TITLE: $\prod_k(x\pm k)$ in binomial basis?
QUESTION [18 upvotes]: Let $x$ be an indeterminate and $n$ a non-negative integer. 

Question. The following seems to be true. Is it?
  $$x\prod_{k=1}^n(k^2-x^2)=\frac1{4^n}\sum_{m=0}^n\binom{n-x}m\binom{n+x}{n-m}(x+2m-n)^{2n+1}.\tag1$$

The problem came out of simplifying some work which reduced to the RHS of (1), but with a GOOD fortune this appeared to factor neatly into the LHS.

REPLY [4 votes]: Here is an argument for the leading coefficient (and more).
We use (formal) generating functions.
Let $f_{n,x}(t):= \sum_{m=0}^n {n-x \choose m } {n+x \choose n-m} e^{(x+2m-n)t}$, we are interested
in the polynomial $q_n(x)=\frac{(2n+1)!}{4^n}\;[t^{2n+1}]\,f_{n,x}(t)$.
Clearly 
\begin{align*}
f_{n,x}(t)&=[z^n] e^{tx}\,\left(1+ze^t\right)^{n-x}\left(1+ze^{-t}\right)^{n+x}\\
          &=[z^n]\left(\frac{e^{t}\,(1+ze^{-t})}{1+ze^t}\right)^x \left(1+ 2z\cosh(t) +z^2\right)^{n}\\
          &=[z^n] (g_{t}(z))^x\,\left(1+ 2z\cosh(t) +z^2\right)^{n}
\end{align*}
where  $g_{t}(z):=\frac{e^{t}\,(1+ze^{-t})}{1+ze^t}$.
 Thus (by the Lagrange-Bürmann theorem)
$$G_x(u):=\sum_{n=0}^\infty f_{n,x} u^n = \frac {(g_t(w))^x}{1-u(2\cosh(t)+2w)}$$
where $w=w(u)$  solves $w=u\left(1+w\cosh(t)+w^2\right)$ with $w(0)=0$.
In the sequel let $c:=\cosh(t), s:=\sinh(t)$.
We find 
\begin{align*}
 w(u)&=\frac{1}{2u}-c -  \sqrt{(c-\frac{1}{2u})^2-1)}\\[0.1cm]
 1-2u(c+w)&=\sqrt{(1-2cu)^2 -4u^2}\\[0.1cm]
         %=\sqrt{1-4cu +4s^2 u^2}\\
    g_t(w)&=c-2s^2u +s \sqrt{1-4cu +4s^2 u^2}
\end{align*}
so that
$$G_x(u)=\frac{\left(c-2s^2u +s \sqrt{1-4cu +4s^2 u^2}\right)^x}{\sqrt{1-4cu +4s^2 u^2}}$$
Observe that the numerator of $G_x$ is of the form $(z+\sqrt{1+z^2})^x$ where $z=s \sqrt{1-4cu +4s^2 u^2}$.
Now consider 
$$f(z)=\left(z+\sqrt{1+z^2}\right)^x=\sum_{n\geq 0}c_nz^n$$
Since $f$ satisfies the linear differential equation
$$(1+z^2)f^{\prime\prime}-x^2f=-zf^\prime$$
with $f(0)=1,\,f^\prime(0)=x$, the coefficients satisfy the recursion $c_0=1, c_1=x$ and 
$$(n+2)(n+1) c_{n+2}=(x^2-n^2) c_n\;.$$
Thus the coefficients $c_n$ are of the form $c_n=\frac {p_n(x)}{n!}$, where
each $p_n(x)$ is a monic polynomial of degree $n$ in $x$, and we may write
$$G_x(u)=\sum_{k\geq 0}\frac{s^{k}p_k(x)}{k!}  \left(\sqrt{1-4cu +4s^2 u^2}\right)^{k-1}\;.$$
Now extract $[u^nt^{2n+1}]G_x(u)=\frac{4^n}{(2n+1)!} q_n(x)$.
Since $c$ and $s^2$ are even functions of $t$ only the summands with odd powers of $s$ can have
non-vanishing coefficient of $t^{2n+1}$, and since the series of $s$ starts with $t$ only summands
with $k\leq 2n+1$ contribute.  Thus the highest power of $x$ in $q_n(x)$ is at most $2n+1$ and can only appear in $p_{2n+1}$,
and since $p_{2n+1}$ is monic the coefficient of $x^{2n+1}$ is 
$$\frac{(2n+1)!}{4^n}[t^{2n+1}u^{n}]\frac{s^{2n+1}}{(2n+1)!} \left(1-4cu +4s^2 u^2\right)^{n}=(-1)^n\;,$$
as desired. 
Now Brendan McKay's argument above shows the desired equality.