Entry 111

Given \(q = e^{2\pi i \tau}\) and define the McKay-Thompson series of Class 10E for Monster (A138516) $$j_{10E}(\tau) =  \frac{\eta(2\tau)\,\eta^5(5\tau)}{\eta(\tau)\,\eta^5(10\tau)} + 1= \left(\frac{\eta^2(2\tau)\,\eta(5\tau)}{\eta(\tau)\,\eta^2(10\tau)}\right)^2$$ On a hunch, I decided to test this since it seems similar to Class 6E of the previous entry. It turns out \(j_{10E}(\tau)\) is also product of fundamental units \(U_n\) for appropriate \(\tau\). (Here is a sample Wolfram calculation for \(U_5\).) Let \(d=20m\) with class number \(h(-d)=4\) for even \(m = 6,14,26,38\) and odd \(m=17\). $$\begin{align}j_{10E}\big(\tfrac{\sqrt{-6/5}}2\big) &= U_2\, U_ {10}\\ j_{10E}\big(\tfrac{\sqrt{-14/5}}2\big) &= U_2^3\, U_{10}\\ j_{10E}\big(\tfrac{\sqrt{-26/5}}2\big) &= U_{13}^3\, U_{65}\\ j_{10E}\big(\tfrac{\sqrt{-38/5}}2\big) &= U_2^4\, U_5^9 \\ j_{10E}\big(\tfrac{1+\sqrt{-17/5}}2\big) &= -U_5^6\, U_{17}\end{align}$$