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Is there any formula that gives the following specific product of two Hypergeometric functions $$_2 F_1(\alpha, \alpha+1 ; m ; z)\, _2 F_1(\alpha, \alpha+1 ; n ; z)\, , $$ for $m, n $ positive integers and $\alpha$ a positive half integer?

I am aware of the general formula in terms of a sum of $_4 F_3$'s, but I am looking for a formula that contains just a single $_p F_q$ which should exist according to "Higher Transcendental Functions, Vol. 1" by A. Erdelyi page 185.

P.S.

There is some physics behind this problem. It is a piece of a kernel in the energy basis arising in the study of a problem in Quantum Mechanics.

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  • $\begingroup$ If $m=n$ you can express the product in terms of the Appell $F_4$ function. See here for more details: arxiv.org/abs/0804.0655 $\endgroup$
    – Aaron Hendrickson
    Commented Mar 11, 2021 at 0:25

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If you allow some relation between $m$ and $n$, then it is possible.

Let $m+n=2\alpha+2$. Then \begin{align} &_2 F_1(\alpha, \alpha+1 ; m ; z)\, _2 F_1(\alpha, \alpha+1 ; n ; z) \\ =&\,\,_4F_3(\alpha,\alpha+1,\alpha+\frac12,\alpha+1;2\alpha+1,m,2\alpha-m;4z(1-z)).\end{align}

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  • $\begingroup$ I still think a more general formula should exist... Also why the downvote?? $\endgroup$
    – Panagiotis Betzios
    Commented Feb 3, 2017 at 14:39

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