Entry 162

There are only four fundamental $d = 4p$ with class number $4$, namely $p=17,73,97,193$.  Given the Kronecker symbol $\big(\frac{d}{m}\big)$, we propose

$$\begin{align}K(k_{17}) &= \frac{\sqrt{2\pi}}{2\sqrt{68}} \frac1{(U_{17})^{1/8}} \left(\sqrt{\frac{-3+\sqrt{17}}8} +\sqrt{\frac{5+\sqrt{17}}8}\right)^{3} \left(\prod_{m=1}^{68}\Big[\Gamma\big(\tfrac{m}{68}\big)\Big]^{\big(\tfrac{-68}{m}\big)}\right)^{\color{red}{1/8}}\\ K(k_{73}) &= \frac{\sqrt{2\pi}}{2\sqrt{292}}  \frac1{(U_{73})^{1/8}} \left(\sqrt{\frac{1+\sqrt{73}}8} +\sqrt{\frac{9+\sqrt{73}}8}\right)^{3} \left(\prod_{m=1}^{292}\Big[\Gamma\big(\tfrac{m}{292}\big)\Big]^{\big(\tfrac{-292}{m}\big)}\right)^{\color{red}{1/8}}\\ K(k_{97}) &= \frac{\sqrt{2\pi}}{2\sqrt{388}} \frac1{(U_{97})^{1/8}} \left(\sqrt{\frac{5+\sqrt{97}}8} +\sqrt{\frac{13+\sqrt{97}}8}\right)^{3}  \left(\prod_{m=1}^{388}\Big[\Gamma\big(\tfrac{m}{388}\big)\Big]^{\big(\tfrac{-388}{m}\big)}\right)^{\color{red}{1/8}}\\ K(k_{193}) &= \frac{\sqrt{2\pi}}{2\sqrt{772}} \frac1{(U_{193})^{1/8}} \left(\sqrt{\frac{22+2\sqrt{193}}8} +\sqrt{\frac{30+2\sqrt{193}}8}\right)^{3} \left(\prod_{m=1}^{772}\Big[\Gamma\big(\tfrac{m}{772}\big)\Big]^{\big(\tfrac{-772}{m}\big)}\right)^{\color{red}{1/8}}\end{align}$$ where the $U_n$ are fundamental units $$U_{17} = 4+\sqrt{17}$$ $$U_{73} = 1068+125\sqrt{73}$$ $$U_{97} = 5604 + 569\sqrt{97}$$ $$U_{193} = {1764132} + 126985\sqrt{193}$$ Going to class number $6$, the red exponent will now be $1/12$.