this random walk starts and ends with a zero, and if we add two more zeros in its end, it will lead to a random walk with a total length of $i_1-n+1=4j-n-2$, starts with a zero and ends with three zeros, with the number of “1” being $s-1$. Therefore, the number of such random walk is $g_{s-1}(\frac{4j-n-2}{4})$ according to the definition. So we have got the total contribution of Case 1 is: $$\begin{aligned}
\label{c11}
{\sum_{\scriptstyle s=1, \cdots, m \atop \scriptstyle j=2, \cdots, k}}\left(\sum_{n=2}^{4j-6}g_{s-1}\left(\frac{4j-n-2}{4}\right)\right)\cdot f_{m-s}(k-j+1)~.\end{aligned}$$

![Illustration of $f_m(k)$.[]{data-label="twoparts"}](twoparts.eps){width="12cm"}

Case 2: We follow the same route as in Case 1. After the first return, the remaining length is $4k-i_2=4k-4j+1$, then we add a zero in front and two zeros in the end, which leads to a random walk of total length $4k-4j+4$, starts with a zero and ends with three zeros. And the number of “1” is $m-s$. By definition, the way of constructing such a random walk is $g_{m-s}(k-j+1)$. Then for the first part, also suppose the first departure from the origin is at time $n-1$ $(n=4,~8,~\cdots, ~i_2-7,~i_2-3)$ and consider the part of the random walk that is above the new x-axes (with the new origin located at $(n,1)$), which is 