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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. $$\sum_{m=0}^n\binom{n-y}m\binom{n+y}{n-m}(y+2m-n)^{2n+1} =y\prod_{k=1}^n4(k^2-y^2).$$

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    $\begingroup$ No need for gamma functions. If $t$ is a nonnegative integer, $\binom st$ is defined as $s(s-1)\cdots(s-t+1)/t!$. I checked the formula for $0\le n\le 300$. Maple 2018 can't do the general sum. $\endgroup$
    – Brendan McKay
    Commented Oct 1, 2018 at 1:40
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    $\begingroup$ More generally, it seems that for $s(n)=\sum_{m=0}^n\binom{n-1/x}m\binom{n+1/x}{n-m}(1+2xm-xn)^{2n+1}$, $s(n)/s(n-1)=4(nx+1)(nx-1)$. $\endgroup$
    – MTyson
    Commented Oct 1, 2018 at 18:31
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    $\begingroup$ It's easy to show by canceling terms that the $s(n)$ in my previous comment has roots $\pm 1/k$ for $n\ge k$. Therefore it suffices to show that $s(n)$ is a degree $2n$ polynomial with the right leading or constant coefficient. $\endgroup$
    – MTyson
    Commented Oct 1, 2018 at 19:04
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    $\begingroup$ For $x=2$ and $x=4$ the oeis has interpretations oeis.org/A000909 and oeis.org/A009564 respectively. $\endgroup$
    – Martin Rubey
    Commented Oct 2, 2018 at 15:18
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    $\begingroup$ For the generalized eq. see mathoverflow.net/questions/262392/… $\endgroup$
    – esg
    Commented Oct 3, 2018 at 10:19

1 Answer 1

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"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.

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