TITLE: A follow up question to: Number of walks on integer lattice with self-edge at zero
QUESTION [5 upvotes]: Let $a(n)$ be the number of lattice paths in  ${\mathbb{Z}^2}$ of length $n$ which start at the origin $(0,0)$ and end up at $(n,0)$ and have only up-steps $U:(i,j) \to (i + 1,j + 1)$, down-steps $D:(i,j) \to (i + 1,j - 1)$ and horizontal steps $H:(i,0) \to (i + 1,0)$ on the $x-$axis.
Is there a direct  combinatorial way (i.e. without use of generating functions) to show that
$a(n) = \sum\limits_{k = 0}^{\left\lfloor {\frac{n}{2}} \right\rfloor } {\frac{{{2^k}(n - 1)!!}}{{k!(n - 1 - 2k)!!}}} $ or of the special case $a(2n+1)=5^n?$

REPLY [2 votes]: The idea of Fedor Petrov led me to the following proof of the above formula. 
In  arXiv:1203.5424  R. Duarte and A. G. de Oliveira  gave a combinatorial proof of the identity
$$S_m(k)=\sum_{i_1+\dots+i_m=k}\binom{2i_1}{i_1}\dots\binom{2i_m}{i_m}=4^k\binom{k+m/2-1}{k}.$$
If we attach a horizontal step $H$ at the beginning then each such path  has a uniquely determined decomposition in the form ${L_{{a_1}}}{L_{{a_2}}} \cdots {L_{{a_m}}}$ where  ${L_a} = H{M_a}$  and ${M_a}$ is a path of length $2a$ starting and ending on height $0$ with $a$ up-steps and $a$ down-steps. Therefore $\sum\limits_{i = 1}^m {\left( {1 + 2{a_i}} \right) = n + 1} $ or if we set $\sum\limits_{i = 1}^m {{a_i} = k} $ then $m = n + 1 - 2k.$ The number of such paths is ${S_{n + 1 - 2k}}(k) = {4^k}\binom{\frac{{n - 1}}{2}}{k}.$ Summing over all $k$ gives the above result.