Let $\phi = \frac{1+\sqrt{5}}{2}$. A base-$\phi$ number $(a_n a_{n-1} \ldots a_1 a_0)_{\phi}$, where $0 \le a_n, a_{n-1}, \ldots, a_0 \le 1$ are integers, is defined by 
\[
(a_n a_{n-1} \ldots a_1 a_0)_{\phi} = a_n \cdot \phi^n + a_{n-1} \cdot \phi^{n-1} + \ldots + a_1 \cdot \phi^1 + a_0.
\]
Compute the number of base-$\phi$ numbers $(b_j b_{j-1} \ldots b_1 b_0)_\phi$ which satisfy $b_j \ne 0$ and
\[
(b_j b_{j-1} \ldots b_1 b_0)_\phi = \underbrace{(100 \ldots 100)_\phi}_{\text{Twenty} \ 100's}.
\]

Let's think step by step and output the final answer within \boxed{}.
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