\text{Case ~2}: & i_2=4j-1 & 2 \leq j\leq k~.
  \end{array}\end{aligned}$$ We partition the random walk into two parts according to the first return, see Figure \[twoparts\].

Case 1: First, we consider the second part, which is of length $4k-i_1=4k-4j+3$, with the number of “1” being $m-s$ in its corresponding characteristic sequence. But this time, the sequence may not start with a “0” (once it returns to the origin, it can depart immediately, and in this case, the sequence starts with a “1”). Therefore, we add a zero in the front artificially, leading to a total length of $4k-4j+4$, which starts and ends with a zero. So the way of constructing such a sequence is $f_{m-s}(k-j+1)$. Then consider the first part, which has a total length of $i_1$. Suppose the first departure from the origin is at time $n-1$, where $n=2,~6,~\cdots, ~i_1-7,~i_1-3$ (also, it means that the first arrival at 1 is at the time $n$), then if we move the axes to the point $(n,1)$ (see the blue axes in Figure \[twoparts\], the original axes are in black) and consider the random walk above the new x-axe (see the red parts in Figure \[twoparts\]), which has the length $i_1-n-1$. Further, 