Entry 163

There are only seven fundamental \(d = 4p\) with class number \(6\), namely \(p=29, 53, 61, 109, 157, 277, 397\).  To prevent clutter, only the first three will be stated. Given the Kronecker symbol \(\big(\frac{d}{m}\big)\), we propose

$$\begin{align}K(k_{29}) &= \frac{\sqrt{2\pi}}{2\sqrt{116}} \frac1{\;(x_1)^{1/3}} \left(\frac{y_1^2+\sqrt{y_1^4+4}}2\right)^{3/4} \left(\prod_{m=1}^{116}\Big[\Gamma\big(\tfrac{m}{116}\big)\Big]^{\big(\tfrac{-116}{m}\big)}\right)^{\color{red}{1/12}}\\ K(k_{53}) &= \frac{\sqrt{2\pi}}{2\sqrt{212}} \frac1{\;(x_2)^{1/3}} \left(\frac{y_2^2+\sqrt{y_2^4+4}}2\right)^{3/4} \left(\prod_{m=1}^{212}\Big[\Gamma\big(\tfrac{m}{212}\big)\Big]^{\big(\tfrac{-212}{m}\big)}\right)^{\color{red}{1/12}}\\ K(k_{61}) &= \frac{\sqrt{2\pi}}{2\sqrt{244}} \frac1{\;(x_3)^{1/3}} \left(\frac{y_3^2+\sqrt{y_3^4+36}}6\right)^{3/4} \left(\prod_{m=1}^{244}\Big[\Gamma\big(\tfrac{m}{244}\big)\Big]^{\big(\tfrac{-244}{m}\big)}\right)^{\color{red}{1/12}}\end{align}$$ where the \(x_n\) are the real roots of the following cubics $$x^3 - 5x^2 - 3x - 1 = 0\\ x^3 - 15x^2 - x - 1= 0\\ x^3 - 27x^2 - 5x - 1= 0$$ while the \(y_n\) are also the real roots of cubics $$y^3 - y^2 - 4y - 4 = 0\\ y^3 - 7y^2 + 13y - 11 = 0\\ y^3 - 6y^2 - 27y - 54 = 0$$ and similarly for the other \(p\).