Entry 65

For level $>119$, there are no more moonshine functions with uppercase index (in Atlas notation). Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$ then for Level 126, $$\begin{align} a &= \left(\frac{d_7^2\,d_{9}^2}{d_3\,d_{14}\,d_{18}\,d_{21}}\right), \quad b = \left(\frac{d_1^2\,d_{63}^2}{d_2\,d_3\,d_{21}\,d_{126}}\right)\\ c &= \left( \frac{d_{14}^2\,d_{18}^2}{d_6\,d_7\,d_{9}\,d_{42}}\right),\;\;\quad d = \left(\frac{d_2^2\,d_{126}^2}{d_1\,d_6\,d_{42}\,d_{63}}\right) \end{align}$$ The quadruple $(a,b,c,d)$ obey $$\begin{align} a-2 &= b\\ c-1 &= d\end{align}$$ and their ratios are cubes $$\begin{align} \left(\frac{a}{b}\right)^2\left(\frac{c}{d}\right) &= \left(\frac{d_7\,d_9}{d_1\,d_{63}}\right)^3\\ \left(\frac{a}{b}\right)\left(\frac{c}{d}\right)^2  &=  \left(\frac{d_{14}\,d_{18}}{d_2\,d_{126}}\right)^3\end{align}$$

Entry 64

Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$ then for Level 42 $$\begin{align} a &= \left(\frac{d_1^2\,d_{14}^2}{d_2\,d_3\,d_{7}\,d_{42}}\right), \quad b = \left(\frac{d_2^2\,d_{7}^2}{d_1\,d_6\,d_{14}\,d_{21}}\right)\\ c &= \left( \frac{d_3^2\,d_{42}^2}{d_1\,d_6\,d_{14}\,d_{21}}\right), \quad d = \left(\frac{d_6^2\,d_{21}^2}{d_2\,d_3\,d_{7}\,d_{42}}\right)\end{align}$$ The quadruple $(a,b,c,d)$ obey $$\begin{align}a\, b\, c\, d &=1\\ a+3 &= b\\ c+1 &= d\end{align}$$ and their ratios are squares $$\begin{align}\frac{a}{d} &= \left(\frac{d_1\,d_{14}}{d_6\,d_{21}}\right)^2\\ \frac{b}{c} &= \left(\frac{d_2\,d_{7}}{d_3\,d_{42}}\right)^2\end{align}$$

Entry 63

For certain even levels divisible by $7$ such as $7\times4,\, 7\times6,\, 7\times18$ or $28, 42, 126$, we may still find an eta quotient $a$ such that $a+m = b$ is also an eta quotient, but only for one integer $m$. Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$ and  $$a=\left(\frac{d_1\,d_7}{d_4\,d_{28}}\right),\quad b = \left(\frac{d_2^3\,d_{14}^3}{d_1\,d_4^2\,d_7\,d_{28}^2}\right)$$ then $$\begin{align}a+2 &= b\\ a+\frac{4}{a}+4 & = \left(\frac{d_2^2\,d_{14}^2}{d_1\,d_4\,d_7\,d_{28}}\right)^3\\ b+\frac{4}{b}-4 &= \left(\frac{d_1\,d_7}{d_2\,d_{14}}\right)^3\end{align}$$ where $(a,b)$ is the McKay-Thompson series of class 28C for the Monster (A161970).

Entry 62

(Under construction.)

Entry 61

(Under construction.)