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Markus Scheuer
Markus Scheuer
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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.\\ &\qquad + \ell \sum_{{\ell_1+\ell_2=\ell}\atop{\ell_1,\ell_2\geq 1}} \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]\\ &\qquad\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.\\ &\qquad + \frac{\ell}{(n_1+1)(n_2+1)} \sum_{{\ell_1+\ell_2=\ell}\atop{\ell_1,\ell_2\geq 1}} \left[(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}\right]\\ &\qquad\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)}}\\ &\qquad\qquad\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*}\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.

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.\\ &\qquad + \ell \sum_{{\ell_1+\ell_2=\ell}\atop{\ell_1,\ell_2\geq 1}} \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]\\ &\qquad\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.\\ &\qquad + \frac{\ell}{(n_1+1)(n_2+1)} \sum_{{\ell_1+\ell_2=\ell}\atop{\ell_1,\ell_2\geq 1}} \left[(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}\right]\\ &\qquad\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)}}\\ &\qquad\qquad\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.

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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Markus Scheuer
Markus Scheuer
  • 516
  • 6
  • 10

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.\\ &\qquad + \ell \sum_{{\ell_1+\ell_2=\ell}\atop{\ell_1,\ell_2\geq 1}} \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]\\ &\qquad\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.\\ &\qquad + \frac{\ell}{(n_1+1)(n_2+1)} \sum_{{\ell_1+\ell_2=\ell}\atop{\ell_1,\ell_2\geq 1}} \left[(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}\right]\\ &\qquad\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)}}\\ &\qquad\qquad\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.