I was trying to find a closed-form for $0
$\frac{\,_2F_1(\frac{1}{m},\,1-\frac{1}{m},\,1,\,1-x)}{\,_2F_1(\frac{1}{m},\,1-\frac{1}{m},\,1,\,x)} = \sqrt{n}$
where $\,_2F_1(a,b,c,z)$ is the hypergeometric function. There are formulas for $m = 2,3,4,6$, so I was wondering if there are for other m as well. However, one thing I observed was that, let,
$q = \exp\left(\frac{-\,\pi\sqrt{n}}{\sin(\pi/m)}\right)$
Conjecture:
$\lim_{n\to \infty}\frac{x}{q} = \text{constant}$
namely,
$\begin{array}{cc} m&\lim_{n\to \infty}\frac{x}{q}\\\\ 2&16\\ 3&27\\ 4&64\\ 5&25\sqrt{5}\left(\frac{1+\sqrt{5}}{2}\right)^\sqrt{5}=163.95\dots\\ 6&432\\ 7&1152.795095384373\dots\\ 8&2^8\left(1+\sqrt{2}\right)^{2\sqrt{2}}=3096.65\dots\\ \end{array}$
and so on. This implies a good approximation to x in,
$\frac{\,_2F_1(\frac{1}{5},\frac{4}{5},\,1,\,1-x)}{\,_2F_1(\frac{1}{5},\frac{4}{5},\,1,\,x)} = \sqrt{n}$
is given by,
$x \approx 25\sqrt{5}\left(\frac{1+\sqrt{5}}{2}\right)^\sqrt{5} \exp\left(\frac{-\,\pi\sqrt{n}}{\sin(\pi/5)}\right)$
(One can numerically solve for x for a given n using Mathematica's FindRoot command.)
Question: Is the conjecture true? And what is the closed-form for the constant $1152.79509\dots$ when $m=7$?
EDIT:
Courtesy of Sasha’s answer below, then the closed-form for m = 7, as radicals raised to radical powers is,
$ (14)^2 \prod_{k=1}^{3} \frac{1}{\sin(\pi k/7)^{4\cos(2\pi k/7)}} = 1152.79509\dots$
In general,
$\lim_{n\to\infty}\frac{x}{q} = (2m)^2 \prod_{k=1}^{\lfloor (m-1)/2 \rfloor} \frac{1}{\sin(\pi k/m)^{4\cos(2\pi k/m)}} $