I was trying to find a closed-form for $0<x<1$ in,

$$\frac{\,_2F_1(\frac{1}{m},\,1-\frac{1}{m},\,1,\,1-x)}{\,_2F_1(\frac{1}{m},\,1-\frac{1}{m},\,1,\,x)} = \sqrt{n}$$

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,

$$\lim_{n\to\infty}\frac{x}{q} = (2m)^2 \prod_{k=1}^{\lfloor (m-1)/2 \rfloor} \frac{1}{\sin(\pi k/m)^{4\cos(2\pi k/m)}} $$