Proving that an alternative sum of factorials is zero

Question

I have to prove that for every $k$ the following differential equation is fulfilled by monomials of even degree less than $2k$: $$ \sum_{j=1}^ka_j^{(k)}x^{j-1}f^{(j)}(x)=0, $$ with $a_j^{(k)}:=\frac{(2k-j-1)!}{(j-1)!\,(k-j)!\,(-2)^{k-j}}$. By linearity I can assume $f(x)=x^{2h}$ for some $h=0,...,k-1$. If I put it into the equation I obtain something equivalent to $$ \sum_{j=1}^{k}(-2)^{j}\frac{(2k-j-1)!}{(j-1)!(k-j)!}(2h)_j=\sum_{j=1}^{min(k,2h)}(-2)^{j}\frac{(2k-j-1)!(2h)!}{(j-1)!(k-j)!(2h-j)!}, $$ where $(x)_n$ means the falling factorial or the Pochhammer symbol, then computing initial cases I get $0$. How to prove that it is always $0$?
My attempt is by induction: the case $h=0$ is obvious. Now, suppose that for some $h=0,...,k-2$ it holds $\sum_{j=1}^{k}(-2)^{j}\frac{(2k-j-1)!}{(j-1)!(k-j)!}(2h)_j=0$, then for $h+1$ we have $$ \begin{split} &\sum_{j=1}^{k}(-2)^{j}\frac{(2k-j-1)!}{(j-1)!(k-j)!}(2h+2)_j, \end{split} $$ but I cannot use my induction hypothesis. Moreover, something has to tell me that for $h=k-1$ the induction should stop working, since for $h=k$ the sum is not zero.
Another way to solve the problem is trying to prove that the sum becomes polynomial in $h$, with $\{0,1,...,k-1\}$ as its roots: $$ \sum_{j=1}^{k}(-2)^{j}\frac{(2k-j-1)!}{(j-1)!(k-j)!}(2h)_j=(-4)^k\prod_{j=0}^{k-1}(h-j)=(-4)^k(h)_{k+1}. $$ For this purpose, I've tried to compute the coefficients of those polynomials: it is easy to show that $a_0=0$ for the coefficient $a_l$ it is enough to derive the formal expression $l$-times and then evaluate in $h=0$. For example $a_1=(-4)^k(k-1)!$ for both, but the $l$-derivative is hard to compute.
However, this second way is preferable, since it proves that the sum is zero if and only if $0<h<k-1$.
If I try to compute the sum with Wolfram alpha it gives me something related to the hypergeometric function (https://en.wikipedia.org/wiki/Hypergeometric_function), but I don't know how to use that information.

Answer 1

I have managed to at least find a bad solution, which is slightly too long for a comment.

• By WolframAlpha the sum is junk times $_2F_1(1-2h, 1-k, 2-2k; 2)$.
• By table lookup, if the argument (last input) to $_2F_1$ is $2$, the third parameter is twice the second, and the first parameter is a negative integer $-n$, then it is more junk times $1+(-1)^n$.
• Since our $n$ is $2h-1$, this is zero.

This method is perhaps even more unsatisfying than using Gosper's algorithm, although it avoids having to learn it if you're not already familiar. (I should say that I didn't bother running Gosper to see that it actually works, but it seems wildly unlikely to me that it wouldn't. The sum is clearly hypergeometric and, as you noted, the output is even polynomial in $h$. So the algorithm certainly succeeds, and it seems unlikely that the output would be very heavily disguised.)