Closed Form for $\sum\limits_{n=-2a}^\infty(n+a){2a\choose-n}^4$

Question

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 $~(4a)!~$

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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 :

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

Answer 1

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.

Answer 2

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 ${n \choose k}={n \choose n-k}$