Entry 46

Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$. Then for Level 16   $$\left(\frac{d_8^3}{d_4\,  d_{16}^2}\right)^{2}-2= \left(\frac{d_1^2\,d_8}{d_2\,d_{16}^2}\right)\quad\\ \left(\frac{d_8^3}{d_4\,  d_{16}^2}\right)^{2}+2 = \left(\frac{d_2^5\,d_8}{d_1^2\,d_4^2\,d_{16}^2}\right)$$ Or more simply as $$a-2 =b\\ a+2=c$$ where $(a,b,c)$ are the McKay-Thompson series of class 16B for Monster (A185338, A208603) and obeys  $$\left(\frac{16a}{bc}+\frac{bc}a\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$ But one of addends is not a moonshine function so this is not one the 9 dependencies found by Conway et al. However, we also have $$\begin{align}a-\frac{4}a &= \left(\frac{d_2}{d_8}\right)^4\\ a+\frac{4}a &= \left(\frac{d_4^3}{d_2\,d_8^2}\right)^4\end{align}$$  Adding the two together yields $$\quad 2\left(\frac{d_8^3}{d_4\,  d_{16}^2}\right)^{2} = \left(\frac{d_2}{d_8}\right)^4+\left(\frac{d_4^3}{d_2\,d_8^2}\right)^4$$ and this is one of the $9$ dependencies found by Conway, Norton, and Atkins such that the moonshine functions span a linear space of $172-9=163$ dimensions.