Entry 49

Define \(d_k=\eta(k\tau)\) with the Dedekind eta function \(\eta(\tau)\). Then for Level 20

$$\left(\frac{d_1\, d_4\, d_{10}}{d_2\,  d_5\, d_{20}}\right)^2 +1 =  \left(\frac{d_1\,d_4\,d_{10}^{10}}{d_2^2\,d_5^5\,d_{20}^5}\right)\quad\\ \left(\frac{d_1\, d_4\, d_{10}}{d_2\,  d_5\, d_{20}}\right)^2 + 5 =  \left(\frac{d_2^{8}}{d_1^3\,d_4^3\,d_5\,d_{20}}\right)$$

This level 20 triple can be derived from a level 10. Expressed as the triple of eta quotients \((a,b,c)\) such that $$a+1 =b\\ a+5=c$$ then \((m,n)=(1,5)\) and $$\begin{align}\frac{ab}c &= \left(\frac{d_1\,d_4\,d_{10}^2}{d_2^2\,d_5\,d_{20}}\right)^6\\ \frac{bc}a &= \left(\frac{d_2^2\, d_{10}^2}{d_1\,  d_4\, d_5\, d_{20}}\right)^4\\ \frac{ac}b &= \left(\frac{d_2^4\,d_5\,d_{20}}{d_1\,d_4\,d_{10}^4}\right)^2\end{align}$$ where \((a,b,c)\) are the McKay-Thompson series of class 20C (A145740). They obey $$\left(\frac{16a}{bc}+\frac{bc}a\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$ but is not one of the \(9\) dependencies found by Conway et al since some of the terms are not moonshine functions.