Entry 42

Define \(d_k=\eta(k\tau)\) with the Dedekind eta function \(\eta(\tau)\). Then for Level 8 $$\left(\frac{d_4^3}{d_2\,  d_8^2}\right)^{4}-4 = \left(\frac{d_1^2\,d_4}{d_2\,d_8^2}\right)^2\quad\\ \left(\frac{d_4^3}{d_2\,  d_8^2}\right)^{4}+4=\left(\frac{d_2^5}{d_1^2\,d_4\,d_8^2}\right)^2$$ Or more simply as $$a-4 =b\\ a+4=c$$ where \((a,b,c)\) are the McKay-Thompson series of class 8E for Monster (A131125) and obeys $$\left(\frac{64a}{bc}+\frac{bc}a\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$ 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 $$\begin{align}a+\frac{16}a &= \left(\frac{d_1}{d_4}\right)^8+8\\ a-\frac{16}a &= \left(\frac{d_2}{d_4}\right)^{12}\end{align}$$ Adding the two together yields $$\quad 2\left(\frac{d_4^3}{d_2\,  d_8^2}\right)^{4} = \left(\frac{d_1}{d_4}\right)^8+\left(\frac{d_2}{d_4}\right)^{12}+8$$ 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.

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