TITLE: Closed Form for $\sum\limits_{n=-2a}^\infty(n+a){2a\choose-n}^4,~a\not\in\mathbb Z$
QUESTION [11 upvotes]: 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.

REPLY [12 votes]: 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...