}_4)$.

    For Item 1, we have that there exists at least one $i\in \{2,\dots, t_1+1\}$ such that $r_i \in \{1,3,5,7\}$. Therefore, since $x_1$ is of order at most two,  $x_1\neq {\mathbf{0}}$.


 For Item 2, we have that $r_i\in 2\mathbb{Z}_8$ for all $i \in \{2,\dots, t_1+1\}$ and $r_i\in \{0,1,2,3\}$ for all $i\in \{t_1+2,\dots, t_1+t_2+1\}$.  Note that, since $x_1$ and $u_1$ are of order at most two,  $x_1\neq {\mathbf{0}}$ if and only if there exists at least one $i$ for $i\in \{t_1+2,\dots, t_1+t_2+1\}$ such that $r_i\in \{1,3\}$, or equivalently, if and only if $u_1\neq {\mathbf{0}}$.


For Item 2, we have that $r_i\in 4\mathbb{Z}_8=\{0,4\}$ for all $i \in \{2,\dots, t_1+1\}$ and $r_i\in \{0,2\}$ for all $i\in \{t_1+2,\dots, t_1+t_2+1\}$. Therefore, since $x_1$ is of order at most two, $x_1={\mathbf{0}}$.
\end{proof}



\begin{lemma}\label{lemma:same_num1}
Let $t_1\geq 1$ be an integer. Let ${\cal{H}}^{t_1,0,1}$ be the $\mathbb{Z}_2\mathbb{Z}_4 \mathbb{Z}_8$-additive code of type $(\alpha_1,\alpha_2,$ $\alpha_3;$ $t_1,0,1)$ generated by the matrix $A^{t_1,0,1}$.
Let $\mathbf{u}=(u_1\mid u_2\mid u_3) \in {\cal{G}}^{t_1,0,1}$. Then,

 \begin{enumerate}
 \item \label{proper-1}if $o(\mathbf{u})=8$, then $u_1$ contains every element of $\mathbb{Z}_2$ the same number of times, $u_2$ is a permutation of $(\mu, \stackrel{m}{\dots}, \mu, 2\mathbb{Z}_{4}, \stackrel{n}{\dots}, 2\mathbb{Z}_{4}, \mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4},\stackrel{r}{\dots},\mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4})$ for some integers $m,n,r\geq 0$ and $\mu \in \{1,3\}$, and $u_3$ is a permutation of $(\mu', \stackrel{m'}{\dots}, \mu', \mu'', \stackrel{m'}{\dots}, \mu'', 2\mathbb{Z}_{8}, \stackrel{n'}{\dots}, 2\mathbb{Z}_{8}, \mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8},\stackrel{r'}{\dots},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8})$ for some integers  $m', n',r'\geq 0$ and $\mu, \mu'\in \{1,3,5,7\}$.
 
  \item \label{proper-2} if $o(\mathbf{u})=4$, then $u_1={\mathbf{0}}$,
         $u_2$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ exactly $\frac{1}{2}(\frac{\alpha_1}{2}+\alpha_2)=4^{t_1-1}$ times and  $\frac{\alpha_2}{2}-\frac{\alpha_1}{4}=4^{t_1-1} -2^{t_1-1}$ times the element  $0$,
   and  $u_3$ is a permutation of $(\mu,\stackrel{m}{\dots}, \mu,4\mathbb{Z}_8,\stackrel{n}{\dots},4\mathbb{Z}_8,2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8, \stackrel{r}{\dots}, 2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8)$ for some integers  $m, n,r\geq 0$ and $\mu \in \{2,6\}$.

  \item \label{proper-3} if $o(\mathbf{u})=2$, then $u_1={\mathbf{0}}$, $u_2={\mathbf{0}}$, and $u_3$ contains the element in $4\mathbb{Z}_8{\setminus}\{0\}=\{4\}$ exactly $\frac{1}{4}(\frac{\alpha_1}{2}+\alpha_2+2\alpha_3)= 8^{t_1-1} $ times and  $\frac{\alpha_3}{2}-\frac{1}{4}(\frac{\alpha_1}{2}+\alpha_2)= 8^{t_1-1} -4^{t_1-1}$ times the element  $0$.
   \end{enumerate}
\end{lemma}
\begin{proof}
We prove this lemma by induction on $t_1\geq 1$. If $t_1=1$,  then by Lemma \ref{relation:t1a1}, $\alpha_1=2$, $\alpha_2=1$, $\alpha_3=1$, and ${\cal{G}}^{1,0,1}= \langle (0,1\mid 1\mid 1)\rangle$. Let $\mathbf{u}=(u_1\mid u_2\mid u_3) \in {\cal{G}}^{1,0,1}$. Then, $\mathbf{u}=\lambda(0,1\mid 1\mid 1)$, where  $ \lambda \in \mathbb{Z}_8$. Thus, we have that $u_1=\lambda(0,1)$, $u_2=(\lambda)$, and $u_3=(\lambda)$. If $o(\mathbf{u})=8$, then $\lambda \in \mathbb{Z}_8{\setminus}2\mathbb{Z}_8$. Therefore, $\mathbf{u}$ satisfies property \ref{proper-1}. If $o(\mathbf{u})=4$, then $\lambda \in \{2,6\}$. In this case, $u_1=(0,0)$, $u_2=(2)$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ exactly $1=\frac{1}{2}(\frac{\alpha_1}{2}+\alpha_2)$ time and  $0=\frac{\alpha_2}{2}-\frac{\alpha_1}{4}$ times the element  $0$, and $u_3=(\lambda)$. Thus, $\mathbf{u}$ satisfies property \ref{proper-2}. If $o(\mathbf{u})=2$, then $\lambda=4$. In this case, $u_1=(0,0)$, $u_2=(0)$, and $u_3=(4)$ contains the element in $4\mathbb{Z}_8{\setminus}\{0\}=\{4\}$ exactly $1=\frac{1}{4}(\frac{\alpha_1}{2}+\alpha_2+2\alpha_3)$ time and  $0=\frac{\alpha_3}{2}-\frac{1}{4}(\frac{\alpha_1}{2}+\alpha_2)$ times the element  $0$. Thus, $\mathbf{u}$ satisfies property \ref{proper-3}. Therefore, the lemma is true for $t_1=1$.
 
 Assume that the lemma holds for the code ${\cal{H}}^{t_1,0,1}$ of type $(\alpha_1, \alpha_2, \alpha_3; t_1,0,1)$ with $t_1\geq 1$.
By Lemma \ref{relation:t1a1}, we have that
 \begin{align}\label{eq:t1_t2_1}
      2^{t_1}=\alpha_1,
    4^{t_1}=\alpha_1+2\alpha_2,~\mbox{and}~
    8^{t_1}=\alpha_1+2\alpha_2+4\alpha_3.
\end{align}
 Now, we have to show that the lemma is also true for the code ${\cal{H}}^{t_1+1,0,1}$.





 
 Let $\mathbf{v}=(v_1\mid v_2 \mid v_3)\in {\cal{G}}^{t_1+1,0,1}$. We can write 
 \begin{align*}
     \mathbf{v}=\mathbf{z}+\lambda \mathbf{w},
 \end{align*}
 where  $\mathbf{z}=(u_1,u_1 \mid x_1,u_2, u_2, u_2, u_2\mid x_2,u_3,\stackrel{8}{\dots},u_3)$, $\mathbf{w}=({\mathbf{0}}, {\mathbf{1}} \mid {\mathbf{1}},{\mathbf{0}}, {\mathbf{1}}, {\mathbf{2}},  \mathbf{3}\mid {\mathbf{1}}, {\mathbf{0}}, \dots, \mathbf{7})$, $\mathbf{u}=(u_1\mid u_2\mid u_3) \in {\cal{G}}^{t_1,0,1}$, $\lambda \in \mathbb{Z}_8$,   $x_1\in (2\mathbb{Z}_4)^{2^{t_1}}$ such that either $x_1={\mathbf{0}}$ or $x_1$ is a permutation of $(2\mathbb{Z}_4, \stackrel{2^{t_1-1}}{\dots}, 2\mathbb{Z}_4)$, and  $x_2\in (2\mathbb{Z}_8)^{4^{t_1}}$ such that either $x_2={\mathbf{0}}$ or $x_2$ is a permutation of $(2\mathbb{Z}_8, \stackrel{4^{t_1-1}}{\dots}, 2\mathbb{Z}_8)$ or $(4\mathbb{Z}_8, \stackrel{2\cdot 4^{t_1-1}}{\dots}, 4\mathbb{Z}_8)$. Then, $v_1=(u_1, u_1)+\lambda({\mathbf{0}}, {\mathbf{1}})$ and,  for $i \in \{2,3\}$, 
\begin{align}\label{eq:vi}
     v_i=(x_{i-1}, u_i,\stackrel{2^i}{\dots}, u_i)+\lambda ({\mathbf{1}}, {\mathbf{0}}, \dots, \mathbf{2^i-1}).
\end{align}
If $\mathbf{z}={\mathbf{0}}$, then $\mathbf{v}=\lambda \mathbf{w}$ and it is easy to see that $\mathbf{v}$ satisfies property \ref{proper-1} if $\lambda\in \mathbb{Z}_8{\setminus} 2\mathbb{Z}_8=\{1,3,5,7\}$, property \ref{proper-2} if $\lambda\in\{2,6\}$, and property \ref{proper-3} if $\lambda=4$. 
Therefore, we focus on the case when $\mathbf{z}\neq {\mathbf{0}}$. 

Case 1: Assume that $o(\mathbf{v})=8$. We have two subcases: when $o(\mathbf{z})$  is arbitrary and $\lambda \in \mathbb{Z}_8{\setminus}2\mathbb{Z}_8$, and  when  $o(\mathbf{z})=8$ and $\lambda \in 2\mathbb{Z}_8$. In both subcases, note that $v_1$ contains every element of $\mathbb{Z}_2$ the same number of times.  For the first subcase, we have that  $(u_i,\stackrel{2^i}{\dots}, u_i)+\lambda ({\mathbf{0}}, \dots, \mathbf{2^i-1})$, for $i \in \{2,3\}$, is a permutation of $(\mathbb{Z}_{2^i},\stackrel{\alpha_i}{\dots},\mathbb{Z}_{2^i})$ by Lemma \ref{lemm:uu+lambda}. 
Thus, from (\ref{eq:vi}), $v_i$ is  a permutation of $(x_{i-1}+\lambda {\mathbf{1}},\mathbb{Z}_{2^i},\stackrel{\alpha_i}{\dots}, \mathbb{Z}_{2^i})$. Since either $x_{i-1}+\lambda {\mathbf{1}}=\lambda {\mathbf{1}}$, or $x_{i-1}+\lambda {\mathbf{1}}$ is a permutation of $(\mathbb{Z}_{2^i}{\setminus}2\mathbb{Z}_{2^i},  \stackrel{2^{(i-1)(t_1-1)}}{\dots}, \mathbb{Z}_{2^i}{\setminus}2\mathbb{Z}_{2^i})$, $\mathbf{v}$ satisfies property \ref{proper-1}.

For the second subcase when $o(\mathbf{v})=8$, that is, when $o(\mathbf{z})=8$ and $\lambda\in2\mathbb{Z}_8$, we have that $o(\mathbf{u})=8$ and, by Item 1 of Lemma \ref{lem:z_t1p1_0_1}, $x_{i-1}$ is a permutation  of $(2\mathbb{Z}_{2^{i}},\stackrel{2^{(i-1)(t_1-1)}}{\dots}, 2\mathbb{Z}_{2^{i}})$ for $i \in \{2,3\}$. 
By induction hypothesis, $\mathbf{u}$ satisfies property \ref{proper-1} and then $u_2$ is a permutation of 
\begin{equation*}\label{eq:v2}
(\mu, \stackrel{m}{\dots}, \mu, 2\mathbb{Z}_{4}, \stackrel{n}{\dots}, 2\mathbb{Z}_{4}, \mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4},\stackrel{r}{\dots},\mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4}),
\end{equation*} where  $m,n,r\geq 0$ and $\mu \in \{1,3\}$, and $u_3$ is a permutation of 
\begin{equation*}\label{eq:v3}
(\mu', \stackrel{m'}{\dots}, \mu', \mu'', \stackrel{m'}{\dots}, \mu'', 2\mathbb{Z}_{8}, \stackrel{n'}{\dots}, 2\mathbb{Z}_{8}, \mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8},\stackrel{r'}{\dots},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8}),
\end{equation*}
where  $m', n',r'\geq 0$ and $\mu', \mu''\in \{1,3,5,7\}$. From (\ref{eq:vi}), $v_2=(x_1, u_2,u_2,u_2,u_2)+\lambda({\mathbf{1}},{\mathbf{0}},{\mathbf{1}},{\mathbf{2}},\mathbf{3}).$
If $\lambda \in \{ 0,4\}$, then $v_2=(x_1, u_2,u_2,u_2,u_2)$  in $\mathbf{v}$ satisfies the same property as $u_2$ in $\mathbf{u}$; that is, property \ref{proper-1}. If $\lambda\in\{2,6\}$, then $v_2=(x_1,u_2,u_2,u_2,u_2)+({\mathbf{2}},{\mathbf{0}}, {\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}})$.
By Item \ref{2z22} of Lemma \ref{Lemma:cases}, we have that $x_1+{\mathbf{2}}$ is a permutation of $(2\mathbb{Z}_{4}, \stackrel{2^{t_1-1}}{\dots},2\mathbb{Z}_{4})$.
Thus, by Lemma \ref{lemm:ord8_u2}, $v_2$ is a permutation of 
$$
(2\mathbb{Z}_{4}, \stackrel{4n+2^{t_1-1}}{\dots},2\mathbb{Z}_{4}, \mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4},\stackrel{4r+2m}{\dots},\mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4}).
$$
Therefore, for $\lambda\in2\mathbb{Z}_8$, $v_2$  satisfies property \ref{proper-1}. Now, we consider the coordinates in $v_3$. From (\ref{eq:vi}), $v_3=(x_2, u_3,\stackrel{8}{\dots},u_3)+\lambda({\mathbf{1}}, {\mathbf{0}}, \dots, \mathbf{7}).$ 
By Item \ref{2z22} of Lemma \ref{Lemma:cases}, we have that, for $\lambda\in 2\mathbb{Z}_8$, $x_2+\lambda {\mathbf{1}}$ is a permutation of $(2\mathbb{Z}_{8}, \stackrel{4^{t_1-1}}{\dots},  2\mathbb{Z}_{8})$. If $\lambda=0$, it is easy to see that $v_3$ satisfies property \ref{proper-1}.
Note that  $\lambda({\mathbf{0}}, \dots, \mathbf{7})$ is a permutation of $({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6},{\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$ if $\lambda \in \{2,6\}$, and a permutation of $({\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}})$ if $\lambda=4$. Thus, by Lemma \ref{lemm:ord8_u3}, $v_3$ satisfies property \ref{proper-1}.  Therefore, if $o(\mathbf{v})=8$, then $\mathbf{v}$ satisfies property \ref{proper-1}.


Case 2: Assume that $o(\mathbf{v})=4$. We have two subcases: when $o(\mathbf{z})=4$ and $\lambda \in 2\mathbb{Z}_8$, and when $o(\mathbf{z})=2$ and $\lambda \in \{2,6\}$. 
For the first subcase, since $o(\mathbf{z})=4$, we have that $o(\mathbf{u})=4$. Moreover, $x_1={\mathbf{0}}$ and $x_2$ is a permutation of $(4\mathbb{Z}_8, \stackrel{2\cdot 4^{t_1-1}}{\dots}, 4\mathbb{Z}_8)$ by Item 2 of Lemma \ref{lem:z_t1p1_0_1}.
By induction hypothesis, $\mathbf{u}$ satisfies property \ref{proper-2}. Then, $u_1={\mathbf{0}}$, $u_2$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ exactly $4^{t_1-1}$ times and $4^{t_1-1}-2^{t_1-1}$ times the element $0$,
   and  $u_3$ is a permutation of 
   \begin{equation*}\label{eq:4v2}
      (\mu,\stackrel{m}{\dots}, \mu,4\mathbb{Z}_8,\stackrel{n}{\dots},4\mathbb{Z}_8,2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8, \stackrel{r}{\dots}, 2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8) 
   \end{equation*}
     for some integers  $m, n,r\geq 0$ and $\mu \in \{2,6\}$.  Since $v_1=(u_1, u_1)+\lambda({\mathbf{0}}, {\mathbf{1}})$, $u_1={\mathbf{0}}$, and $\lambda\in 2\mathbb{Z}_8$, we have that $v_1={\mathbf{0}}$.  From (\ref{eq:vi}), $v_2=(x_1, u_2,u_2,u_2,u_2)+\lambda({\mathbf{1}}, {\mathbf{0}}, {\mathbf{1}}, {\mathbf{2}}, \mathbf{3}).$
    If $\lambda \in \{0,4\}$, then $v_2=(x_1, u_2,u_2,u_2,u_2)$. Since $x_1={\mathbf{0}}$ is of length $2^{t_1}$, it is easy to see that $v_2$ in $\mathbf{v}$ satisfies the same property as $u_2$ in $\mathbf{u}$; that is, property \ref{proper-2}. If $\lambda \in \{2,6\}$, then 
    $v_2=(x_1, u_2, u_2,u_2,u_2)+({\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}}),$ where $x_1={\mathbf{0}}$ is of length $2^{t_1}$.
    Note that $u_2+{\mathbf{2}}$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ as many times as $u_2$ contains the element $0$, and the element $0$ as many times as $u_2$ contains the element $2$. Thus, $v_2$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ exactly 
    $2^{t_1} + 2 (4^{t_1-1}) + 2( 4^{t_1-1}-2^{t_1-1})=4^{t_1}$ times
   
   
    and  
    $2 (4^{t_1-1}) + 2( 4^{t_1-1}-2^{t_1-1})=4^{t_1} -2^{t_1}$ times
   
    the element $0$. Therefore, for $\lambda \in 2\mathbb{Z}_8$, $v_2$  satisfies property \ref{proper-2}. Now, we consider the coordinates in $v_3$. From (\ref{eq:vi}), $v_3=(x_2, u_3,\stackrel{8}{\dots},u_3)+\lambda({\mathbf{1}}, {\mathbf{0}}, \dots, \mathbf{7})$.
    If $\lambda=0$, it is easy to see that $v_3$ satisfies property \ref{proper-2}.
    For $\lambda=4$, $x_2+\lambda {\mathbf{1}}$ is a permutation of  $(4\mathbb{Z}_8, \stackrel{2\cdot 4^{t_1-1}}{\dots}, 4\mathbb{Z}_8)$, and 
    for $\lambda \in \{2,6\}$, it is a permutation of  
    $$
    (2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8, \stackrel{2\cdot 4^{t_1-1}}{\dots}, 2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8).
    $$
    Note that  $\lambda({\mathbf{0}}, \dots, \mathbf{7})$ is a permutation of $({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6},{\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$ if $\lambda \in \{2,6\}$, and a permutation of $({\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}})$ if $\lambda=4$.
    Hence, by Lemma \ref{lemm:ord4_u3}, $v_3$ also satisfies property \ref{proper-2}, and so does $\mathbf{v}$.
    
Now, we consider the second subcase, that is, when $o(\mathbf{z})=2$  and $\lambda \in \{2,6\}$. Since $o(\mathbf{z})=2$, we have that $o(\mathbf{u})=2$. Then, by Item 3 of Lemma \ref{lem:z_t1p1_0_1}, $x_1={\mathbf{0}}$ and $x_2={\mathbf{0}}$.  
By induction hypothesis, $\mathbf{u}$ satisfies property \ref{proper-3}, so
 $u_1={\mathbf{0}}$,  $u_2={\mathbf{0}}$, and  $u_3$ contains the element in $4\mathbb{Z}_8{\setminus}\{0\}=\{4\}$ exactly $m=8^{t_1-1}$ times and   $m'=8^{t_1-1}-4^{t_1-1} $ times the element $0$. Since $v_1=(u_1, u_1)+\lambda({\mathbf{0}}, {\mathbf{1}})$, $u_1={\mathbf{0}}$, and  $\lambda \in \{2,6\}$, we have that $v_1={\mathbf{0}}$. From (\ref{eq:vi}), $v_2=(x_1, u_2,u_2,u_2,u_2)+({\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}})$. Since $x_1={\mathbf{0}}$ and $u_2={\mathbf{0}}$, of length $\alpha_1$ and $\alpha_2$, respectively, we have that $v_2=({\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}}).$ Therefore, $v_2$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ exactly $\alpha_1+2\alpha_2=4^{t_1}$ times and  $2\alpha_2= 4^{t_1}-2^{t_1}$ times  the element $0$, by (\ref{eq:t1_t2_1}). Therefore, $v_2$  satisfies property \ref{proper-2}. Now, we consider the coordinates in $v_3$. From (\ref{eq:vi}), $v_3=(x_2, u_3,\stackrel{8}{\dots},u_3)+\lambda({\mathbf{1}}, {\mathbf{0}}, \dots, \mathbf{7}).$ Since $x_2={\mathbf{0}}$,  $x_2+\lambda {\mathbf{1}}=(\lambda,\stackrel{4^{t_1}}{\dots},\lambda)$. Note that $u_3$ is a permutation of 
 $$
 (4,\stackrel{m-m'}{\dots}, 4, 4\mathbb{Z}_8,\stackrel{m'}{\dots}, 4\mathbb{Z}_8).
 $$ 
Moreover, since $\lambda \in \{2,6\}$, $\lambda({\mathbf{0}},\dots, \mathbf{7})$ is a permutation of $({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6}, {\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$. Thus, by Item \ref{2z22} of  Lemma \ref{Lemma:cases}, $(u_3,\stackrel{8}{\dots},u_3)+({\mathbf{0}},{\mathbf{2}}, {\mathbf{4}}, \mathbf{6},{\mathbf{0}},{\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$ is a permutation of $$
 (2\mathbb{Z}_8,\stackrel{2(m-m')+4m'}{\dots}, 2\mathbb{Z}_8).
 $$
 Thus, $v_3$ is a permutation of $(\lambda,\stackrel{4^{t_1}}{\dots},\lambda, 2\mathbb{Z}_8,\stackrel{2(m-m')+4m'}{\dots}, 2\mathbb{Z}_8)$ with $\lambda \in \{2,6\}$, and hence $v_3$ also  satisfies property \ref{proper-2} and so does $\mathbf{v}$. Therefore, if $o(\mathbf{v})=4$, then $\mathbf{v}$ satisfies property \ref{proper-2}.
 
 
Case 3: Assume that $o(\mathbf{v})=2$. Then,  $o(\mathbf{z})=2$ and $\lambda \in \{0,4\}$.
Since $o(\mathbf{z})=2$, then $o(\mathbf{u})=2$.  Moreover, $x_1= {\mathbf{0}}$ and $x_2={\mathbf{0}}$ by Item 3 of Lemma \ref{lem:z_t1p1_0_1}. By induction hypothesis, $\mathbf{u}$ satisfies property \ref{proper-3}, and then $u_1={\mathbf{0}}$, $u_2={\mathbf{0}}$, and $u_3$ contains the element in $4\mathbb{Z}_8{\setminus}\{0\}=\{4\}$ exactly $8^{t_1-1}$ times and  $8^{t_1-1}-4^{t_1-1} $ times the element  $0$. Since $v_1=(u_1, u_1)+\lambda({\mathbf{0}}, {\mathbf{1}})$, $v_1={\mathbf{0}}$. From (\ref{eq:vi}), $v_2=(x_1, u_2,u_2,u_2,u_2)+\lambda({\mathbf{1}}, {\mathbf{0}}, {\mathbf{1}}, {\mathbf{2}}, \mathbf{3})$, where $x_1={\mathbf{0}}$ and $u_2={\mathbf{0}}$, so $v_2={\mathbf{0}}$.
  From (\ref{eq:vi}), $v_3=(x_2, u_3,\stackrel{8}{\dots},u_3)+\lambda({\mathbf{1}},{\mathbf{0}}, \dots, \mathbf{7})$, where $x_2={\mathbf{0}}$ is of length $4^{t_1}$. If $\lambda=0$, it is easy to see that $v_3$ satisfies property \ref{proper-3}. If $\lambda=4$, $v_3=(x_2, u_3,\stackrel{8}{\dots},u_3)+({\mathbf{4}},{\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}},{\mathbf{0}}, {\mathbf{4}},{\mathbf{0}}, {\mathbf{4}})$. Note that $u_3+{\mathbf{4}}$ contains the element in $4\mathbb{Z}_8{\setminus}\{0\}=\{4\}$ as many times as $u_3$ contains the element $0$, and the element $0$ as many times as $u_3$ contains the element $4$.

Then, $v_3$ contains the element $4$ exactly 
$4^{t_1}+4( 8^{t_1-1}) + 4(8^{t_1-1} - 4^{t_1-1})=8^{t_1} $ times and 
$4( 8^{t_1-1}) + 4(8^{t_1-1} - 4^{t_1-1})=8^{t_1}-4^{t_1} $
the element 0. Therefore, $\mathbf{v}$ satisfies property \ref{proper-3}. This completes the proof.
\end{proof}






 





\begin{lemma}\label{lemma:same_num2}
Let $t_1\geq 1$ and $t_2\geq 0$ be integers. Let ${\cal{H}}^{t_1,t_2,1}$ be the $\mathbb{Z}_2\mathbb{Z}_4 \mathbb{Z}_8$-additive code of type $(\alpha_1,\alpha_2,\alpha_3; t_1,t_2,1)$ generated by the matrix $A^{t_1,t_2,1}$. 
Let $\mathbf{u}=(u_1\mid u_2\mid u_3) \in {\cal{G}}^{t_1,t_2,1}$.

 \begin{enumerate}
 \item \label{proper_1} If $o(\mathbf{u})=8$, then $\mathbf{u}$ has the following property:
 \begin{enumerate}
     \item \label{proper_1a}$u_1$ contains every element of $\mathbb{Z}_2$ the same number of times, $u_2$ is a permutation of $(\mu, \stackrel{m}{\dots}, \mu, 2\mathbb{Z}_{4}, \stackrel{n}{\dots}, 2\mathbb{Z}_{4}, \mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4},\stackrel{r}{\dots},\mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4})$ for some integers  $m,n,r\geq 0$ and $\mu \in \{1,3\}$, and $u_3$ is a permutation of $(\mu', \stackrel{m'}{\dots}, \mu', \mu'', \stackrel{m'}{\dots}, \mu'', 2\mathbb{Z}_{8}, \stackrel{n'}{\dots}, 2\mathbb{Z}_{8}, \mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8},\stackrel{r'}{\dots},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8})$ for some integers  $m', n',r'\geq 0$ and $\mu, \mu'\in \{1,3,5,7\}$.
 \end{enumerate}
  \item \label{proper_2} If $o(\mathbf{u})=4$, then $\mathbf{u}$ has one of the following properties:
  \begin{enumerate}
      \item \label{proper_2a}$u_1={\mathbf{0}}$, $u_2$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ exactly $\frac{1}{2}(\frac{\alpha_1}{2}+\alpha_2)=4^{t_1+t_2-1}$ times and  $\frac{\alpha_2}{2}-\frac{\alpha_1}{4}=4^{t_1+t_2-1}-2^{t_1+t_2-1}$ times the element $0$,
   and  $u_3$ is a permutation of $(\mu,\stackrel{m}{\dots}, \mu,4\mathbb{Z}_8,\stackrel{n}{\dots},4\mathbb{Z}_8,2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8, \stackrel{r}{\dots}, 2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8)$ for some integers  $m, n,r\geq 0$ and $\mu \in \{2,6\}$.
    \item \label{proper_2b} $u_1$ contains every element of $\mathbb{Z}_2$ the same number of times,  $u_2$ is a permutation of  $(\mu,\stackrel{m}{\dots},\mu,2\mathbb{Z}_4,\stackrel{n}{\dots},2\mathbb{Z}_4,\mathbb{Z}_4{\setminus}2\mathbb{Z}_4, \stackrel{r}{\dots}, \mathbb{Z}_4{\setminus}2\mathbb{Z}_4)$ for some integers  $m, n,r\geq 0$ and $\mu \in \{1,3\}$, 
   
    and
    $u_3$ is a permutation of $(4\mathbb{Z}_{8}, \stackrel{t}{\dots}, 4\mathbb{Z}_{8}, 2\mathbb{Z}_{8}{\setminus}4\mathbb{Z}_{8},\stackrel{t'}{\dots},2\mathbb{Z}_{8}{\setminus}4\mathbb{Z}_{8})$ for some integers $t, t'\geq 0$.
    
  \end{enumerate}
  \item \label{proper_3} If $o(\mathbf{u})=2$, then $\mathbf{u}$ has one of the following properties:
  \begin{enumerate}
   \item \label{proper_3a}$u_1={\mathbf{0}}$, $u_2={\mathbf{0}}$, and $u_3$ contains the element in $4\mathbb{Z}_8{\setminus}\{0\}=\{4\}$ exactly $\frac{1}{4}(\frac{\alpha_1}{2}+\alpha_2+2\alpha_3)=8^{t_1-1}4^{t_2}$ times and  $\frac{\alpha_3}{2}-\frac{1}{4}(\frac{\alpha_1}{2}+\alpha_2)=8^{t_1-1}4^{t_2}-4^{t_1+t_2-1}$ times the element  $0$.
  \item \label{proper_3b}$u_1={\mathbf{0}}$,  $u_2$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ exactly $\frac{1}{2}(\frac{\alpha_1}{2}+\alpha_2)=4^{t_1+t_2-1}$ times and  $\frac{\alpha_2}{2}-\frac{\alpha_1}{4}=4^{t_1+t_2-1}-2^{t_1+t_2-1}$ times the element  $0$, and $u_3$ is a permutation of $(4\mathbb{Z}_8,\stackrel{m}{\dots},4\mathbb{Z}_8)$ for some $m\geq 0$.
  \end{enumerate}
   \end{enumerate}
\end{lemma}
\begin{proof}
We prove this lemma by induction on $t_2\geq 0$. The lemma is true for the code  ${\cal{H}}^{t_1,0,1}$ by Lemma \ref{lemma:same_num1}.
Assume that the lemma holds for the code ${\cal{H}}^{t_1,t_2,1}$ of type $(\alpha_1, \alpha_2, \alpha_3; t_1,t_2,1)$ with $t_1\geq 1$ and $t_2\geq 0$.
By Lemma \ref{relation:t1a1}, we have that
 \begin{align}\label{eq:t1t2}
      2^{t_1+t_2}=\alpha_1, 
    4^{t_1+t_2}=\alpha_1+2\alpha_2,~\mbox{and}~ 
    8^{t_1}4^{t_2}=\alpha_1+2\alpha_2+4\alpha_3.
\end{align}
 Now, we have to show that the lemma is also true for the code ${\cal{H}}^{t_1,t_2+1,1}$. 





 
  Let $\mathbf{v}=(v_1\mid v_2 \mid v_3)\in {\cal{G}}^{t_1,t_2+1,1}$. We can write 
 \begin{align*}
     \mathbf{v}=\mathbf{z}+\lambda \mathbf{w},
 \end{align*}
 where  $\mathbf{z}=(u_1,u_1 \mid x_1,u_2, u_2, u_2, u_2\mid u_3,u_3, u_3, u_3)$, $\mathbf{w}=({\mathbf{0}}, {\mathbf{1}} \mid {\mathbf{1}}, {\mathbf{0}}, {\mathbf{1}}, {\mathbf{2}}, \mathbf{3}\mid  {\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$, $\mathbf{u}=(u_1\mid u_2\mid u_3) \in {\cal{G}}^{t_1,t_2,1}$, $\lambda \in \{0,1,2,3\}$, and $x_1\in (2\mathbb{Z}_4)^{2^{t_1+t_2}}$ such that either $x_1={\mathbf{0}}$ or  a permutation  of $(2\mathbb{Z}_4,\stackrel{2^{t_1+t_2-1}}{\dots}, 2\mathbb{Z}_4)$. Then,  
\begin{equation}\label{eq:vi_}
\begin{split}
   v_1&=(u_1, u_1)+\lambda({\mathbf{0}}, {\mathbf{1}}),\\
     v_2&=(x_1,u_2,u_2,u_2,u_2)+\lambda ({\mathbf{1}}, {\mathbf{0}}, {\mathbf{1}}, {\mathbf{2}}, \mathbf{3}),\\
     v_3&=(u_3,u_3,u_3,u_3)+\lambda({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6}). 
\end{split}
\end{equation}
If $\mathbf{z}={\mathbf{0}}$, then $\mathbf{v}=\lambda \mathbf{w}$. It is easy to see that $\mathbf{v}$ satisfies property \ref{proper_2b} if $\lambda\in \{1,3\}$ and  property \ref{proper_3b} if $\lambda=2$.
Therefore, we focus on the case  when $\mathbf{z}\neq {\mathbf{0}}$. 

Case 1: Assume that $o(\mathbf{v})=8$. Then, $o(\mathbf{z})=8$ and $\lambda \in \{0,1,2,3\}$. We have that $o(\mathbf{u})=8$ and,  by Item 1 of Lemma \ref{lem:z_t1t2p1_1}, $x_1$ is a permutation of $(2\mathbb{Z}_4,\stackrel{2^{t_1+t_2-1}}{\dots}, 2\mathbb{Z}_4)$. By induction hypothesis, $\mathbf{u}$ satisfies property \ref{proper_1a}. Then, $u_1$ contains every element of $\mathbb{Z}_2$ the same number of times, $u_2$ is a permutation of 
\begin{equation*}\label{eq:v2_}
(\mu, \stackrel{m}{\dots}, \mu, 2\mathbb{Z}_{4}, \stackrel{n}{\dots}, 2\mathbb{Z}_{4}, \mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4},\stackrel{r}{\dots},\mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4}),
\end{equation*} where  $m,n,r\geq 0$ and $\mu \in \{1,3\}$, and $u_3$ is a permutation of 
\begin{equation*}\label{eq:v3_}
(\mu', \stackrel{m'}{\dots}, \mu', \mu'', \stackrel{m'}{\dots}, \mu'', 2\mathbb{Z}_{8}, \stackrel{n'}{\dots}, 2\mathbb{Z}_{8}, \mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8},\stackrel{r'}{\dots},\mathbb{Z}_{8}{\setminus}2\mathbb{Z}_{8}),
\end{equation*}
where  $m', n',r'\geq 0$ and $\mu', \mu''\in \{1,3,5,7\}$. First, since $ v_1=(u_1, u_1)+\lambda({\mathbf{0}}, {\mathbf{1}})$, $v_1$ contains every element of $\mathbb{Z}_2$ the same number of times, for any $\lambda \in \{0,1,2,3\}$. Second, from (\ref{eq:vi_}), $v_2=(x_1, u_2,u_2,u_2,u_2)+\lambda({\mathbf{1}},{\mathbf{0}},{\mathbf{1}},{\mathbf{2}},\mathbf{3}).$ If $\lambda=0$, then $v_2$ clearly satisfies \ref{proper_1a}. 
If $\lambda \in \{1,3\}$, then  we have that $(u_2,u_2,u_2,u_2)+\lambda({\mathbf{0}},{\mathbf{1}},{\mathbf{2}},\mathbf{3})$ is a permutation of $(\mathbb{Z}_4,\stackrel{\alpha_2}{\dots}, \mathbb{Z}_4)$ by Lemma \ref{lemm:uu+lambda}. For $\lambda \in \{1,3\}$, since $x_1+\lambda {\mathbf{1}}$ is a permutation of $(\mathbb{Z}_4{\setminus}2\mathbb{Z}_4, \stackrel{2^{t_1+t_2-1}}{\dots},\mathbb{Z}_4{\setminus}2\mathbb{Z}_4)$ by Item  \ref{2z2_} of Lemma \ref{Lemma:cases}, we have that
$v_2$  satisfies property \ref{proper_1a}. 
If $\lambda=2$, $v_2=(x_1, u_2,u_2,u_2,u_2)+({\mathbf{2}},{\mathbf{0}},{\mathbf{2}},{\mathbf{0}},{\mathbf{2}}).$ By Item  \ref{2z22} of Lemma \ref{Lemma:cases}, we have that $x_1+{\mathbf{2}}$ is a permutation of $(2\mathbb{Z}_{4}, \stackrel{2^{t_1+t_2-1}}{\dots},2\mathbb{Z}_{4})$. Therefore, by Lemma \ref{lemm:ord8_u2}, $v_2$ is a permutation of $(2\mathbb{Z}_{4}, \stackrel{4n+2^{t_1+t_2-1}}{\dots}, 2\mathbb{Z}_{4}, \mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4},\stackrel{4r+2m}{\dots},\mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4})$ and then $v_2$  satisfies property \ref{proper_1a}. Finally, we consider the coordinates in $v_3$. From (\ref{eq:vi_}), $v_3=(u_3,u_3, u_3, u_3)+\lambda({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6}).$ If $\lambda=0$, then $v_3$ clearly satisfies \ref{proper_1a}. Note that $\lambda({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})=({\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}})$ if $\lambda=2$ and $\lambda({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$ is a permutation of $({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$ if $\lambda \in \{1,3\}$. Therefore, by Lemma \ref{lemm:ord8_u3}, $v_3$  satisfies property  \ref{proper_1a}, and so does $\mathbf{v}$.

Case 2: Assume that $o(\mathbf{v})=4$. We have two subcases: when $o(\mathbf{z})=4$ and $\lambda \in \{0,1,2,3\}$, and when $o(\mathbf{z})=2$ and $\lambda \in \{1,3\}$. For the first subcase, since $o(\mathbf{z})=4$, $o(\mathbf{u})=4$. By induction hypothesis, $\mathbf{u}$ satisfies property \ref{proper_2a} or \ref{proper_2b}. Assume that $\mathbf{u}$ satisfies property \ref{proper_2a}. Then, $u_1={\mathbf{0}}$, $u_2$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ exactly $ 4^{t_1+t_2-1}$ times and $ 4^{t_1+t_2-1} - 2^{t_1+t_2-1}$ times the element $0$,
   and  $u_3$ is a permutation of 
   \begin{equation*}\label{eq:4v2_}
      (\mu,\stackrel{m}{\dots}, \mu,4\mathbb{Z}_8,\stackrel{n}{\dots},4\mathbb{Z}_8,2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8, \stackrel{r}{\dots}, 2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8) 
   \end{equation*}
     for some integers  $m, n,r\geq 0$ and $\mu \in \{2,6\}$. Note that, in this case,  $x_1={\mathbf{0}}$ by Item 2 of Lemma \ref{lem:z_t1t2p1_1}. 
    
     If $\lambda=0$, then it is easy to see that $\mathbf{v}$ satisfies property \ref{proper_2a}. If $\lambda=2$, we show that $\mathbf{v}$ satisfies property \ref{proper_2a}.  Since $v_1=(u_1, u_1)+\lambda({\mathbf{0}}, {\mathbf{1}})$, $u_1={\mathbf{0}}$, and $\lambda=2$, we have that $v_1={\mathbf{0}}$.
     From (\ref{eq:vi_}), $v_2=(x_1, u_2,u_2,u_2,u_2)+({\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}})$, where $x_1={\mathbf{0}}$ is of length $2^{t_1+t_2}$.
     Note that $u_2+{\mathbf{2}}$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ as many times as $u_2$ contains the element $0$, and the element $0$ as many times as $u_2$ contains the element $2$. Thus, $v_2$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ exactly 
     $2^{t_1+t_2} + 2(4^{t_1+t_2-1}) + 2(4^{t_1+t_2-1} - 2^{t_1+t_2-1} ) =4^{t_1+t_2}$ times
    
    
     and  $2(4^{t_1+t_2-1}) + 2(4^{t_1+t_2-1} - 2^{t_1+t_2-1} ) =4^{t_1+t_2}-2^{t_1+t_2}$ times
    
    
     the element $0$, 
     so $v_2$ satisfies property \ref{proper_2a}.
     From (\ref{eq:vi_}), $v_3=( u_3,u_3,u_3,u_3)+( {\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}})$. By Item \ref{lemm:ord4_case2} of Lemma \ref{lemm:ord4_u3}, $v_3$ is a permutation of 
     $$
     (4\mathbb{Z}_8,\stackrel{4n}{\dots},4\mathbb{Z}_8,2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8, \stackrel{4r+2m}{\dots}, 2\mathbb{Z}_8{\setminus}4\mathbb{Z}_8).
     $$
     Therefore, for $\lambda =2$, $\mathbf{v}$ satisfies property \ref{proper_2a}. Finally, if $\lambda \in \{1,3\}$, we show that $\mathbf{v}$ satisfies property \ref{proper_2b}. 
     Since $v_1=(u_1, u_1)+\lambda({\mathbf{0}}, {\mathbf{1}})$, $u_1={\mathbf{0}}$, and $\lambda \in \{1,3\}$, we have that $v_1$ contains every element of $\mathbb{Z}_2$ the same number of times. 
     From (\ref{eq:vi_}), $v_2=(x_1, u_2,u_2,u_2,u_2)+\lambda({\mathbf{1}}, {\mathbf{0}}, {\mathbf{1}}, {\mathbf{2}}, \mathbf{3})$, where $x_1={\mathbf{0}}$ is of length $2^{t_1+t_2}$. Since $\lambda \in \{1,3\}$, by Lemma \ref{lemm:uu+lambda}, we have that $v_2$ is a permutation of $(\lambda, \stackrel{2^{t_1+t_2}}{\dots},\lambda,\mathbb{Z}_4,\stackrel{\alpha_2}{\dots}, \mathbb{Z}_4)$.
     From (\ref{eq:vi_}), $v_3=(u_3,u_3, u_3, u_3)+\lambda({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6}).$ Note that, for $\lambda \in \{1,3\}$,  $\lambda({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$ is a permutation of $({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$. Thus, by Item \ref{lemm:ord4_case1} of Lemma \ref{lemm:ord4_u3}, $v_3$  satisfies property  \ref{proper_2b}, and so does $\mathbf{v}$. Therefore, if $o(\mathbf{u})=4$ and $\mathbf{u}$ satisfies property 
     \ref{proper_2a}, we have that $\mathbf{v}$ satisfies either property \ref{proper_2a} or \ref{proper_2b}. 
     
     We continue with the first subcase, when $o(\mathbf{z})=4$ and $\lambda \in \{0,1,2,3\}$. Again, we have that  $o(\mathbf{u})=4$. Now, we assume that $\mathbf{u}$ satisfies property 
     \ref{proper_2b}.
     Then, $u_1$ contains every element of $\mathbb{Z}_2$ the same number of times,  $u_2$ is a permutation of 
     $$
     (\mu,\stackrel{m}{\dots},\mu,2\mathbb{Z}_4,\stackrel{n}{\dots},2\mathbb{Z}_4,\mathbb{Z}_4{\setminus}2\mathbb{Z}_4, \stackrel{r}{\dots}, \mathbb{Z}_4{\setminus}2\mathbb{Z}_4)
     $$ 
    for some integers  $m, n,r\geq 0$ and $\mu \in \{1,3\}$,
    and $u_3$ is a permutation of 
    $(4\mathbb{Z}_{8}, \stackrel{t}{\dots}, 4\mathbb{Z}_{8}, 2\mathbb{Z}_{8}{\setminus}4\mathbb{Z}_{8},\stackrel{t'}{\dots},2\mathbb{Z}_{8}{\setminus}4\mathbb{Z}_{8})$
    for some integers $t, t'\geq 0$. 
   Note that, in this case,  $x_1$ is a permutation of $(2\mathbb{Z}_4, \stackrel{2^{t_1+t_2-1}}{\dots}, 2\mathbb{Z}_4)$ by Item 2 of Lemma \ref{lem:z_t1t2p1_1}.
  
    Now, we show that $\mathbf{v}$ satisfies property \ref{proper_2b}.  
    Since $v_1=(u_1, u_1)+\lambda({\mathbf{0}}, {\mathbf{1}})$ and $u_1$ contains every element of $\mathbb{Z}_2$ the same number of times, we have that $v_1$ contains every element of $\mathbb{Z}_2$ the same number of times, for any $\lambda \in \{0,1,2,3\}$. 
    From (\ref{eq:vi_}), $v_2=(x_1, u_2,u_2,u_2,u_2)+\lambda({\mathbf{1}}, {\mathbf{0}}, {\mathbf{1}}, {\mathbf{2}}, \mathbf{3})$. 
If $\lambda=0$, it is clear that $v_2$ satisfies property \ref{proper_2b}. Note that $x_1+\lambda {\mathbf{1}}$ is a permutation of $(2\mathbb{Z}_4, \stackrel{2^{t_1+t_2-1}}{\dots}, 2\mathbb{Z}_4)$ if $\lambda=2$, and  a permutation of $(\mathbb{Z}_4{\setminus}2\mathbb{Z}_4, \stackrel{2^{t_1+t_2-1}}{\dots}, \mathbb{Z}_4{\setminus}2\mathbb{Z}_4)$ if $\lambda \in \{1,3\}$.   If $\lambda=2$, then by Lemma \ref{lemm:ord8_u2}, $(u_2,u_2,u_2,u_2)+({\mathbf{0}}, {\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}})$ is a permutation of 
    $$
    (2\mathbb{Z}_{4}, \stackrel{4n}{\dots}, 2\mathbb{Z}_{4}, \mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4},\stackrel{4r+2m}{\dots},\mathbb{Z}_{4}{\setminus}2\mathbb{Z}_{4}).
    $$
    If $\lambda \in \{1,3\}$, then by Lemma \ref{lemm:uu+lambda}, $(u_2,u_2,u_2,u_2)+\lambda({\mathbf{0}}, {\mathbf{1}}, {\mathbf{2}}, \mathbf{3})$ is a permutation of $(\mathbb{Z}_4,\stackrel{\alpha_2}{\dots}, \mathbb{Z}_4)$.
    Therefore, $v_2$ satisfies property \ref{proper_2b}. From (\ref{eq:vi_}), $v_3=(u_3,u_3, u_3, u_3)+\lambda({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6}).$ If $\lambda=0$, it is clear that $v_3$ satisfies property \ref{proper_2b}. Note that   $\lambda({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$ is a permutation of $({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$ if $\lambda \in \{1,3\}$, and $\lambda({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})=({\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}})$ if $\lambda=2$. Therefore, by Lemma \ref{lemm:ord4_u3}, $v_3$  satisfies property \ref{proper_2b}, and so does $\mathbf{v}$.
    
    Now, we consider the second subcase when $o(\mathbf{v})=4$, that is, when $o(\mathbf{z})=2$ and $\lambda\in \{1,3\}$. Since $o(\mathbf{z})=2$, $o(\mathbf{u})=2$. By induction hypothesis, $\mathbf{u}$ satisfies property \ref{proper_3a} or \ref{proper_3b}. Assume that $\mathbf{u}$ satisfies property \ref{proper_3a}. Then, $u_1={\mathbf{0}}$,  $u_2={\mathbf{0}}$, and  $u_3$ contains the element in $4\mathbb{Z}_8{\setminus}\{0\}=\{4\}$ exactly
    $m=8^{t_1-1}4^{t_2}$
    times and   
   
    $m'=8^{t_1-1}4^{t_2}-4^{t_1+t_2-1}$
    times the element $0$.
     By Item 3 of Lemma \ref{lem:z_t1t2p1_1}, we have that  $x_1={\mathbf{0}}$. 
      Since $v_1=(u_1, u_1)+\lambda({\mathbf{0}}, {\mathbf{1}})$, $u_1={\mathbf{0}}$, and $\lambda \in \{1,3\}$, we have that $v_1$ contains every element of $\mathbb{Z}_2$ the same number of times.  From (\ref{eq:vi_}), $v_2=(x_1, u_2,u_2,u_2,u_2)+\lambda({\mathbf{1}}, {\mathbf{0}}, {\mathbf{1}}, {\mathbf{2}}, \mathbf{3})$, where $x_1={\mathbf{0}}$ is of length $2^{t_1+t_2}$.
    By Lemma \ref{lemm:uu+lambda}, we have that $v_2$ is a permutation of 
    $$
    (\lambda, \stackrel{2^{t_1+t_2}}{\dots},\lambda,\mathbb{Z}_4,\stackrel{\alpha_2}{\dots}, \mathbb{Z}_4),
    $$ where $\lambda \in \{1,3\}$.
    From (\ref{eq:vi_}), $v_3=(u_3,u_3, u_3, u_3)+\lambda({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6}).$ Note that $u_3$ is a permutation of $(4,\stackrel{m-m'}{\dots}, 4, 4\mathbb{Z}_8,\stackrel{m'}{\dots}, 4\mathbb{Z}_8)$ and, since $\lambda \in \{1,3\}$,  $\lambda({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$ is a permutation of $({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$. Thus, by Item \ref{2z22} of Lemma \ref{Lemma:cases}, $v_3=(u_3,u_3, u_3, u_3)+({\mathbf{0}},{\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$ is a permutation of $(2\mathbb{Z}_8,\stackrel{m+m'}{\dots}, 2\mathbb{Z}_8)$, so  $v_3$ satisfies property \ref{proper_2b}, and so does $\mathbf{v}$. Therefore, if $o(\mathbf{u})=2$ and $\mathbf{u}$ satisfies property 
     \ref{proper_3a}, we have that $\mathbf{v}$ satisfies property \ref{proper_2b}. 
     
 We continue with the second subcase, when $o(\mathbf{z})=2$ and $\lambda \in \{1,3\}$. Again, we have that  $o(\mathbf{u})=2$. Now, we assume that $\mathbf{u}$ satisfies property    
     \ref{proper_3b}.
    Then, $u_1={\mathbf{0}}$,  $u_2$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ exactly $4^{t_1+t_2-1}$ 
   
    times and  $4^{t_1+t_2-1}-2^{t_1+t_2-1}$
    times the element  $0$, and $u_3$ is a permutation of $(4\mathbb{Z}_8,\stackrel{m}{\dots},4\mathbb{Z}_8)$ for some $m\geq 0$. By Item 3 of Lemma \ref{lem:z_t1t2p1_1}, we have that  $x_1={\mathbf{0}}$. 
    Since $v_1=(u_1, u_1)+\lambda({\mathbf{0}}, {\mathbf{1}})$, $u_1={\mathbf{0}}$, and $\lambda \in \{1,3\}$, we have that $v_1$ contains every element of $\mathbb{Z}_2$ the same number of times. 
    From (\ref{eq:vi_}), $v_2=(x_1, u_2,u_2,u_2,u_2)+\lambda({\mathbf{1}}, {\mathbf{0}}, {\mathbf{1}}, {\mathbf{2}}, \mathbf{3})$, where $x_1={\mathbf{0}}$ is of length $2^{t_1+t_2}$. By Lemma \ref{lemm:uu+lambda}, we have that $v_2$ is a permutation of $$(\lambda, \stackrel{2^{t_1+t_2}}{\dots},\lambda,\mathbb{Z}_4,\stackrel{\alpha_2}{\dots}, \mathbb{Z}_4),$$ where $\lambda \in \{1,3\}$.
    From (\ref{eq:vi_}), $v_3=(u_3,u_3, u_3, u_3)+\lambda({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6}).$ Since $\lambda \in \{1,3\}$,  $\lambda({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$ is a permutation of $({\mathbf{0}}, {\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$. 
    Thus, by Item \ref{2z22} of Lemma \ref{Lemma:cases}, $v_3=(u_3,u_3, u_3, u_3)+({\mathbf{0}},{\mathbf{2}}, {\mathbf{4}}, \mathbf{6})$ is a permutation of $(2\mathbb{Z}_8,\stackrel{2m}{\dots}, 2\mathbb{Z}_8)$.  Therefore,  $v_3$  satisfies property \ref{proper_2b}, and so does $\mathbf{v}$. 
    
Case 3: Assume that $o(\mathbf{v})=2$. Then,  $o(\mathbf{z})=2$ and $\lambda \in \{0,2\}$.  Since $o(\mathbf{z})=2$, we have that  $o(\mathbf{u})=2$ and, by Item 3 of Lemma \ref{lem:z_t1t2p1_1}, $x_1={\mathbf{0}}$. By induction hypothesis, $\mathbf{u}$ satisfies property \ref{proper_3a} or \ref{proper_3b}. Assume that $\mathbf{u}$ satisfies property \ref{proper_3a}. Then, $u_1={\mathbf{0}}$,  $u_2={\mathbf{0}}$, and  $u_3$ contains the element in $4\mathbb{Z}_8{\setminus}\{0\}=\{4\}$ exactly 
$m=8^{t_1-1}4^{t_2}$
times and   
$m'=8^{t_1-1}4^{t_2}-4^{t_1+t_2-1}$
times the element $0$.
If $\lambda=0$, then $\mathbf{v}=( {\mathbf{0}} \mid {\mathbf{0}} \mid v_3)$ satisfies property \ref{proper_3a}, since $v_3$ contains $4m$ times the element $4$ and $4m'$ the element $0$. 
Now, we assume that $\lambda=2$. Since $v_1=(u_1, u_1)+\lambda({\mathbf{0}}, {\mathbf{1}})$, $u_1={\mathbf{0}}$, and $\lambda=2$, we have that $v_1={\mathbf{0}}$. 
From (\ref{eq:vi_}),  $v_2=(x_1, u_2,u_2,u_2,u_2)+({\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}})$, where $x_1={\mathbf{0}}$ is of length $2^{t_1+t_2}$ and $u_2={\mathbf{0}}$. Therefore, $v_2$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ exactly $\alpha_1+2\alpha_2=4^{t_1+t_2}$ times  and  $2\alpha_2=4^{t_1+t_2}-2^{t_1+t_2} $ times the element $0$, by (\ref{eq:t1t2}). From (\ref{eq:vi_}), $v_3=(u_3,u_3, u_3, u_3)+({\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}}).$ Note that $u_3$ is a permutation of 
$$
(4,\stackrel{m-m'}{\dots}, 4, 4\mathbb{Z}_8,\stackrel{m'}{\dots}, 4\mathbb{Z}_8).
$$ 
Thus, by  Item \ref{2z22} of Lemma \ref{Lemma:cases}, $v_3$ is a permutation of $(4\mathbb{Z}_8,\stackrel{2m+2m'}{\dots}, 4\mathbb{Z}_8)$, so $v_3$ satisfies property \ref{proper_3b}, and so does $\mathbf{v}$.   Therefore, if $o(\mathbf{u})=2$ and $\mathbf{u}$ satisfies property \ref{proper_3a}, we have that $\mathbf{v}$ satisfies property \ref{proper_3b}. 

We continue with the case when $o(\mathbf{z})=2$ and $\lambda \in \{0,2\}$. Again, we have that $o(\mathbf{u})=2$ and $x_1={\mathbf{0}}$. Now, we assume that $\mathbf{u}$ satisfies property \ref{proper_3b}.
Then, $u_1={\mathbf{0}}$,  $u_2$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ exactly $4^{t_1+t_2-1}$ times and  $4^{t_1+t_2-1}-2^{t_1+t_2-1}$ times the element  $0$, and $u_3$ is a permutation of $(4\mathbb{Z}_8,\stackrel{m}{\dots},4\mathbb{Z}_8)$ for some $m\geq 0$. 
If $\lambda=0$, then it is easy to see that $\mathbf{v}$ satisfies property \ref{proper_3b}. Now, we assume that $\lambda=2$.
Since $v_1=(u_1, u_1)+\lambda({\mathbf{0}}, {\mathbf{1}})$, $u_1={\mathbf{0}}$, and $\lambda=2$, we have that $v_1={\mathbf{0}}$.  From (\ref{eq:vi_}), $v_2=(x_1, u_2,u_2,u_2,u_2)+({\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}}, {\mathbf{0}}, {\mathbf{2}})$, where $x_1={\mathbf{0}}$ is of length $2^{t_1+t_2}$. Note that $u_2+{\mathbf{2}}$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ as many times as $u_2$ contains the element $0$, and the element $0$ as many times as $u_2$ contains the element $2$. Therefore, $v_2$ contains the element in $2\mathbb{Z}_4{\setminus}\{0\}=\{2\}$ exactly
 $2^{t_1+t_2} + 2(4^{t_1+t_2-1}) + 2(4^{t_1+t_2-1} - 2^{t_1+t_2-1} ) =4^{t_1+t_2}$ times
     and  $2(4^{t_1+t_2-1}) + 2(4^{t_1+t_2-1} - 2^{t_1+t_2-1} ) =4^{t_1+t_2}-2^{t_1+t_2}$ times
     the element $0$.
From (\ref{eq:vi_}), $v_3=(u_3,u_3, u_3, u_3)+({\mathbf{0}}, {\mathbf{4}}, {\mathbf{0}}, {\mathbf{4}}).$ By Item \ref{2z22} of Lemma \ref{Lemma:cases},
$v_3$ is a permutation of $(4\mathbb{Z}_8,\stackrel{4m}{\dots}, 4\mathbb{Z}_8)$.  Therefore,  $v_3$ satisfies property \ref{proper_3b}, and so does $\mathbf{v}$. This completes the proof.
\end{proof}
 


 
 \begin{proposition}\label{Th:H1}
Let $t_1\geq 1$ and $t_2\geq 0$ be integers. The $\mathbb{Z}_2\mathbb{Z}_4 \mathbb{Z}_8$-additive code 
${\cal{H}}^{t_1,t_2,1}$, generated by the matrix $A^{t_1,t_2,1}$, is a $\mathbb{Z}_2\mathbb{Z}_4 \mathbb{Z}_8$-additive Hadamard code. 
\end{proposition}
\begin{proof}
Let ${\cal{H}}^{t_1,t_2,1}$ be the $\mathbb{Z}_2\mathbb{Z}_4 \mathbb{Z}_8$-additive code of type $(\alpha_1, \alpha_2, \alpha_3;t_1, t_2,1)$ and  $H^{t_1,t_2,1}$ be the corresponding  $\mathbb{Z}_2\mathbb{Z}_4 \mathbb{Z}_8$-linear code of length $N$. We have that $N=\alpha_1+2\alpha_2+4\alpha_3$. The cardinality of $H^{t_1, t_2,1}$ is $8^{t_1}\cdot 4^{t_2}\cdot 2=2(\alpha_1+2\alpha_2+4\alpha_3)=2N$ by Lemma \ref{relation:t1a1}. By Proposition \ref{disweight}, the minimum distance of $H^{t_1, t_2,1}$ is equal to the minimum weight of $H^{t_1, t_2,1}$. Therefore, we just need to prove that the minimum weight of $H^{t_1, t_2,1}$ is $N/2$. 

We can write that ${\cal{H}}^{t_1,t_2,1}={{\cal G}}^{t_1,t_2,1} \cup ({{\cal G}}^{t_1,t_2,1}+({\mathbf{1}} \mid {\mathbf{2}} \mid {\mathbf{4}}))$. By Corollary \ref{lemma2}, $H^{t_1,t_2,1}=\Phi({{\cal G}}^{t_1,t_2,1})\cup (\Phi({{\cal G}}^{t_1,t_2,1})+{\mathbf{1}})$. Let $\mathbf{u}=(u_1 \mid u_2 \mid u_3) \in {\cal{H}}^{t_1,t_2,1}{\setminus} \{{\mathbf{0}}, ({\mathbf{1}}\mid {\mathbf{2}}\mid {\mathbf{4}})\}$. We show that ${\rm wt}_H(\Phi(\mathbf{u}))=N/2$. First, consider $\mathbf{u} \in {{\cal G}}^{t_1,t_2,1}{\setminus}\{{\mathbf{0}}\}$. 
If $o(\mathbf{u})=8$, then by Lemma \ref{lemma:same_num2}, $u_1$ contains every element of $\mathbb{Z}_2$ the same number of times, and for $i \in \{2,3\}$, $u_i$ contains every element of $2^{i-1}\mathbb{Z}_{2^i}$ exactly $s_i$ times, $s_i\geq 0$, and the remaining $\alpha_i - 2s_i$ coordinates of $u_i$ are from $\mathbb{Z}_{2^i}{\setminus}2^{i-1}\mathbb{Z}_{2^i}$. Thus, from the definition of $\Phi$, we have that ${\rm wt}_H(\Phi(\mathbf{u}))=\alpha_1/2+2s_2+(\alpha_2-2s_2)\cdot 1+ 4s_3+(\alpha_3-2s_3)\cdot 2=\alpha_1/2+\alpha_2+2\alpha_3=N/2$. If $o(\mathbf{u})=4$, then $\mathbf{u}$ satisfies property \ref{proper_2a} 
or \ref{proper_2b} given in Lemma \ref{lemma:same_num2}. If $\mathbf{u}$ satisfies property \ref{proper_2a}, then $u_3$ contains every element of $4\mathbb{Z}_8$ exactly $m$ times, $m\geq 0$, and the remaining coordinates of $u_3$ are from $\mathbb{Z}_8{\setminus}4\mathbb{Z}_8$. Thus, ${\rm wt}_H(\Phi(\mathbf{u}))=\alpha_1/2+\alpha_2+4m+(\alpha_3-2m)\cdot 2=\alpha_1/2+\alpha_2+2\alpha_3=N/2$. Otherwise, if $\mathbf{u}$ satisfies property \ref{proper_2b}, then ${\rm wt}_H(\Phi(\mathbf{u}))=\alpha_1/2+2n+(\alpha_2-2n)\cdot 1+4t+(\alpha_3 -2t)\cdot 2=N/2$. If $o(\mathbf{u})=2$, then $\mathbf{u}$ satisfies property \ref{proper_3a}  or \ref{proper_3b}  given in Lemma \ref{lemma:same_num2}. If $\mathbf{u}$ satisfies property \ref{proper_3a}, then ${\rm wt}_H(\Phi(\mathbf{u}))=\frac{1}{4}(\alpha_1/2+\alpha_2+2\alpha_3)\cdot4=N/2$. Otherwise, if $\mathbf{u}$ satisfies property \ref{proper_3b}, then ${\setminus}=2\cdot \frac{1}{2}(\alpha_1/2+\alpha_2)+4m+(\alpha_3-2m)\cdot 2=N/2$.

Finally, note that ${\rm wt}_H(\Phi(\mathbf{u})+{\mathbf{1}})=N/2$. Therefore, we have that the weight of every element of $H^{t_1,t_2,1}{\setminus}\{{\mathbf{0}}, {\mathbf{1}}\}$ is $N/2$, that is, the minimum weight of $H^{t_1,t_2,1}$ is $N/2$.
\end{proof}

\begin{proposition}\label{Th:H2}
Let $t_1\geq 1$, $t_2\geq 0$, and $t_3\geq1$ be integers. If ${\cal{H}}^{t_1,t_2,t_3}$ is a $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-additive Hadamard code of type $(\alpha_1, \alpha_2,\alpha_3; t_1,t_2,t_3)$, then, by applying  construction (\ref{eq:recGenMatrix3}), ${\cal{H}}^{t_1,t_2,t_3+1}$ is a $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-additive Hadamard code of type $(2\alpha_1,2\alpha_2,2\alpha_3; t_1,t_2,t_3+1)$.
\end{proposition}
\begin{proof}
By construction (\ref{eq:recGenMatrix3}), ${\cal{H}}^{t_1,t_2,t_3+1}$ is a $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-additive code of type $(\alpha'_1, \alpha'_2, \alpha'_3; t_1, t_2,t_3+1)$, where $\alpha'_1=2\alpha_1$, $\alpha'_2=2\alpha_2$, and  $\alpha'_3=2\alpha_3$. 

Since $H^{t_1,t_2,t_3}$ is a Hadamard code of length $N=\alpha_1+2\alpha_2+4\alpha_3$, then its minimum distance is $N/2$ and $|H^{t_1,t_2,t_3}|=2N$. Note that $H^{t_1,t_2,t_3+1}$ is a $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear code of length $N'=\alpha'_1+2\alpha'_2+4\alpha'_3=2N$ and $|H^{t_1,t_2,t_3+1}|=8^{t_1}4^{t_2}2^{t_3+1}=2|H^{t_1,t_2,t_3}|=2\cdot 2N=2N'$. By Proposition \ref{disweight}, the minimum distance of $H^{t_1, t_2,t_3+1}$ is equal to the minimum weight of $H^{t_1, t_2,t_3+1}$. Now, we only have to prove that the minimum weight of $H^{t_1,t_2,t_3+1}$ is $N'/2$. Let ${\cal{H}}^{t_1,t_2,t_3}=({\cal{H}}_1\mid {\cal{H}}_2\mid {\cal{H}}_3)$. Note that 
\begin{align*}
    {\cal{H}}^{t_1,t_2,t_3+1}=\bigcup_{\lambda \in  \{0,1\}}(({\cal{H}}_1, {\cal{H}}_1\mid {\cal{H}}_2, {\cal{H}}_2\mid {\cal{H}}_3, {\cal{H}}_3)+\lambda({\mathbf{0}}, {\mathbf{1}} \mid {\mathbf{0}}, {\mathbf{2}} \mid {\mathbf{0}}, {\mathbf{4}})).
\end{align*}
By Corollaries \ref{coro4} and  \ref{lemma2}, 
\begin{align}\label{eq:Ht3p1_}
    H^{t_1,t_2,t_3+1}&=\bigcup_{\lambda \in \{0,1\}}(\Phi({\cal{H}}_1, {\cal{H}}_1\mid {\cal{H}}_2, {\cal{H}}_2\mid {\cal{H}}_3, {\cal{H}}_3)+\lambda({\mathbf{0}}, {\mathbf{1}},  {\mathbf{0}}, {\mathbf{1}},  {\mathbf{0}}, {\mathbf{1}}))\nonumber\\
    &=A_0 \cup A_1,
\end{align}
where $A_\lambda=\Phi({\cal{H}}_1, {\cal{H}}_1\mid {\cal{H}}_2, {\cal{H}}_2\mid {\cal{H}}_3, {\cal{H}}_3)+\lambda({\mathbf{0}}, {\mathbf{1}},  {\mathbf{0}}, {\mathbf{1}},  {\mathbf{0}}, {\mathbf{1}})$, ${\lambda \in \{0,1\}}$.
Next, we show that the minimum weight of $A_\lambda$ is $N'/2$.  Any element in $A_\lambda$ is of the form $\Phi(u_1, u_1\mid u_2, u_2 \mid u_3, u_3)+\lambda({\mathbf{0}}, {\mathbf{1}},  {\mathbf{0}}, {\mathbf{1}},  {\mathbf{0}}, {\mathbf{1}})$, for $\mathbf{u}=(u_1\mid u_2 \mid u_3) \in ({\cal{H}}_1\mid {\cal{H}}_2\mid {\cal{H}}_3)$. Let $\mathbf{u}=(u_1\mid u_2 \mid u_3) \in ({\cal{H}}_1\mid {\cal{H}}_2\mid {\cal{H}}_3){\setminus}\{{\mathbf{0}}\}$. 
When $\lambda =0$, we have that ${\rm wt}_H(\Phi(u_1, u_1\mid u_2, u_2 \mid u_3, u_3))=2{\rm wt}_H(\Phi(\mathbf{u}))$. Thus, the minimum weight of $A_0$
is $2\cdot N/2=N'/2$. Otherwise, when $\lambda=1$, we have that ${\rm wt}_H(\Phi(u_1, u_1\mid u_2, u_2 \mid u_3, u_3)+({\mathbf{0}}, {\mathbf{1}},  {\mathbf{0}}, {\mathbf{1}},  {\mathbf{0}}, {\mathbf{1}}))= {\rm wt}_H(\Phi(\mathbf{u}))+\alpha_1-{\rm wt}_H(u_1)+2\alpha_2-{\rm wt}_H(\Phi_2(u_2))+ 4\alpha_3-{\rm wt}_H(\Phi_3(u_3))={\rm wt}_H(\Phi(\mathbf{u})) +\alpha_1+2\alpha_2+4\alpha_3-{\rm wt}_H(\Phi(\mathbf{u}))=N=N'/2$. Thus, the minimum weight of $A_1$ 
is $N'/2$. Therefore, from (\ref{eq:Ht3p1_}), the minimum weight of $H^{t_1,t_2,t_3+1}$ is $N'/2$.
\end{proof}

 \begin{theorem}\label{Th:H3}
Let $t_1\geq 1$, $t_2\geq 0$, and $t_3\geq 1$ be integers. The $\mathbb{Z}_2\mathbb{Z}_4 \mathbb{Z}_8$-additive code 
${\cal{H}}^{t_1,t_2,t_3}$, generated by the matrix $A^{t_1,t_2,t_3}$, is a $\mathbb{Z}_2\mathbb{Z}_4 \mathbb{Z}_8$-additive Hadamard code. 
\end{theorem}
\begin{proof}
It follows from Propositions \ref{Th:H1} and \ref{Th:H2}.
\end{proof}


\begin{example}\label{ex:linear101}
The $\mathbb{Z}_2\mathbb{Z}_4 \mathbb{Z}_8$-additive code ${\cal{H}}^{1,0,1}$ generated by the  matrix $A^{1,0,1}$, given in (\ref{eq:recGenMatrix0}),
is a $\mathbb{Z}_2\mathbb{Z}_{4}\mathbb{Z}_8$-additive Hadamard code of type $(2,1,1;1,0,1)$.  
We can write ${\cal{H}}^{1,0,1}=\bigcup_{\alpha \in\mathbb{Z}_2} ({{\cal A}} + \alpha \textbf{1})$, where ${{\cal A}}=\{\lambda (0, 1\mid 1\mid 1): \lambda \in \mathbb{Z}_8\}$.
Thus, $H^{1,0,1}=\Phi({\cal{H}}^{1,0,1})= \bigcup_{\alpha \in\mathbb{Z}_2} (\Phi({{\cal A}}) + \alpha \textbf{1})$, where $\Phi({{\cal A}})$ consists of all the rows of the Hadamard matrix 
\begin{equation*}
    H(2,4)=
    \left(\begin{array}{cccccccc}
0 &0 &0 &0 &0 &0 &0 &0  \\
0 &1 &0 &1 &0 &1 &0 &1   \\
0 &0 &1 &1 &0 &0 &1 &1  \\
0 &1 &1 &0 &0 &1 &1 &0 \\
0 &0 &0 &0 &1 &1 &1 &1 \\
0 &1 &0 &1 &1 &0 &1 &0 \\
0 &0 &1 &1 &1 &1 &0 &0 \\
0 &1 &1 &0 &1 &0 &0 &1 \\
\end{array}\right).
\end{equation*}
Note that $\Phi({{\cal A}})$ is linear and the minimum distance of $\Phi({{\cal A}})$ is $4$, so $H^{1,0,1}$ is a binary linear Hadamard code of length $8$. 
\end{example}


\begin{proposition}\label{HadamardEq}
 Let $t_1\geq 1$, $t_2\geq 0$, and $t_3\geq 1$ be integers. Let $H^{t_1,t_2,t_3}$ be a $\mathbb{Z}_2\mathbb{Z}_4 \mathbb{Z}_8$-linear Hadamard code of length $2^t$. Then, $t+1=3t_1+2t_2+t_3$. 
\end{proposition}
\begin{proof}
Since $H^{t_1,t_2,t_3}$ is a binary Hadamard code of length $2^t$, we have that $|H^{t_1,t_2,t_3}|=2\cdot 2^t=2^{t+1}$. Note that $|H^{t_1,t_2,t_3}|=2^{3t_1+2t_2+t_3}$, and hence $t+1=3t_1+2t_2+t_3$. 
\end{proof}


\medskip
 Now, we recall the following theorem in order to compare the $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear Hadamard codes (with $\alpha_1\neq 0$, $\alpha_2\neq 0$ and $\alpha_3\neq 0$) with the $\mathbb{Z}_2\mathbb{Z}_4$-linear Hadamard codes (with $\alpha_1\neq 0$, $\alpha_2\neq 0$).

\begin{theorem}\cite{PRV06}\label{rankZ2Z4}
Let $t\geq 3$ and $t_2 \in \{0,\dots,\lfloor t/2 \rfloor\}$. Let $H^{t_1,t_2}$ be the nonlinear $\mathbb{Z}_2\mathbb{Z}_4$-linear Hadamard code of length $2^t$ and type $(\alpha_1, \alpha_2;t_2,t_3)$, where $\alpha_1=2^{t-t_1}$, $\alpha_2=2^{t-1}-2^{t-t_1-1}$, and $t_2=t+1-2t_1$. Then, $$\operatorname{rank}(H^{t_1,t_2})=t_2+2t_1+ \binom{t_1}{2} \ \mbox{ and } \ \operatorname{ker}(H^{t_1,t_2})=t_1+t_2.$$
\end{theorem}

Also, we recall the construction of the $\mathbb{Z}_{2^s}$-linear Hadamard codes with $s\geq 2$ studied in \cite{KernelZ2s}, and the following theorem given in \cite{EquivZ2s}, in order to compare these codes with the $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear Hadamard codes having $\alpha_1\neq 0$, $\alpha_2\neq 0$, and $\alpha_3\neq 0$. Let $T_i=\lbrace j\cdot 2^{i-1}\, :\, j\in\lbrace0,1,\dots,2^{s-i+1}-1\rbrace \rbrace$ for all $i \in \{1,\ldots,s \}$.
Note that $T_1=\lbrace0,\dots,2^{s}-1\rbrace$. Let $t_1$, $t_2$,\dots,$t_s$ be non-negative integers with $t_1\geq1$. Consider the matrix $\bar{A}^{t_1,\dots,t_s}$ whose columns are exactly all the vectors of the form $\mathbf{z}^T$, $\mathbf{z}\in\lbrace1\rbrace\times T_1^{t_1-1}\times T_{2}^{t_2}\times\cdots\times T_s^{t_s}$. Let $\bar{\mathcal{H}}^{t_1,\dots,t_s}$ be the $\mathbb{Z}_{2^s}$-additive code of type $(n;t_1,\dots,t_s)$  generated by the matrix $\bar{A}^{t_1,\dots,t_s}$.
 Let $\bar{H}^{t_1,\dots,t_s}=\Phi(\bar{\mathcal{H}}^{t_1,\dots,t_s})$ be the corresponding $\mathbb{Z}_{2^s}$-linear Hadamard code.

\begin{theorem}\cite{EquivZ2s}\label{theo:equi}
Let $\bar{H}^{t_1,\dots,t_s}$ be the $\mathbb{Z}_{2^s}$-linear Hadamard code, with $s\geq 2$ and $t_s\geq 1$.
Then,  for all $\ell\in\{1,\dots,t_s\}$, $\bar{H}^{t_1,\dots,t_s}$  is permutation equivalent to the
$\mathbb{Z}_{2^{s+\ell}}$-linear Hadamard code
$\bar{H}^{1,{\mathbf{0}}^{\ell-1},t_1-1,t_2,\dots,t_{s-1},t_s-\ell}.$
\end{theorem}

For $5\leq t\leq 11$, Tables 1 and 3 given in \cite{KernelZ2s}  show all possible values of ($t_1,\dots,t_s$) corresponding to nonlinear $\mathbb{Z}_{2^s}$-linear Hadamard codes, with $s\geq 2$, of length $2^t$. For each of them, the values $(r,k)$ are shown, where $r$ is the rank  and $k$ is the dimension of the kernel.  Note that if two codes have different values $(r,k)$, they are not equivalent. 
 The following example shows that all the nonlinear $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear Hadamard codes of length $2^{11}$, with $\alpha_1 \neq 0$, $\alpha_2 \neq 0$, and $\alpha_3 \neq 0$,  are not equivalent to any  $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear Hadamard codes of any other type, any $\mathbb{Z}_2\mathbb{Z}_4$-linear Hadamard code, with $\alpha_1 \neq 0$ and $\alpha_2 \neq 0$, and any $\mathbb{Z}_{2^s}$-linear Hadamard code, with $s\geq 2$, of the same length $2^{11}$.

\begin{example}\label{ex:classi}
   Consider $t=11$. By solving equation $t+1=3t_1+2t_2+t_3$ given in Proposition \ref{HadamardEq}, all $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear Hadamard codes  of length $2^{11}$ are the ones  in $$T=\{H^{1,0,9}, H^{1,1,7}, H^{1,2,5}, H^{1,3,3}, H^{1,4,1}, H^{2,0,6}, H^{2,1,4}, H^{2,2,2}, H^{3,0,3}, H^{3,1,1}\}.$$  By using Magma, their corresponding values of $(r,k)$,  where $r$ is the rank  and $k$ is the dimension of the kernel, are  $(12,12)$, $(14,9)$, $(17,8)$, $(21,7)$, $(26,6)$, $(17,8)$, $(22,7)$, $(28,6)$, $(28,6)$, and $(37,5)$, respectively. The code $H^{1,0,9}$ is the only linear code in $T$ since it has the same rank and the dimension of the kernel. By using Magma, we can check that the following codes in each pair are nonequivalent to each other: $(H^{1,2,5}, H^{2,0,6})$, $(H^{2,2,2}, H^{3,0,3})$. Therefore, none of the $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear Hadamard codes of length $2^{11}$ is equivalent to another $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear Hadamard code of any other type. 
   
   Let $\Bar{T}=T \setminus \{H^{1,0,9}\}$.  Similarly, by  solving equation $t+1=2t_1+t_2$ given in Theorem \ref{rankZ2Z4}, all nonlinear $\mathbb{Z}_2\mathbb{Z}_4$-linear Hadamard codes  of length $2^{11}$ are $H^{2,8}$, $H^{3,6}$, $H^{4,4}$ and $H^{5,2}$, and by Theorem \ref{rankZ2Z4}, their corresponding values of $(r,k)$ are $(13,10)$, $(15,9)$, $(18,8)$, and $(22,7)$, respectively. Note that if two codes have different values $(r,k)$, they are not equivalent.  By using Magma, we can check that $H^{2,1,4}$ and $H^{5,2}$ are nonequivalent. Therefore, all the codes in $\Bar{T}$ are nonequivalent to any $\mathbb{Z}_2\mathbb{Z}_4$-linear Hadamard code of length $2^{11}$. 
   
   Finally, note that all the codes in $\Bar{T}$, except $H^{1,1,7}$ and $H^{2,1,4}$,  are not equivalent to any $\mathbb{Z}_{2^s}$-linear Hadamard code, with $s\geq 2$, of length $2^{11}$, since they have different values of $(r,k)$. The $\mathbb{Z}_{2^s}$-linear Hadamard codes of length $2^{11}$, having the same values $(r,k)=(14,9)$ as $H^{1,1,7}$, are $\Bar{H}^{2,0,6}$, $\Bar{H}^{1,1,0,5}$, $\Bar{H}^{1,0,1,0,4}$, $\Bar{H}^{1,0,0,0,1,0,2}$, and $\Bar{H}^{1,0,0,0,0,0,1,0,0}$, which are equivalent to each other by Theorem \ref{theo:equi}. The $\mathbb{Z}_4$-linear Hadamard code $\Bar{H}^{6,0}$ is the only $\mathbb{Z}_{2^s}$-linear Hadamard code of length $2^{11}$, having the same values $(r,k)=(22,7)$ as $H^{2,1,4}$. However, by using Magma, we can check that the following codes in each pair are nonequivalent to each other: $(H^{1,1,7},\Bar{H}^{2,0,6})$, $(H^{2,1,4}, \Bar{H}^{6,0})$. 
   
   Therefore, all nonlinear $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear Hadamard codes  of length $2^{11}$  are  not equivalent to any $\mathbb{Z}_2\mathbb{Z}_4\mathbb{Z}_8$-linear Hadamard code of any other type, any $\mathbb{Z}_2\mathbb{Z}_4$-linear and $\mathbb{Z}_{2^s}$-linear Hadamard code, with $s\geq 2$, of length $2^{11}$.
\end{example}


 