Entry 179

Continuing from Entry 178, we select some \(d=4m\) with class number \(h(-d) = 2\). It is known for \(m = 5,13,17\) 

$$\begin{align}G_{5} &= \left(\frac{1+\sqrt{5}}2\right)^{1/4}\\ G_{13} &= \left(\frac{3+\sqrt{13}}2\right)^{1/4}\\ \,G_{37} &=\, \big(6+\sqrt{37}\big)^{1/4}\end{align}$$

and calculate the j-function at the points,

$$\begin{align}j\big(\tfrac{1+\sqrt{-5}}2\big) &= 2^3(25-13\sqrt{5})^3\\ j\big(\tfrac{1+\sqrt{-13}}2\big) &= 30^3(31-9\sqrt{13})^3\\ j\big(\tfrac{1+\sqrt{-37}}2\big) &= 60^3(2837-468\sqrt{37})^3\end{align}$$

which are all negative values just like in the previous entry. The proposed relation has a bound \(m\geq7\), so the first value can't be used. But we conjecture 

$$K(k_{13}) = \frac{\pi}2  \frac{\big(3+\sqrt{13}\big)^{1/2}}{\big({-30^3}(31-9\sqrt{13})^3 \big)^{1/12}}\, \sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big(30^3(31-9\sqrt{13})^3\big)^n}}$$

$$K(k_{37}) = \frac{\pi}2  \frac{2^{3/4}\big(6+\sqrt{37}\big)^{3/4}}{\big({-60^3}(2837-468\sqrt{37} \big)^{1/12}}\, \sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big({60^3(2837-468\sqrt{37}}\big)^n}}$$

and so on for other \(j\big(\tfrac{1+\sqrt{-m}}2\big)\) with \(m\geq7\).