Entry 55

Define \(d_k = \eta(k\tau)\) with Dedekind eta function \(\eta(\tau)\). Then for Level 60,

$$\begin{align}\frac{d_2\,d_3\,d_5\,d_{12}\,d_{20}\,d_{30}}{d_1\,d_4\,d_6\,d_{10}\,d_{15}\,d_{60}}+1 &= \frac{(d_2\,d_3\,d_5\,d_{12}\,d_{20}\,d_{30})(d_2\, d_{30})^3}{(d_1\,d_4\,d_6\,d_{10}\,d_{15}\,d_{60})^2}\\ \frac{d_2\,d_3\,d_5\,d_{12}\,d_{20}\,d_{30}}{d_1\,d_4\,d_6\,d_{10}\,d_{15}\,d_{60}}-1 &= \frac{(d_6\,d_{10})^3}{d_2\,d_3\,d_5\,d_{12}\,d_{20}\,d_{30}}\end{align}$$

Or more simply $$a+1=b \qquad \\ a-1=c \qquad $$ we find \((a,b,c)\) as the McKay-Thompson series of class 60C for Monster (A145725). They obey,

$$\left(\frac{4a}{bc}+\frac{bc}a\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$

If we express \(a=\dfrac{p}{q}\) as

$$\begin{align}
p &= d_2\,d_3\,d_5\,d_{12}\,d_{20}\,d_{30}\\
q &= d_1\,d_4\,d_6\,d_{10}\,d_{15}\,d_{60}
\end{align}$$

then we have the simple cube,$$\frac{pq+q^2}{p}=\big(d_2\,d_{30})^3$$ where, of course, Level \(60 = 2\times30\).