Entry 110

Given \(q = e^{2\pi i \tau}\). Define the McKay-Thompson series of Class 6E for Monster (A128633) $$\begin{align}j_{6E}(\tau) &=\frac1{\big(\text{K}_6(q)\big)^3}+1 \\ &= \left(\frac{\eta(2\tau)\,\eta^3(3\tau)}{\eta(\tau)\,\eta^3(6\tau)}\right)^3 + 1 = \left(\frac{\eta^2(2\tau)\,\eta(3\tau)}{\eta(\tau)\,\eta^2(6\tau)}\right)^4\end{align}$$ where \(\text{K}_6(q)\) is the cubic continued fraction. Just like the modular lambda function \(\lambda(\tau)\), it seems \(j_{6E}(\tau)\) is also a product of fundamental units \(U_n\) for appropriate \(\tau\). (Here is a sample Wolfram calculation for \(U_5\).) We choose \(d=12m\) with class number \(h(-d)=4\) for even \(m = 10, 14, 26, 34\) (also found in the previous entry)

$$\begin{align}j_{6E}\big(\tfrac{\sqrt{-10/3}}2\big) &= U_2^2\,U_5^6 \\ j_{6E}\big(\tfrac{\sqrt{-14/3}}2\big) &= U_6\,U_{14} \\ j_{6E}\big(\tfrac{\sqrt{-26/3}}2\big) &= U_2^4\,U_{26}^2 \\ j_{6E}\big(\tfrac{\sqrt{-34/3}}2\big) &= U_2^6\,U_{17}^2\end{align}$$ and odd \(m = 7, 11, 19, 31, 59\), 

$$\begin{align}j_{6E}\big(\tfrac{1+\sqrt{-7/3}}2\big) &= U_3\sqrt{U_{21}^3} \\ j_{6E}\big(\tfrac{1+\sqrt{-11/3}}2\big) &= U_6\sqrt{U_{33}} \\ j_{6E}\big(\tfrac{1+\sqrt{-19/3}}2\big) &= U_3^3\sqrt{U_{57}} \\ j_{6E}\big(\tfrac{1+\sqrt{-31/3}}2\big) &= U_3^3\sqrt{U_{93}^3} \\ j_{6E}\big(\tfrac{1+\sqrt{-59/3}}2\big) &= U_2^6\sqrt{U_{177}} \end{align}$$ Why this factors nicely I don't know. Also, some fundamental units \(U_n\) for composite \(n\) may actually be squares. For example, $$\begin{align}U_{21} &= \tfrac{5+\sqrt{21}}2=\Big(\tfrac{\sqrt{3}+\sqrt{7}}2\Big)^2\\ U_{33} &= 23+4\sqrt{33}=\big(2\sqrt{3}+\sqrt{11}\big)^2\end{align}$$ and so on, hence these \(\sqrt{U_n}\) simplifies a bit.