# Closed form for $\sum_{j=0}^{\infty}{\alpha \choose j} {\beta \choose j}x^j$

As stated, I wonder if there is a closed form for the generating function $$F_{\alpha,\beta}(x):=\sum_{j=0}^{\infty}{\alpha \choose j} {\beta \choose j}x^j$$ where $$\alpha,\beta \in\mathbb{N}$$. Calling this a generating function is slightly misleading since $${\alpha \choose j}=0$$ when $$j>\alpha$$ so this is really a finite sum. A few cases are known already: In the case $$x=1$$ we have that

$$F_{\alpha,\beta}(1)=\sum_{j=0}^{\infty}{\alpha \choose j} {\beta \choose j}={\alpha+\beta \choose \alpha}$$

by Chu-Vandermore. The main approach I would take for this sort of problem would be to use the recurrence relations on binomial coefficients to get a differential question. In this case, this yields the equation

$$\left(x^{2}-x\right)F''\left(x\right)-\left(1+\left(\alpha+\beta-1\right)x\right)F'\left(x\right)+\alpha\beta F\left(x\right)=0$$

This equation is really similar to an Euler differential equation, but the fact that the terms are polynomials with terms of varying degrees messes it up. I can't solve it, and Wolfram Alpha gives a useless answer in terms of a function that takes in $$9$$ arguments and a seperarte function that takes $$4$$. This feels like the sort of problem which would have a nice solution, but fixing $$\alpha$$ and $$\beta$$ and looking at the polynomials generated they do not seem to be very simple nor do they have roots at rational numbers.

• Have you tried the WZ method on a specialization (say $x=2$)? If you don't find a hypergeometric form for the result, that's a pretty compelling argument that there isn't a 'good' formula. Commented Nov 16, 2020 at 2:34
• @StevenStadnicki I've heard of the WZ method, but I've never actually seen it applied... If you think that it is the correct tool for this problem then I will take a stab at it. Commented Nov 16, 2020 at 2:37
• It's certainly the first tool I'd try out, especially since it guarantees finding a closed hypergeometric form for a summation if one exists. In particular, you might want to look at en.wikipedia.org/wiki/Petkov%C5%A1ek%27s_algorithm ... Commented Nov 16, 2020 at 3:36
• Not what was asked, but we do have the generating function $\sum_{\alpha,\beta\geq 0}F_{\alpha,\beta}(x)q^\alpha t^\beta = 1/(1-q-t+qt-qtx)$. Thus the Carnevale-Voll conjecture is about the coefficients of $1/(1-q-t+2qt)$. Commented Nov 16, 2020 at 20:38
• You can use Petkovšek's algorithm (en.wikipedia.org/wiki/Petkovšek%27s_algorithm) to prove that there is no closed form. Of course it depends on what you mean by "closed form". Commented Nov 16, 2020 at 20:51

WolframAlpha immediately gives hypergeometric form $$F_{\alpha, \beta}(x) = {}_2 F_1(-\alpha, -\beta; 1; x)$$.

• That is one solution, but the general one involves the Meijer G function. Why are you assuming the constant multiple of the Meijer G is 0? Commented Nov 16, 2020 at 6:04
• We don't in fact need to solve any DEs, just writing down the definition for ${}_2 F_1$ with these arguments gives formally identical sum. Commented Nov 16, 2020 at 6:15
• Writing this as a hypergeometric series is just restating the formula. It just says that $\binom{\alpha}{j}\binom{\beta}{j} = (-\alpha)_j (-\beta)_j/j!^2$, where $(u)_j$ is the rising factorial $u(u+1)\cdots (u+j-1)$. You don't need Wolfram Alpha to get this. Commented Nov 16, 2020 at 20:45

Here is a "metaproof" that no simple closed form exists.

A conjecture by Carnevale and Voll states that:

For nonnegative integers $$\alpha,\beta$$ with $$\alpha>\beta$$, we have that $$F_{\alpha,\beta}(-1)\neq 0.$$

As far as I know, the conjecture is still open!