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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.

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A partial answer.

The coefficient of $z^u$ in $\phantom{}_2 F_1(\alpha,\alpha+1;m;z)$ is given by: $$ \frac{(-1+m)! (-1+a+u)! (a+u)!}{(-1+a)! a! u! (-1+m+u)!} $$ hence the coefficient of $z^u$ in $\phantom{}_2 F_1(\alpha,\alpha+1;m;z)\;\phantom{}_2 F_1(\alpha,\alpha+1;n;z)$ is given by:

$$ \sum_{k=0}^{u}\frac{(-1+m)! (-1+a+k)! (a+k)!}{(-1+a)! a! k! (-1+m+k)!}\cdot \frac{(-1+n)! (-1+a+u-k)! (a+u-k)!}{(-1+a)! a! (u-k)! (-1+n+u-k)!}$$ that should be manageable through Vandermonde's or Dixon's identity.

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    I was reluctant to use this since you get 8 terms that contain $k$ which is to be summed over, but maybe I can merge them to less terms. Anyhow I will give it a shot... – 2017-02-02