$$\begin{align}\left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2-1 &= \left(\frac{d_1^3\,d_6^2\, d_9^3}{d_2^3\,d_3^2 d_{18}^3}\right)\quad\\ \left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2+3 &=\left(\frac{d_3^6}{d_1\,d_2\,d_6^2\,d_9\,d_{18}}\right)\end{align}$$ Expressed as the triple of eta quotients \((a,b,c)\) such that $$a-1 =b\\ a+3=c$$ so \((m,n)=(-1,3)\). Then
$$\begin{align}\frac{ab}{c} &= \left(\frac{d_1\, d_6^2\, d_9}{d_2\, d_3^2\, d_{18}}\right)^6\\ \frac{bc}{a} &= \left(\frac{d_3}{d_6}\right)^8\\ \frac{ac}{b} &= \left(\frac{d_3^2\, d_6^2}{d_1\, d_2\, d_9\, d_{18}}\right)^{2}\end{align}$$ where \((a,b,c)\) are the McKay-Thompson series of class 18C (A215412). They obey $$\left(\frac{16a}{bc}+\frac{bc}a\right)+2a = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$ but \(\dfrac{ab}{c}\) is a non-monster function with expansion (A128512).
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