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I wonder if there is a closed formula for the following generalized hypergeometric function at $z=1$:

$${}_4F_3\left(a,a,a,a;a+1,a+1,2a;1\right)$$

Specifically, I would like to have a formula in terms of gamma functions and its derivatives, analogous to the identity

$${}_3F_2\left(a,a,a;a+1,2a;1\right) = \frac{\Gamma(2a+1)}{4\Gamma(a)^2}\left[\psi'\left(\frac{a}{2}\right)-\psi'\left(\frac{a+1}{2}\right)\right]\,,$$

where $\psi(z)=\Gamma'(z)/\Gamma(z)$.

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    $\begingroup$ at integer $a$ it evaluates to $\alpha+\beta\pi^2+\gamma\zeta(3)$, with rational $\alpha,\beta,\gamma$. $\endgroup$
    – Carlo Beenakker
    Nov 3, 2017 at 9:33
  • $\begingroup$ @Carlo Beenakker: Indeed, and this is one reason I think it should be expressible using $\Gamma$ and $\psi^{(n)}$ with $n\leq 2$. $\endgroup$
    – D M
    Nov 3, 2017 at 14:29
  • $\begingroup$ I see no reason that: projecteuclid.org/download/pdf_1/euclid.bams/1183507351 Eq(15) as indicated by section 7 p=q+1 doesn't reduce your equation to 3F2(a,a,a;a+1,2a;1) ?? If that fails I have another paper (somewhere) that deals with things like a_k-b_j= integer. – $\endgroup$
    – rrogers
    Jun 7, 2019 at 1:20

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