How can I decide upon the parity of the expression $N_{n}=\sum_{k=1}^n\sum_{s=0}^k\binom{n}{s}\begin{Bmatrix} n-s\\ k-s \end{Bmatrix}$? Here, $n,k\in \mathbb{N}$ and $\begin{Bmatrix} n\\ k \end{Bmatrix}$ denotes Stirling number of the second kind.
I checked that $N_1, N_4$ are even and $N_2,N_3,N_5,N_6$ are odd.
$$N_{n}=\sum_{k=1}^n\sum_{s=0}^k\binom{n}{s}{n-s\brace k-s}=\sum_{s=0}^n\binom{n}{s}\sum_{k=s}^n{n-s\brace k-s}=\sum_{s=0}^n\binom{n}{s}\sum_{k=0}^{n-s}{n-s\brace k} =\sum_{s=0}^n\binom{n}{s}B_{n-s}=\sum_{s=0}^n\binom{n}{s}B_{s}=B_{n+1}$$ where $B_n$ is the Bell number. Then it says on the Wikipedia page for Bell numbers that these numbers repeat the pattern odd-odd-even with period three, and gives reference for that.