Number of cyclic paths in a rectangular grid? 
If we start at the down-left corner or equivalently the origin and move only in the first quadrant, using only 4-directional moves, what are the number of ways to make a cyclic routes back to origin in exactly n-moves?

I think that if we denote $n_i$ as move in $i$-direction. then we have $n_u\ge n_d$ and $n_r\ge n_l$ anytime and also $n_u+n_l+n_r+n_d=n$, finally. So (let $n_u=k$):
$$\sum_{k=0}^{n/2}\binom n{2k}C_{k}C_{n/2-k}=\sum_{k=0}^{n/2}\frac1{(n+1)(n+2)}\binom{n+2}{n/2+1}\binom{n/2+1}{k}\binom{n/2+1}{k+1}=\frac1{(n+2)(n+1)}\binom{n+2}{n/2+1}\binom{n+2}{n/2}$$
Question: Am I right? Is there any other easy or combinatoric way?
 A: It does appear to be correct, though perhaps not the simplest approach.
It’s clear that $n$ must be even, so for convenience let $n=2m$. In any such path the number of horizontal moves must be even, so it must be $2k$ for some $k=0,\ldots,m$. For each of these values of $k$ there are $\binom{n}{2k}$ possible sets of positions for the horizontal moves. Within this set of positions there are $C_k$ possible sequences that avoid negative $x$-coordinates, and the there are $C_{m-k}$ possible sequences for the $n-2k$ vertical moves that avoid negative $y$-coordinates. The total is therefore
$$\begin{align*}
\sum_{k=0}^m\binom{n}{2k}C_kC_{m-k}&=\sum_{k=0}^m\binom{n}{2k}\binom{2k}k\binom{n-2k}{m-k}\frac1{(k+1)(m-k+1)}\\
&=\sum_{k=0}^m\binom{2m}{k,k,m-k,m-k}\frac1{(k+1)(m-k+1)}\;.
\end{align*}$$
From this it’s not hard to calculate the values for $m=0,1,2$, and $3$: they are $1,2,10$, and $70$. Searching for this in the On-Line Encyclopedia of Integer Sequences, we find that the first return is OEIS A005568, one of whose comments identifies it as the sequence that we want. Better yet, it turns out that these are just the products of two consecutive Catalan numbers:
$$\sum_{k=0}^m\binom{n}{2k}C_kC_{m-k}=C_mC_{m+1}=\frac1{(m+1)(m+2)}\binom{2m}m\binom{2m+2}{m+1}\;.$$
