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Jun 29, 2019 at 2:46 answer added Ira Gessel timeline score: 5
Jun 28, 2019 at 14:06 vote accept Max Alekseyev
Jun 28, 2019 at 12:40 answer added Johannes Trost timeline score: 5
Jun 28, 2019 at 11:32 comment added Max Alekseyev @JohannesTrost: Nice catch! Please add this as an answer - I will accept it. Do they give any proof idea?
Jun 28, 2019 at 9:09 comment added Johannes Trost typo in my comment : the second $_2F_0$ has argument $-z$.
Jun 28, 2019 at 8:53 comment added Johannes Trost In "Higher Trancendental Functions", Vol 1, by A. Erdelyi (ed.), on page 86 equation (4) says $_2F_0(\alpha,\beta;z) \ _2F_0(\alpha,\beta;z) = _4F_1(\alpha, \beta, \frac{1}{2} (\alpha+\beta), \frac{1}{2} (\alpha+\beta+1); \alpha+\beta; 4 z^2)$, from which your formula can be derived by setting $\alpha=-n$, $\beta=n+1$.
Jun 28, 2019 at 3:41 history edited user64494 CC BY-SA 4.0
A typo in the title is corrected.
Jun 27, 2019 at 21:54 answer added Robert Israel timeline score: 3
S Jun 27, 2019 at 21:26 history suggested Ali Taghavi
I add a tag.
Jun 27, 2019 at 21:26 review Suggested edits
S Jun 27, 2019 at 21:26
Jun 27, 2019 at 21:20 history asked Max Alekseyev CC BY-SA 4.0