of total length $i_2-n-1$. Then we add two more zeros in the end, results in a random walk of total length $i_2-n+1=4j-n$, starts with a zero and ends with three zeros, and the number of “1” is $s-1$. The way of constructing such a random walk is $g_{s-1}(\frac{4j-n}{4})$. And combine these two parts, the contribution of Case 2 is: $$\begin{aligned}
\label{c21}
{\sum_{\scriptstyle s=1, \cdots, m \atop \scriptstyle j=2, \cdots, k}}\left(\sum_{n=4}^{4j-4}g_{s-1}\left(\frac{4j-n}{4}\right)\right)\cdot g_{m-s}(k-j+1)~.\end{aligned}$$

Overall, combine and leads to the following recursion: $$\begin{aligned}
\label{fm}
f_{m}(k)&={\sum_{\scriptstyle s=1, \cdots, m \atop \scriptstyle j=2, \cdots, k}}\left(\sum_{n=2}^{4j-6}g_{s-1}\left(\frac{4j-n-2}{4}\right)\right)\cdot f_{m-s}(k-j+1)\nonumber\\
&\quad~+{\sum_{\scriptstyle s=1, \cdots, m \atop \scriptstyle j=2, \cdots, k}}\left(\sum_{n=4}^{4j-4}g_{s-1}\left(\frac{4j-n}{4}\right)\right)\cdot g_{m-s}(k-j+1)\nonumber\\
&={\sum_{\scriptstyle s=1, \cdots, m \atop \scriptstyle j=2, \cdots, k}}\left(\sum_{n=4}^{4j-4}g_{s-1}\left(\frac{4j-n}{4}\right)\right)\cdot \left[f_{m-s}(k-j+1)+g_{m-s}(k-j+1)\right]~.\end{aligned}$$

### Master recursion $2$

Then we start with $g_m(k)$. Also suppose the first return time is $i$, where $i=5,7,\cdots,4k-5,4k-3$ (in the definition of $g_m(k)$, the characteristic sequence ends with three zeros, so the maximum value of $i$ is $4k-3$ here). We divide these $i$ into two parts: $$\begin{aligned}
  \begin{array}{lll}
    \text{Case ~1}: & i_1=4j-5 & 3 \leq j\leq k ~,\\
    \text{Case ~2}: & i_2=4j-3 & 2 \leq j\leq k~.
  \end{array}\end{aligned}$$ As before, we suppose the number of “1” is $s$ in the first part.

Case 1: First for the second 