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I have checked the following identity (1) below for $n\leq 40$ with a computer. Let $(n)_k$ denote the falling factorial $n(n-1)\ldots (n-k+1)$, let $Z_n=\sum_{k=0}^n (n)_k x^{n-k}$, and finally let $D_n=x^2Z_{n-1}Z_{n-2}-2nxZ_nZ_{n-2}+Z_nZ_{n-1}$. Then, I conjecture that

$$ D_n=\sum_{0 \leq i \leq j \leq n-2} (n-j)!(n-j-1)!(j-i)! \binom{n-2}{j}\binom{j}{i}\binom{2n+1-i-j}{j-i}x^{i+j}\tag{1} $$

I tried my usual tools (induction, WS method, using similar identities) and failed. Any help appreciated.

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  • $\begingroup$ Did you try to find coefficients of $D_n$ in oeis.org ? $\endgroup$ Mar 31, 2015 at 6:54
  • $\begingroup$ @AlexeyUstinov Which coefficients ? $D_n$ is a sum on two indexes $i$ and $j$, and the summand involves extra parameters $n$ and $x$ in addition to $i$ and $j$. $\endgroup$ Mar 31, 2015 at 7:06
  • $\begingroup$ You have just a sequence of polynomials: $D_2=2$, $D_3=2(6+6x+x^2)$,... And they have coefficients 2, 2, 12, 12, ... I've checked. The answer is "no". $\endgroup$ Mar 31, 2015 at 7:25
  • $\begingroup$ But this triangle of numbers is good enough. At least first columns and dioganals are simple. $\endgroup$ Mar 31, 2015 at 7:40
  • $\begingroup$ @AlexeyUstinov Every column, row and diagonal is a polynomial. $\endgroup$ Mar 31, 2015 at 10:36

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