Experiments support the below identity.



> 
> **Question.** Is this true? Combinatorial proof preferred if possible.
>  $$\sum\_{m=0}^n\binom{n-\frac13}m\binom{n+\frac13}{n-m}(1+6m-3n)^{2n+1}
> =\left(\frac43\right)^n\frac{(3n+1)!}{n!}.$$
> 
> 
> 


In View of MTyson's suggestion (see below), a generalized question can be asked:



> 
> **Question.** Is this true? Combinatorial proof preferred if possible.
>  $$\sum\_{m=0}^n\binom{n-y}m\binom{n+y}{n-m}(y+2m-n)^{2n+1}
> =y\prod\_{k=1}^n4(k^2-y^2).$$
> 
> 
> 



A proof required for this identity
**"show that the a priori degree $\le n$ polynomial $\sum\_{i+j=n}{n−x\choose i}{n+x\choose j}(i−j)^k$ is only degree $k$".**


That is fairly obvious when $k<2n+1$ because it suffices to check it for integer $x\in\{-n,\dots,n\}$. We will just show that for every $a,b\ge 0$ with $a+b\le k$, the sum $\sum\_{i+j=n}{n−x\choose i}{n+x\choose j}{i\choose a}{j\choose b}$ is a polynomial of degree $\le a+b$ in $x$. When $|x|\le n$, this sum has a simple combinatorial meaning: choose some $n$ balls out of $2n$ ($i$ out of first $n-x$, $j$ out of last $n+x$) and color $a$ in the first group red and $b$ in the last group blue. But we can then compute it as ${n-x\choose a}{n+x\choose b}{2n-a-b\choose n-a-b}$ (choose colored balls first and add the remaining ones afterwards). 


That the leading coefficient is right is not immediately obvious to me but, perhaps, someone can see it too.


