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I have reduced solving this question to proving the following identity, for $n, \ell \ge 0$: $$ (n-2\ell+1)^{n-1} \binom{n}{\ell-1} = \\ \frac{1}{2} \sum_{n_1+n_2=n-1}\left[ (n_1+1)^{n_1-1} (n_2-2\ell+1)^{n_2} \binom{n_2}{\ell-1} + \\ \ell \sum_{\ell_1+\ell_2=\ell} \left[(n_1-2\ell_1+1)^{n_1} \binom{n_1}{\ell_1-1} \frac{1}{\ell_1} (n_2-2\ell_2+1)^{n_2} \binom{n_2}{\ell_2-1} \frac{1}{\ell_2}\right] + (n_2+1)^{n_2-1} (n_1-2\ell+1)^{n_1} \binom{n_1}{\ell-1} \right] $$ where $n_1,n_2$ and non-negative integers and $\ell_1, \ell_2$ are positve integers.

I have checked this computationally for small values of $n$ and $\ell$, and it seems to be good.

It is not hypergeometric, but I think it may be "Abel-type" in the sense of this paper.

How could I prove an identity like this?

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Here is a starter. We can simplify the RHS somewhat.

We obtain \begin{align*} \frac{1}{2}& \sum_{{n_1+n_2=n-1}\atop{n_1,n_2\geq 0}}\left[ (n_1+1)^{n_1-1} (n_2-2\ell+1)^{n_2} \binom{n_2}{\ell-1}\right.\\ &\quad + \ell \sum_{{\ell_1+\ell_2=\ell}\atop{\ell_1,\ell_2\geq 1}} (n_1-2\ell_1+1)^{n_1} \binom{n_1}{\ell_1-1} \frac{1}{\ell_1} (n_2-2\ell_2+1)^{n_2} \binom{n_2}{\ell_2-1} \frac{1}{\ell_2}\\ &\quad\left. + (n_2+1)^{n_2-1} (n_1-2\ell+1)^{n_1} \binom{n_1}{\ell-1} \right]\\ &=\frac{1}{2} \sum_{{n_1+n_2=n-1}\atop{n_1,n_2\geq 0}}\left[ (n_1+1)^{n_1-1} (n_2-2\ell+1)^{n_2} \binom{n_2+1}{\ell}\frac{\ell}{n_2+1}\right.\\ &\quad + \frac{\ell}{(n_1+1)(n_2+1)} \sum_{{\ell_1+\ell_2=\ell}\atop{\ell_1,\ell_2\geq 1}}(n_1-2\ell_1+1)^{n_1} \binom{n_1+1}{\ell_1} (n_2-2\ell_2+1)^{n_2} \binom{n_2+1}{\ell_2}\\ &\quad\left. + (n_2+1)^{n_2-1} (n_1-2\ell+1)^{n_1} \binom{n_1+1}{\ell}\frac{\ell}{n_1+1} \right]\\ &\color{blue}{=\frac{\ell}{2} \sum_{{n_1+n_2=n-1}\atop{n_1,n_2\geq 0}}\frac{1}{(n_1+1)(n_2+1)}}\\ &\quad\quad\color{blue}{\cdot\sum_{{\ell_1+\ell_2=\ell}\atop{\ell_1,\ell_2\geq 0}}(n_1-2\ell_1+1)^{n_1} \binom{n_1+1}{\ell_1} (n_2-2\ell_2+1)^{n_2} \binom{n_2+1}{\ell_2}}\\ \end{align*} In the first step we use the binomial identity $\binom{n}{k-1}\frac{1}{k}=\binom{n+1}{k}\frac{1}{n+1}$. In the second step we collect all terms and start the inner sum with indices $\ell_1,\ell_2\geq 0$.

Note: I could not verify the equality of LHS and RHS for all small values. In case of $n=3,l=2$ the LHS $$(n-2\ell+1)^{n-1} \binom{n}{\ell-1}=(3-4+1)^2\binom{3}{1}=0$$ while the RHS is $2$ if I'm not mistaken.

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  • $\begingroup$ The table of lhs-rhs shows growing discrepancy,$$ \begin{array}{ccccc} -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & -2 & 0 & 2 & 0 \\ 2 & -8 & 12 & -8 & -30 \\ 39 & -61 & 6 & -122 & 195 \\ \end{array}$$ $\endgroup$ Commented Jul 16, 2017 at 21:35
  • $\begingroup$ I omitted a factor of $\binom{n-1}{n_1}$ right after the first summation. That is probably causing the discrepancy. Sorry about that! I did get an answer to my original (linked) question, so that also answers this one. $\endgroup$ Commented Jul 25, 2017 at 22:55
  • $\begingroup$ @Drew: No worries. Good that your question could be answered and thanks for the info. $\endgroup$ Commented Jul 30, 2017 at 19:19

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