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Do either $~S_4^+(a)~=~\displaystyle\sum_{n=0}^\infty(n+a){2a\choose n}^4~$ or $~S_4^-(a)~=~\displaystyle\sum_{n=-2a}^\infty(n+a){2a\choose-n}^4~$ possess a meaningful closed form expression in terms of the general parameter a ?

Ramanujan provided the following result : $~S_4^-\Big(-\tfrac18\Big)~=~\dfrac1{\bigg[\Big(-\tfrac14\Big){\large!}\bigg]^2~\sqrt{8\pi}}~,~$ which would point to a possible closed form expression in terms of $~(2a)!~$ and/or $~(4a)!~$


For lesser values of the exponent, we have Dixon's identity :

$$\sum_{n=0}^\infty(-1)^n{2a\choose n}^3 ~=~ \sum_{n=-2a}^\infty(-1)^n{2a\choose-n}^3 ~=~ \cos(a\pi)~{3a\choose a,a},$$

and Vandermonde's identity :

$$\sum_{n=0}^\infty(-1)^n{2a\choose n}^2 ~=~ \sum_{n=-2a}^\infty(-1)^n{2a\choose-n}^2 ~=~ \cos(a\pi)~{2a\choose a},$$

$$\sum_{n=0}^\infty{a\choose n}^2 ~=~ \sum_{n=-a}^\infty{a\choose-n}^2 ~=~ {2a\choose a},$$

as well as the binomial theorem :

$$\sum_{n=0}^\infty{a\choose n}^1x^n ~=~ (1+x)^a,\qquad\sum_{n=-a}^\infty{a\choose-n}^1x^n ~=~ \Big(1+\tfrac1x\Big)^a.$$

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    $\begingroup$ First of all, good to see you back here ! I suppose that you do not want hypergeometric functions. Right ? Cheers $\endgroup$ – Claude Leibovici Oct 28 '19 at 7:28
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    $\begingroup$ @ClaudeLeibovici: Same here. :-) And no, since the latter are simply an exercise in mathematical taxonomy, so to say. $\endgroup$ – Lucian Oct 28 '19 at 7:32
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    $\begingroup$ I was sure of that. $\endgroup$ – Claude Leibovici Oct 28 '19 at 7:35
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    $\begingroup$ Now posted to MO, mathoverflow.net/questions/347906/… $\endgroup$ – Gerry Myerson Dec 9 '19 at 0:49
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When $2a$ is a non-negative integer: $$S=\sum_{n=-2a}^{\infty} (n+a) {2a \choose -n}^4= \sum_{n=0}^{2a}(-n+a) {2a \choose n}^4=\sum_{n=0}^{2a}~(-(2a-n)+a)~{2a\choose n}^4$$ $$=\sum_{n=0}^{2a} (n-a) {2a \choose n}^4~ =-S.$$ So $$S=-S \implies S=0$$.Here we have used : ${n \choose-k}= 0~$ if $k\in I^+$, $\sum_{k=0}^{n} f(k) =\sum_{k=0}^n f(n-k)$, and ${n \choose k}={n \choose n-k}$

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    $\begingroup$ For some mysterious reason, this answer seems to assume that the two parameters are integers. As can be clearly inferred from the question's second paragraph, this is not necessarily the case. $\endgroup$ – Lucian Oct 28 '19 at 9:44
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    $\begingroup$ @Lucian OK, I have reduced ny claim now in the edit. $\endgroup$ – Z Ahmed Oct 28 '19 at 9:49
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    $\begingroup$ He solved a case he could handle. $\endgroup$ – marty cohen Nov 25 '19 at 20:08
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I am posting this answer, so as to remove this question from the ever-growing Unanswered Queue, where, thanks to the substantial contribution of the mathematical community at Math Overflow, it no longer belongs.

$$S_4^-(a)~=~S_3^-(a)\cdot{4a\choose2a}~=~-\frac{\sin(2a\pi)}{2\pi}\cdot{4a\choose2a}$$

See Dougall's very well-poised summation formula; also, notice that both sides satisfy the same recursion formula.

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