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This is the generating function version of Darij's answer. Define $$ S_p(n) = \sum_{k=0}^n\frac{k^p}{k!}\sum_{l=0}^{n-k}\frac{(-1)^l}{l!} = \sum_{t=0}^n \frac{1}{t!} \sum_{k=0}^t \binom{t}{k}k^p(-1 …

It isn't true. Choose $n,a,k$ such that $n+a-k>0$, $n+a-2k<0$ and $k>1$. The sum has only one nonzero term which is not equal to the right side.

Let $F(x)$ be the right side. You need that $F(x)=0$ for [corrected] $x\in\{-n,-n+1,\ldots,n-1,n\}$. Assume we have one of those $x$s. The summation can stop at $m=n-x$ as later terms are zero. For …

The divisibility is easy to prove and a more general phenomenon. Let $Y=1-X$, then $$F_n(X) = \sum_{k=0}^n (-1)^k \binom nk (1-Y)^{k(n-k)}=\sum_{r\ge 0} (-1)^r a_{r,n}Y^r$$ where $$a_{r,n} = \sum_{k= …

Let $L_i$ and $R_i$ be the $i$-th rows of the left and right matrices. Then $$\begin{align} R_n &= L_n \\ R_{n-1} &= L_n + L_{n-1} \\ R_{n-2} &= L_n + 2L_{n-1} + L_{n-2} \\ R_{n-3} …