Multiplying three factorials with three binomials in polynomial identity

Question

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.

This identity appears naturally in trying to answer this other MSE question.

Answer 1

It seems there is a flaw in the stated problem. Maybe you want to check it although some time passed by since it was posted.

The polynomials $Z_n$ with \begin{align*} Z_n=\sum_{k=0}^n n^{\underline{k}}x^{n-k} \end{align*} start with \begin{align*} Z_0&=0^{\underline{0}}x^0=1\\ Z_1&=1^{\underline{0}}x^1+1^{\underline{1}}x^0=x+1\\ Z_2&=2^{\underline{0}}x^2+2^{\underline{1}}x^1+2^{\underline{2}}x^0=x^2+2x+4 \end{align*}

We obtain for $n=2$ \begin{align*} x^2&Z_{n-1}Z_{n-2}-2nxZ_nZ_{n-2}+Z_nZ_{n-1}\\ &=x^2Z_1Z_0-4xZ_2Z_0+Z_2Z_1\\ &=x^2(x+1)-4x(x^2+2x+4)+(x^2+2x+4)(x+1)\\ &=-2x^3-4x^2-10x+4\tag{2} \end{align*}

On the other hand

we obtain in (1) for $n=2$ \begin{align*} D_2&=\sum_{j=0}^{n-2}\sum_{i=0}^j(n-j)!(n-j-1)!(j-i)!\binom{n-2}{j}\binom{j}{i}\binom{2n+1-i-j}{j-i}x^{i+j}\\ &=\sum_{j=0}^{0}\sum_{i=0}^j(2-j)!(1-j)!(j-i)!\binom{0}{j}\binom{j}{i}\binom{5-i-j}{j-i}x^{i+j}\\ &=2!1!0!\binom{0}{0}\binom{0}{0}\binom{5}{0}x^0\\ &=2 \end{align*} which does not coincide with (2).