Entry 113

This continues Entry 112. Recall the modular lambda function \(\lambda(\tau)\) $$\lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac12\tau\big)\,\eta^2\big(2\tau\big)}{\eta^3(\tau)}\right)^8$$ We chose fundamental discriminants \(d=4m\) with class number \(h(-d)=4\) for even \(m\) with eight divisors, hence only \(m = 30, 42, 78, 102\), and \(m =70, 130, 190\) which are evenly divisible by \(3\) and \(5\), respectively.  $$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-30})}} &= (2+\sqrt3)(5+2\sqrt6)(4+\sqrt{15})\sqrt{11+2\sqrt{30}}\\  \frac1{\sqrt{\lambda(\sqrt{-42})}} &= (2+\sqrt3)^2(1+\sqrt2)^2(8+3\sqrt{7})\sqrt{13+2\sqrt{42}}\\ \frac1{\sqrt{\lambda(\sqrt{-78})}} &= (2+\sqrt3)^3(5+2\sqrt6)(25+4\sqrt{39})\sqrt{53+6\sqrt{78}}\\ \frac1{\sqrt{\lambda(\sqrt{-102})}} &=(2+\sqrt3)^2(5+2\sqrt6)^2(50+7\sqrt{51})\sqrt{101+10\sqrt{102}}\end{align}$$ as well as $$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-70})}} &= (8+3\sqrt7)(15+4\sqrt{14})(6+\sqrt{35})\sqrt{251+30\sqrt{70}}\\  \frac1{\sqrt{\lambda(\sqrt{-130})}} &= (1+\sqrt2)^4\,(3+\sqrt{10})^2\,(5+\sqrt{26})\,(57+5\sqrt{130})\\ \frac1{\sqrt{\lambda(\sqrt{-190})}} &= (170+39\sqrt{19})(37+6\sqrt{38})(39+4\sqrt{95})\sqrt{52021+3774\sqrt{190}}\end{align}$$

For the last, note that \(U_{190}=52021+3774\sqrt{190} = (51\sqrt{10}+37\sqrt{19})^2\) and similarly for other \(U_n\) with composite \(n\) but I chose to retain the \(a+b\sqrt{n}\) form. Most of these do not appear in Mathworld's list. Arranged by class number, one can see some patterns like the \(m\) have eight divisors and \(\lambda(\tau)\) is a product of four \(U_n\). Next are \(m\) that have sixteen divisors and \(\lambda(\tau)\) is a product of eight \(U_n\).