Entry 183

The more famous value of \(j_2(\tau)\) is

$$j_2\big(\tfrac12\sqrt{-58}\big) = 396^4$$

So for even \(d\) and since \(K(k) = \frac{\pi}2\,_2F_1\big(\tfrac12,\tfrac12;1,k^2\big)\), we propose a slightly different formula,

$$\,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-d})\big) = \frac{(2^{1/4}g_d)^3}{\;\big(j_2(\tau)\big)^{1/8}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big(j_2(\tau)\big)^n}}$$

with modular lambda function \(\lambda(z)\), Ramanujan g-function \(g_d\) for integer \(d\geq4\) and \(\tau = \frac12{\sqrt{-d}}\),

$$2^{1/4}g_n = \frac{\eta\big(\frac12\sqrt{-d}\big)}{\eta(\sqrt{-d})}$$

For Ramanujan's \(G_d\) and \(g_d\)-functions, he tabulated them for odd and even \(d\), respectively and found,

$$\begin{align}g_6 &=  (1+\sqrt2)^{1/6}\\ g_{10} &=  \left(\tfrac{1+\sqrt{5}}2\right)^{1/2}\\ g_{22} &=  (1+\sqrt2)^{1/2}\\ g_{58} & = \left(\tfrac{5+\sqrt{29}}2\right)^{1/2}\end{align}$$ Therefore we conjecture

$$\begin{align}\,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-6})\big) &= \frac{(2^{1/4}g_{6})^3}{\;\sqrt{4\sqrt3}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{(4\sqrt3)^{4n}}}\\ \,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-10})\big) &= \frac{(2^{1/4}g_{10})^3}{\;\sqrt{12}}\; \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{12^{4n}}}\\ \,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-22})\big) &= \frac{(2^{1/4}g_{22})^3}{\;\sqrt{12\sqrt{11}}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{(12\sqrt{11})^{4n}}}\\ \,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-58})\big) &= \frac{(2^{1/4}g_{58})^3}{\;\sqrt{396}}\; \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{396^{4n}}}\end{align}$$