Entry 45

Level 12 (Part 2).  For Part 2, one term of the linear dependency is not a moonshine function. Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$. Then $$\left(\frac{d_4^2\,d_6}{d_2\,d_{12}^2}\right)^2-1=\left(\frac{d_1\,d_4^2\,d_6^9}{d_2^3\,d_3^3\,d_{12}^6}\right)\\ \quad\left(\frac{d_4^2\,d_6}{d_2\,d_{12}^2}\right)^2+3=\left(\frac{d_2^7\,d_3}{d_1^3\,d_4^2\,d_6\,d_{12}^2}\right)$$ Expressed as the triple of eta quotients $(a,b,c)$ such that $$a-1 =b\\ a+3=c$$ so $(m,n)=(-1,3)$. Then  $$\begin{align}\frac{ab}{c} &= \left(\frac{d_1\, d_4^2\,d_6^3}{d_2^3\,d_3\, d_{12}^2}\right)^{4}\\ \frac{bc}{a} &= \left(\frac{d_2^3\, d_6^3}{d_1\,d_3\,d_4^2  d_{12}^2}\right)^2\\ \frac{ac}{b} &=  \left(\frac{d_2^2\, d_3}{d_1\,d_6^2}\right)^4\\ d \, &= \, \left(\frac{d_1\, d_3}{d_2\,d_6}\right)^6 \end{align}$$ where $(a,b,c)$ are still the McKay-Thompson series of class 12I (A187144) but $d$ is the McKay-Thompson series of class 6C (A121666). They obey $$\left(\sqrt{\frac{16a}{bc}}\color{red}-\sqrt{\frac{bc}{a}}\right)^2=d$$ $$d+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}\color{red}-8$$ in contrast to Part 1 which used the positive sign. Equivalently, $$\left(\frac{16a}{bc}+\frac{bc}{a}\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}\quad$$ However, this is not one of the 9 linear dependencies by Conway et al since the $\left(\frac{ab}c\right)$ term (which is A193522) is not a moonshine function.