I. Some results
In a previous post, we considered the complete elliptic integral of the first kind $K(k_{58})$ which is not in Mathworld's limited list of $K(k_d)$. Their list is short, and there is no ordering that gives a pattern. But there is a pattern. For example, we provide $K(k_{23})$ and $K(k_{31})$,
$$\big(2\,K(k_{23})\big)^3 = \frac{P^4}{23^{5/4}(2\pi)^4} \prod_{n=1}^{11}\Gamma\left(\frac{a_n}{23}\right) $$
$$\big(2\,K(k_{31})\big)^3 = \frac{S^4}{31^{5/4}(2\pi)^6} \prod_{n=1}^{15}\Gamma\left(\frac{b_n}{31}\right) $$
where the numerators are,
$$a_n = (1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18)\\ b_n = (1, 2, 4, 5, 7, 8, 9, 10, 14, 16, 18, 19, 20, 25, 28)$$
with plastic constant $P$ and supergolden ratio $S$, the unique real roots of,
$$P^3-P-1=0\\ S^3-S^2-1=0$$
II. Definitions
To find a pattern, we start with prime discriminant $d=4m+3$. Of the two kinds, $8m+3$ and $8m+7$, begin with the latter since it is more well-behaved. Given the class number $n = h(-d),$ an algebraic number $x_d$ of deg $n$, and a version of the Weber modular function,
$$w_d = e^{\pi\,i/24}\frac{\;\eta(\tau)}{\sqrt2\,\eta(2\tau)}$$
where $\tau = \tfrac{1+\sqrt{-d}}2.$ Like in the previous post, consider the product of gamma functions,
$$\prod_{m=1}^{d}\Gamma\left(\frac{m}{d}\right)^{\big(\frac{-d}{\; m}\big)}$$
and the exponent $\big(\frac{-d}{\; m}\big)$ is the Kronecker symbol. (A sample by Wolfram.)
III. Class Number 1
There is only one, namely $d=7$,
$$\prod_{m=1}^{7}\Gamma\left(\frac{m}{7}\right)^{\big(\frac{-7}{\; m}\big)} = x_{7}^{-4}\times\left(K(k_{7})\sqrt{\frac{2\cdot7}{\pi}}\right)^\color{red}2$$
So $x_7$ is an algebraic number of degree $n=1$. In fact, it is just $x_7 = w_7 = 1$. Notice the red exponent is twice the class number $n$.
IV. Class Number 3
There are two, namely $d=23,31,$
$$\prod_{m=1}^{23}\Gamma\left(\frac{m}{23}\right)^{\big(\frac{-23}{\; m}\big)} = x_{23}^{-4}\times\left(K(k_{23})\sqrt{\frac{2\cdot23}{\pi}}\right)^\color{red}6$$
$$\prod_{m=1}^{31}\Gamma\left(\frac{m}{31}\right)^{\big(\frac{-31}{\; m}\big)} = x_{31}^{-4}\times\left(K(k_{31})\sqrt{\frac{2\cdot31}{\pi}}\right)^\color{red}6$$
So $x_{23}$ and $x_{31}$ have degree $n=3$. Surprising, they are squares of well-known cubic roots,
$$x_{23} = (w_{23})^2 = P^2\\ x_{31} = (w_{31})^2 = S^2$$
where $P$ is the plastic constant and $S$ is the supergolden ratio and are the real roots of,
$$P^3-P-1=0\\ S^3-S^2-1=0$$
V. Class Number 5
There are four $d = 47, 79, 103, 127$. For the first two,
$$\prod_{m=1}^{47}\Gamma\left(\frac{m}{47}\right)^{\big(\frac{-47}{\; m}\big)} = x_{47}^{-4}\times\left(K(k_{47})\sqrt{\frac{2\cdot47}{\pi}}\right)^\color{red}{10}$$
$$\prod_{m=1}^{79}\Gamma\left(\frac{m}{79}\right)^{\big(\frac{-79}{\; m}\big)} = x_{79}^{-4}\times\left(K(k_{79})\sqrt{\frac{2\cdot79}{\pi}}\right)^\color{red}{10}$$
As expected, $x_{47}$ and $x_{79}$ have deg $n=5$. But this is where the problem starts since, other than sharing the same number field, there seems to be no easy relation between $x_d$ and $w_d,$
$$x_{47} \neq (w_{47})^\alpha\\ x_{79} \neq (w_{79})^\beta$$
for some rational $(\alpha, \beta).$ In fact, $x_{47}$ and $x_{79}$ are the real roots of,
$$-1 - 6x - 13x^2 - 10x^3 - 2x^4 + x^5 =0\\ -1 - 5x - 2x^2 + 17x^3 - 11x^4 + x^5 =0$$
respectively, while $w_{47}$ and $w_{79}$ are the real roots of,
$$-1 - 2w - 2w^2 - w^3 + w^5 = 0\\ -1 + w - w^2 + 2w^3 - 3w^4 + w^5 =0$$
All are solvable in radicals. For class number $7$, there are five $d = 71, 151, 223, 463, 487.$ And so on.
V. Question
Given prime $d=8k+7$ with class number $n=h(-d)$ and the proposed equation,
$$\prod_{m=1}^{d}\Gamma\left(\frac{m}{d}\right)^{\big(\frac{-d}{\; m}\big)} = x_{d}^{-4}\times\left(K(k_{d})\sqrt{\frac{2d}{\pi}}\right)^{2n}$$
The equation of degree $n$ for the algebraic number $x_d$ can be found using computers. But does $x_d$ in fact have a closed-form in terms of special functions, just like $(x_7, x_{23}, x_{31})$?
