I want to show: for any $1\leq i \leq k-1$, and $1\leq j\leq k-1$, $$c_{ij} = 0,$$ where: $$ c_{ij} = k \sum_{\ell=1}^{k-1} \frac{1}{\ell} \left(-1\right)^{\ell-i}\sum_{m=1}^{\min\{\ell,k-\ell\}}\binom{\ell}{m}\binom{k-\ell-1}{m-1}\sum_{c=0}^m\binom{m}{c} \binom{k-2m}{i + j -2c -\ell}\binom{m}{\ell+c-i}. $$ I think this is equivalent to: $$ c_{ij} = k\sum_{\ell=1}^{k-1}\sum_{m=0}^{\ell}\sum_{c=0}^m\frac{1}{\ell} \left(-1\right)^{\ell-i} \binom{\ell}{m}\binom{k-\ell-1}{m-1}\binom{m}{c} \binom{k-2m}{i + j -2c -\ell}\binom{m}{\ell+c-i}. $$ This should be numerically correct (verified by numerical method.) But I have been stuck for a month to prove this. I have tried induction on $i,j,k$, and hypergeometric functions. I think Egorychev method should help, but this equation involves 5 binomial coefficients, which I don't know how to solve. I really appreciate your help!
More details: $c_{i,j}$ is the coefficient of a polynomial.
Method 1-Egorychev method(I tried): Using Egorychev Method we have: $$ \binom{k-\ell-1}{m-1} = \frac{1}{2\pi \mathbf{i}} \int_{|w_2|=\epsilon}\frac{\left(1+w_2\right)^{k-\ell-1}}{w_2^{m}}d w_2 $$ $$ \binom{k-2m}{i + j -2c -\ell}= \frac{1}{2\pi \mathbf{i}} \int_{|z_2|=\epsilon}\frac{\left(1+z_2\right)^{k-2m}}{z_2^{i + j -2c -\ell+1}}dz_2 $$ $$ \binom{m}{\ell+c-i} = \frac{1}{2\pi \mathbf{i}} \int_{|z_3|=\epsilon}\frac{\left(1+z_3\right)^m}{z_3^{\ell+c-i+1}}dz_3 $$ Then, I can re-write $c_{ij}$ as: \begin{align*} c_{ij} &= k \sum_{\ell=1}^{k-1} \frac{1}{\ell} \left(-1\right)^{\ell-i} \sum_{m=1}^{\ell} \binom{\ell}{m}\binom{k-\ell-1}{m-1}\sum_{c=0}^m\binom{m}{c} \binom{k-2m}{i + j -2c -\ell}\binom{m}{\ell+c-i}\\ & = \frac{1}{2\pi \mathbf{i}} \int_{|z_2|=\epsilon}\frac{\left(1+z_2\right)^{k}}{z_2^{i + j+1}} \frac{1}{2\pi \mathbf{i}} \int_{|z_3|=\epsilon}\frac{1}{z_3^{-i+1}}\sum_{\ell=1}^{k-1} \left( \frac{z_2}{z_3}\right)^{\ell}\left(-1\right)^{\ell} \\ & ~~~~ \sum_{m=0}^{\ell} \frac{k}{\ell}\binom{\ell}{m}\binom{k-\ell-1}{m-1} \left(\frac{1+z_3}{\left(1+z_2\right)^{2}}\left(1+\frac{z_2^2}{z_3}\right)\right)^m dz_2 dz_3\\ & = \sum_{\ell=1}^{k-1} \frac{k}{\ell} \frac{1}{2\pi \mathbf{i}} \int_{|z_2|=\epsilon}\frac{\left(1+z_2\right)^{k}}{z_2^{i + j+1}} \frac{1}{2\pi \mathbf{i}} \int_{|z_3|=\epsilon}\frac{1}{z_3^{-i+1}} \\ & ~~~~ \frac{1}{2\pi \mathbf{i}} \int_{|w_2|=\epsilon}\left(1+w_2\right)^{k-\ell-1}\left(1+\frac{1}{w_2} \frac{z_2^2+z_3+z_2^2z_3+z_3^2}{z_3\left(1+z_2\right)^{2}} \right)^{\ell}\left( \frac{z_2}{z_3}\right)^{\ell}\left(-1\right)^{\ell} d w_2 d z_2 d z_3\\ & = \sum_{\ell=1}^{k-1} \frac{k}{\ell} \frac{1}{2\pi \mathbf{i}} \int_{|z_2|=\epsilon}\frac{\left(1+z_2\right)^{k}}{z_2^{i + j+1}} \frac{1}{2\pi \mathbf{i}} \int_{|z_3|=\epsilon}\frac{1}{z_3^{-i+1}} \\ & ~~~~ \frac{1}{2\pi \mathbf{i}} \int_{|w_2|=\epsilon}\left(1+w_2\right)^{k-\ell-1}\left(-\frac{z_2}{z_3}-\frac{z_2}{w_2} \frac{z_2^2+z_3+z_2^2z_3+z_3^2}{z_3^2\left(1+z_2\right)^{2}} \right)^{\ell} d w_2 d z_2 d z_3\\ \end{align*} Not sure how to do it next.
Method 2: Induction on $i,j,k$ $$ h_{m,\ell,i,j,k} = \sum_{c=0}^m \binom{m}{c} \binom{k-2m}{i + j -2c -\ell}\binom{m}{\ell+c-i} $$ And $$ h_{m,\ell,i,j,k+1} =h_{m,\ell,i,j,k}+h_{m,\ell,i,j-1,k} $$ When $\ell > m$, we have the: $$ h_{m,\ell,i+1,j} = h_{m,\ell-1,i,j} $$ Thus, we have: \begin{align*} c_{i,j,k} & = k\sum_{\ell=1}^{k-1}\frac{1}{\ell} \sum_{m=0}^{\ell} h_{m,\ell,i,j,k}\binom{\ell}{m}\binom{k-\ell-1}{m-1} \end{align*} \begin{align*} c_{i+1,j,k} & = k\sum_{\ell=1}^{k-1}\frac{1}{\ell} \sum_{m=0}^{\ell} h_{m,\ell-1,i,j}\binom{\ell}{m}\binom{k-\ell-1}{m-1} \\ & = k\sum_{\ell=1}^{k-1}\frac{1}{\ell} \sum_{m=0}^{\ell} h_{m,\ell-1,i,j}\binom{\ell}{m}\binom{k-\ell-1}{m-1} \\ \end{align*} More complicated.... if we look into the difference between the two equations.
I really appreciate the help!!!!
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