Entry 161

Entry 160 dealt with discriminants \(d\) with class number \(1\). Going higher, there are only three fundamental \(d = 4p\) with class number \(2\), namely \(p=5,13,17\).  Given the Kronecker symbol \(\big(\frac{d}{m}\big)\), we propose

$$\begin{align}K(k_5) &= \frac{\sqrt{2\pi}}{2\sqrt{20}}\left(\frac{1+\sqrt5}2\right)^{3/4}\left(\prod_{m=1}^{20}\Big[\Gamma\big(\tfrac{m}{20}\big)\Big]^{\big(\tfrac{-20}{m}\big)}\right)^{\color{red}{1/4}}\\ K(k_{13}) &= \frac{\sqrt{2\pi}}{2\sqrt{52}}\left(\frac{3+\sqrt{13}}2\right)^{3/4}\left(\prod_{m=1}^{52}\Big[\Gamma\big(\tfrac{m}{52}\big)\Big]^{\big(\tfrac{-52}{m}\big)}\right)^{\color{red}{1/4}}\\ K(k_{37}) &= \frac{\sqrt{2\pi}}{2\sqrt{148}}\left(6+\sqrt{37}\right)^{3/4}\left(\prod_{m=1}^{148}\Big[\Gamma\big(\tfrac{m}{148}\big)\Big]^{\big(\tfrac{-148}{m}\big)}\right)^{\color{red}{1/4}}\end{align}$$Going to class number \(4\), the red exponent will now be \(1/8\).