Entry 112

Regarding the previous two entries, the more famous function with evaluations that involve fundamental units \(U_n\) is the modular lambda function \(\lambda(\tau)\). Define the McKay-Thompson series of class 4C for the Monster (A007248) $$j_{4C}(\tau) = \left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^8+16 = \left(\frac{\eta^3(2\tau)}{\eta(\tau)\,\eta^2(4\tau)}\right)^8$$ compare the RHS to $$\lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac12\tau\big)\,\eta^2\big(2\tau\big)}{\eta^3(\tau)}\right)^8$$ which solves $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)} = -\tau\,i$$ We chose fundamental discriminants \(d=4m\) with class number \(h(-d)=2\) for even \(m = 6, 10, 22, 58\), hence $$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-6})}} &= U_3\sqrt{U_6} = (2+\sqrt3)\sqrt{5+2\sqrt6}\\ \frac1{\sqrt{\lambda(\sqrt{-10})}} &= U_2^3\,U_{10} = (1+\sqrt2)^2(3+\sqrt{10})\\ \frac1{\sqrt{\lambda(\sqrt{-22})}} &= U_{11}\sqrt{U_{22}}= (10+3\sqrt{11})\sqrt{197+42\sqrt{22}}\\ \frac1{\sqrt{\lambda(\sqrt{-58})}} &=\, U_2^6\,U_{58}\; =\; (1+\sqrt2)^6(99+13\sqrt{58})\end{align}$$ Note that these \(m\) have four divisors and \(\lambda(\tau)\) is a product of two \(U_n\). Also $$\begin{align}U_6 &= 5+2\sqrt6 = (\sqrt2+\sqrt3)^2\\ U_{22} &= 197+42\sqrt{22} = (7\sqrt2+3\sqrt{11})^2\end{align}$$ can be expressed as squares, which simplify things. For the next entry, the \(m\) have eight divisors and \(\lambda(\tau)\) is a product of four \(U_n\).