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Inspired by this question and by using Mathematica the following conjecture seems to be true for all nonzero complex $t$ number: $${_4F_3}\left(\begin{array}c1,1,\tfrac54,\tfrac74\\\tfrac32,2,2\end{array}\middle|\,-t\right) \stackrel{?}{=} \frac{16}{3t}\ln\left(\tfrac14\sqrt{1+\sqrt{1+t}}\left(\sqrt{1+\sqrt{1+t}}+\sqrt{2}\right)\right),$$ where ${_4F_3}$ is a generalized hypergeometric function.

How could we prove this conjectured identity?

Some special cases:

$$\begin{align} {_4F_3}\left(\begin{array}c1,1,\tfrac54,\tfrac74\\\tfrac32,2,2\end{array}\middle|\,-4\right) &\stackrel{?}{=} \frac{4}{3}\ln\left(\frac{\sqrt{\varphi}+\varphi}{2}\right),\\ {_4F_3}\left(\begin{array}c1,1,\tfrac54,\tfrac74\\\tfrac32,2,2\end{array}\middle|\,-8\right) &\stackrel{?}{=} \frac{2}{3}\ln\left(\frac{\sqrt2 + 2}{2}\right),\\ {_4F_3}\left(\begin{array}c1,1,\tfrac54,\tfrac74\\\tfrac32,2,2\end{array}\middle|\,-15\right) &\stackrel{?}{=} \frac{16}{45}\ln\left(\frac{\sqrt{10} + 5}{4}\right),\\ {_4F_3}\left(\begin{array}c1,1,\tfrac54,\tfrac74\\\tfrac32,2,2\end{array}\middle|\,-35\right) &\stackrel{?}{=} \frac{16}{105}\ln\left(\frac{\sqrt{14} + 7}{4}\right),\\ {_4F_3}\left(\begin{array}c1,1,\tfrac54,\tfrac74\\\tfrac32,2,2\end{array}\middle|\,-48\right) &\stackrel{?}{=} \frac{1}{9}\ln 3, \end{align}$$ where $\varphi$ is the golden ratio.

Specially for all $n \neq 1$ nonnegative integers

$${_4F_3}\left(\begin{array}c1,1,\tfrac54,\tfrac74\\\tfrac32,2,2\end{array}\middle|\,1-n^2\right) \stackrel{?}{=} \frac{16}{3n^2-3}\ln\left(\frac{\sqrt{2n+2}+(n+1)}{4}\right).$$

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  • $\begingroup$ The following expansion may be useful: $$hypergeom([1, 1, 5/4, 7/4], [3/2, 2, 2], -t)= \sum _{k=0}^{\infty }1/3\,{\frac { \left( -1 \right) ^{k} \left( 4\,k+ 1 \right) !\, \left( 3+4\,k \right) {16}^{-k}{t}^{k}}{ \left( 2\,k+1 \right) \left( \left( 2\,k \right) ! \right) ^{2} \left( k+1 \right) ^{2}}} $$ and $$ \ln \left( 1/4\,\sqrt {1+\sqrt {1+t}} \right)=-3/2\,\ln \left( 2 \right) +\sum _{k=0}^{\infty }1/8\,{\frac { \left( -1 \right) ^{k} \left( 2\,k+1 \right) !\,{4}^{-k}{t}^{k+1}}{ \left( k! \right) ^{2} \left( k+1 \right) ^{2}}} $$ $\endgroup$ – user64494 Aug 17 '15 at 18:29
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If the parameters $\{b_k\}$ of $_4F_3$ were a little more generic ($b_k\notin \mathbb{Z}$ and $b_j-b_k\notin\mathbb{Z}$) the proof would be very simple: it would suffice to check that the appropriate generalized hypergeometric equation is satisfied and $t\to 0 $ behaviors of both sides match.

In our non-generic case the equation is easily verified but the leading asymptotics does not necessarily fix the solution uniquely. However, we can use that $$\frac{d}{dt}\left(t\,{}_{p+1}F_q\Bigl[\begin{array}{c} 1,a_1,\ldots,a_p\\ b_1,\ldots,b_q \end{array};-t\Bigr]\right)={}_{p+1}F_q\Bigl[\begin{array}{c} 2,a_1,\ldots,a_p\\ b_1,\ldots,b_q \end{array};-t\Bigr].$$ Thus if we denote $\displaystyle f(t):={}_{4}F_3\biggl[\begin{array}{c} 1,1,\frac54,\frac74\\ 2,2,\frac32\end{array};-t\biggr]$, the last equation implies that $$\frac{d}{dt}\left(t\frac{d}{dt}tf(t)\right)={}_2F_1\left( \frac54,\frac74; \frac32;-t\right).\tag{$\spadesuit$}$$ Now let us denote $g(t):=\frac{16}{3t}\ln\left(\tfrac14\sqrt{1+\sqrt{1+t}}\left(\sqrt{1+\sqrt{1+t}}+\sqrt{2}\right)\right)$ and apply the procedure described in the very beginning to show that $$\frac{d}{dt}\left(t\frac{d}{dt}tg(t)\right)={}_2F_1\left( \frac54,\frac74; \frac32;-t\right).\tag{$\clubsuit$}$$ (This is straightforward, as here we have no problem with integer critical exponents. Also, note that the logarithm is killed by the derivatives, so the left side is algebraic).

We have thereby shown that $$t\frac{d}{dt}tf(t)=t\frac{d}{dt}tg(t)+C_1\quad \Longrightarrow\quad f(t)=g(t)+\frac{C_1\ln t+C_2}{t}$$ with some constants $C_1$ and $C_2$. However since both $f(t)$ and $g(t)$ are regular at $t=0$, it follows that $C_1=C_2=0$ and thus $f(t)=g(t)$.

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