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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 expression1 in terms of the general parameter $a\not\in\mathbb Z$ ?

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


For somewhat similar expressions with 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 :

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


This question has already been asked on Mathematics Stack Exchange, where, despite having been posted a considerable amount of time ago, and also put up for a bounty, it nonetheless still failed to receive a satisfying answer.

1 As opposed to, say, a mere rewriting in terms of $($generalized$)$ hypergeometric functions, which, as already pointed out in the comment section to the original post, constitutes nothing more than a formal exercise in mathematical taxonomy.

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    $\begingroup$ Formally, $S_4^-(a)$ is equal to $-a\,{}_5F_4(-a+1, -2a, -2a, -2a, -2a; -a, 1, 1, 1;1)$. This can be summed by Dougall's very well-poised $_5F_4$ summation formula (dlmf.nist.gov/16.4#E9) to give $$-\frac{a\Gamma(1+4a)}{\Gamma(1+2a)^3 \Gamma(1-2a)}.$$ $\endgroup$
    – Ira Gessel
    Dec 9 '19 at 17:07
  • $\begingroup$ Unfortunately, the command of Maple restart;s1:=sum((k + a)*binomial(2*a, -k)^4, k = -2*a .. infinity) assuming a>0, a::Not(integer); produces an incorrect result and the command of Mathematica Sum[(n + a)*Binomial[2*a, -n]^4, {n, -2*a, Infinity}, Assumptions -> a [Element] Reals && a [NotElement] Integers] performs the answer in terms of a hypergeometric function. $\endgroup$
    – user64494
    Dec 9 '19 at 17:50
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    $\begingroup$ In other words, $~S_3^-(a) = -\dfrac{\sin(2a\pi)}{2\pi}~$ and $~S_4^-(a) = -\displaystyle\frac{\sin(2a\pi)}{2\pi}\cdot{4a\choose 2a}~$ $\endgroup$
    – Lucian
    Dec 9 '19 at 17:51
  • $\begingroup$ @IraGessel: Feel free to post an answer both here and on the original post. $\endgroup$
    – Lucian
    Dec 10 '19 at 3:39
  • $\begingroup$ Just to be clear ... what is summation from $n=-2a$ when $a \notin \mathbb Z$? $\endgroup$ Dec 15 '19 at 13:10
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Some experimentation suggests that the second series is given by $$ S_4^-(a) = \sum_{n=-2a}^\infty (n+a) \binom{2a}{-n}^4 = \frac{1}{4\cos(2\pi a) \,\Gamma(2a+1)^2\,\Gamma(-4a)}, $$ which agrees with Ramanujan's at $a = -1/8$, but I have no proof...

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  • $\begingroup$ Despite having installed Maple on my computer ages ago, for precisely this very reason (i.e., to double-check expressions Mathematica was not able to express in closed form), it somewhat never dawned on me, during all this time, to pass the summation through it. Seems so silly, in retrospect, to have committed such an obvious blunder. $\endgroup$
    – Lucian
    Dec 9 '19 at 17:26
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    $\begingroup$ The RHS satisfies $f(a+1) = \frac{f(a) (-4a)}{(2a+1)^2(2a+2)^2}$, check the LHS $g(a)$ satisfies the same, thus $h(a)=g(a)\Gamma(2a+1)^2\Gamma(-4a)-\frac1{4\cos(2\pi a)}$ is $1$-periodic, check it is entire function, show $\frac{h(a)}{a}$ vanishes as $a\to i\infty$ which implies $h(a)=O(a)$ thus $h(a)=h(0)+ah'(0)=h(0)$. @Lucian $\endgroup$
    – reuns
    Dec 9 '19 at 22:01
  • $\begingroup$ @reuns: Feel free to post an answer both here and on the original post. $\endgroup$
    – Lucian
    Dec 10 '19 at 3:40

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