Entry 54

Define \(d_k = \eta(k\tau)\) with Dedekind eta function \(\eta(\tau)\). Then for Level 36,

$$\begin{align}\left(\frac{d_2\,d_3^2\,d_{12}^2\, d_{18}}{d_1\,  d_4\,d_6^2\, d_9\, d_{36}}\right)^2+1 &= \left(\frac{d_3\,d_{12}}{d_6^2}\right)^2\left(\frac{d_2^2\,d_{18}^2}{d_1\,d_4\,d_9\,d_{36}}\right)^3 \\ \left(\frac{d_2\,d_3^2\,d_{12}^2\, d_{18}}{d_1\,  d_4\,d_6^2\, d_9\, d_{36}}\right)^2-3 &= \left(\frac{d_6^2}{d_3\,d_{12}}\right)^2 \left(\frac{d_1\,d_4\,d_6^{12}\,d_9\,d_{36}}{\big(d_2\,d_3\,d_{12}\,d_{18}\big)^4}\right)\end{align}$$

Or more simply $$a+1=b \qquad \\ a-3=c \qquad $$ and we find \((a,b,c)\) as the McKay-Thompson series of class 36A for Monster (A227585). They obey,

$$a-\frac3{a}-2=\frac{bc}{a}=\left(\frac{d_6^2}{d_3\,d_{12}}\right)^8\quad$$

$$\left(\frac{16a}{bc}+\frac{bc}a\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}\;$$

but this is not one of the 9 linear dependencies by Conway et al.