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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.

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    $\begingroup$ 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. $\endgroup$
    – Steven Stadnicki
    Nov 16, 2020 at 2:34
  • $\begingroup$ @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. $\endgroup$
    – Milo Moses
    Nov 16, 2020 at 2:37
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    $\begingroup$ 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 ... $\endgroup$
    – Steven Stadnicki
    Nov 16, 2020 at 3:36
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    $\begingroup$ 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 $1/(1-q-t+2qt)$. $\endgroup$
    – Richard Stanley
    Nov 16, 2020 at 20:38
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    $\begingroup$ 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". $\endgroup$
    – Ira Gessel
    Nov 16, 2020 at 20:51

2 Answers 2

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

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  • $\begingroup$ 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? $\endgroup$
    – Milo Moses
    Nov 16, 2020 at 6:04
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    $\begingroup$ 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. $\endgroup$
    – Mikhail Tikhomirov
    Nov 16, 2020 at 6:15
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    $\begingroup$ 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. $\endgroup$
    – Ira Gessel
    Nov 16, 2020 at 20:45
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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!

For recent work in this direction see this article by Habsieger.

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  • $\begingroup$ That's fascinating (+1). Thank you very much for bringing this up. $\endgroup$
    – Milo Moses
    Nov 16, 2020 at 20:45
  • $\begingroup$ No pbm. For another conjecture in similar vein (probably harder) see the end of this article arxiv.org/abs/1903.11147 $\endgroup$
    – Abdelmalek Abdesselam
    Nov 16, 2020 at 20:55
  • $\begingroup$ Thanks for the reference! Would you mind clarifying what sort of non-existence results does the conjecture imply? $\endgroup$
    – Mikhail Tikhomirov
    Nov 16, 2020 at 21:19
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    $\begingroup$ @Mikhail: loosely speaking it implies that there’s no known closed form that would make it easy to answer this question. E.g. no expression as a product (unlike, say, the Vandermonde determinant). $\endgroup$
    – Qiaochu Yuan
    Nov 17, 2020 at 0:05
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    $\begingroup$ @MikhailTikhomirov: My answer was just an attempt at making a, hopefully instructive, joke. This is not a rigorous nonexistence proof. Just saying that if a simple expression was know, then the conjecture would have been solved (trivially), which is not the case... $\endgroup$
    – Abdelmalek Abdesselam
    Nov 17, 2020 at 0:12

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