Entry 168

For class number $5$, there are only four $d$ of the kind $d \equiv 7\,\text{mod}\,8$, namely $d = 47, 79, 103, 127$. We illustrate only the first two and propose, $$\begin{align}K(k_{47}) &=\frac{\sqrt{2\pi}}{2\sqrt{47}}\,x^{2/3} \left(\prod_{m=1}^{47}\Big[\Gamma\big(\tfrac{m}{47}\big)\Big]^{\big(\tfrac{-47}{m}\big)}\right)^{\color{red}{1/10}}\\ K(k_{79}) &=\frac{\sqrt{2\pi}}{2\sqrt{79}}\,y^{2/3} \left(\prod_{m=1}^{79}\Big[\Gamma\big(\tfrac{m}{79}\big)\Big]^{\big(\tfrac{-79}{m}\big)}\right)^{\color{red}{1/10}}\end{align}$$ where $(x,y)$ are the real roots of the quintics solvable in radicals $$x^5 - 2x^4 - 10x^3 - 13x^2 - 6x - 1=0\\ y^5 - 11y^4 + 17y^3 - 2y^2 - 5y - 1=0$$ And similarly for $d= 103,127$. But for the next discriminant or $d=131$ which is of kind $d \equiv 3\,\text{mod}\,8$, one now has to deal with an algebraic number of degree $3\times5 = 15$. And it seems difficult to find the eta quotients responsible for $(x,y)$.