Entry 169

Recall Ramanujan's $G$-function $$2^{1/4}G_n=\frac{\eta^2(\tau)}{\eta(\tfrac{\tau}2)\,\eta(\tau)}$$ where $\tau=\sqrt{-n}$. There are only three fundamental $d=4p$ with class number $2$ for prime $p$, namely $p = 5,13,37$. Hence $$\begin{align}G_5 &= \left(\frac{1+\sqrt5}2\right)^{1/4}\\ G_{13} &= \left(\frac{3+\sqrt{13}}2\right)^{1/4}\\ G_{37} &=\, \big(6+\sqrt{37}\big)^{1/4}\end{align}$$ Going higher, there are only four fundamental $d=4p$ with class number $4$ for prime $p$, namely $p = 17,73,97,193$.

$$\begin{align}G_{17} &= \sqrt{\frac{-3+\sqrt{17}}8} +\sqrt{\frac{5+\sqrt{17}}8} \\ G_{73} &= \sqrt{\frac{1+\sqrt{73}}8} +\sqrt{\frac{9+\sqrt{73}}8} \\ G_{97} &= \sqrt{\frac{5+\sqrt{97}}8} +\sqrt{\frac{13+\sqrt{97}}8} \\ G_{193} &=  \sqrt{\frac{22+2\sqrt{193}}8} +\sqrt{\frac{30+2\sqrt{193}}8} \end{align}$$ all of which were already known to Ramanujan. But as was shown in Entry 161 and Entry 162, it turns out these also appear in the closed-form of the complete elliptic integral of the first kind $K(k_n)$. The next entry will be for class number $6$ where one had to extract $4$th roots again.