Entry 114

This continues Entry 113. Recall the modular lambda function \(\lambda(\tau)\) $$\lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac12\tau\big)\,\eta^2\big(2\tau\big)}{\eta^3(\tau)}\right)^8$$calculated in Mathematica as ModularLambda[tau]. We now chose fundamental discriminants \(d=4m\) with class number \(h(-d)=8\) for even \(m\) with 16 divisors. There are only three, namely \(m = 210, 330, 462\), hence $$\begin{align}210 &= 2\times3\times5\times7\\ 330 &= 2\times3\times5\times11\\ 462 &= 2\times3\times7\times11\end{align}$$ Ramanujan found \(m=210\), though I'm not sure he did for the other two, so I went ahead and found them $$\small\begin{align}\frac1{\sqrt{\lambda(\sqrt{-210})}} &= (1+\sqrt{2})^2 (2+\sqrt{3}) (8+3\sqrt{7}) (3+\sqrt{10})^2 (4+\sqrt{15})^2 (6+\sqrt{35}) (\sqrt{6}+\sqrt{7})^2 (\sqrt{14}+\sqrt{15}) \\ \frac1{\sqrt{\lambda(\sqrt{-330})}} &= (1+\sqrt{2})^2(2+\sqrt{3})^3(3+\sqrt{10})^2(10+3\sqrt{11})(4+\sqrt{15})(65+8\sqrt{66}) (\sqrt{44}+\sqrt{45})^2(\sqrt{54}+\sqrt{55}) \\ \frac1{\sqrt{\lambda(\sqrt{-462})}} &= (2+\sqrt{3})^2(5+2\sqrt{6})^2 (8+3\sqrt{7})^2(10+3\sqrt{11})(15+4\sqrt{14})(76+5\sqrt{231})(7\sqrt{2}+3\sqrt{11})^2(\sqrt{21}+\sqrt{22})\end{align}$$ They are products of eight fundamental units \(U_n\). To find products of sixteen fundamental units \(U_n\), I checked class number \(h(-d)=16\) for even \(m\) with 32 divisors from the class number list. Surprisingly, there are none. So it seems the pattern of \(\lambda(\sqrt{-m})\) as a product of \(\color{blue}{2^k}\) quadratic units \(U_n\) where class number \(h(-4m)=\color{blue}{2^k}\) stops at \(8\). Note that $$\quad e^{\pi\sqrt{462}} = \frac{16}{\lambda\big(\sqrt{-462}\big)}-8.0000000000000000000000000000094\dots$$ or for 29 zeros.