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 $m=17$.  $$\begin{align}j_{10E}\big(\tfrac{\sqrt{-6/5}}2\big) &= U_2\, U_ {10}\\ j_{10E}\big(\tfrac{\sqrt{-14/5}}2\big) &= U_2^3\, U_{10}\\ j_{10E}\big(\tfrac{\sqrt{-26/5}}2\big) &= U_{13}^3\, U_{65}\\ j_{10E}\big(\tfrac{\sqrt{-38/5}}2\big) &= U_2^4\, U_5^9 \\ j_{10E}\big(\tfrac{1+\sqrt{-17/5}}2\big) &= -U_5^6\, U_{17}\end{align}$$