Levels 126 & 252

I. Non-moonshine functions. There are no moonshine functions with uppercase index (in Atlas notation) for level $>119$. However, surprisingly we can still find trinomial identities for level 144 (some a consequence of level 72) and as high as level 252 (a consequence of level 126). For the latter, define the two pairs of eta quotients, $$\begin{align} a(\tau) &= \left(\frac{d_7^2\,d_{9}^2}{d_3\,d_{14}\,d_{18}\,d_{21}}\right), \quad b(\tau) = \left(\frac{d_1^2\,d_{63}^2}{d_2\,d_3\,d_{21}\,d_{126}}\right)\\ c(\tau) &= \left( \frac{d_{14}^2\,d_{18}^2}{d_6\,d_7\,d_{9}\,d_{42}}\right),\;\quad d(\tau) = \left(\frac{d_2^2\,d_{126}^2}{d_1\,d_6\,d_{42}\,d_{63}}\right)\end{align}$$ Given one moonshine function of level 21, namely $j_{21B}(\tau) = \left(\frac{d_1\,d_3}{d_7\,d_{21}}\right)$ and define, $$e(\tau) = \frac{j_{21B}(\tau)}{j_{21B}(3\tau)} = \left(\frac{d_1\,d_{63}}{d_7\,d_9}\right)$$ then ratios of the pairs are simply, $$\begin{align} \frac{a}{b} &= \frac{e(2\tau)}{e^2(\tau)}\\ \frac{c}{d} &= \frac{e(\tau)}{e^2(2\tau)}\\\ \end{align}$$ They obey, $$\begin{align} a(\tau)-2 &= b(\tau)\\ c(\tau)-1 &= d(\tau)\end{align}$$ $$\begin{align} a\big(\tfrac12+\tau\big)-2 &= b\big(\tfrac12+\tau\big)\\ c\big(\tfrac12+\tau\big)-1 &= d\big(\tfrac12+\tau\big)\end{align}$$ which is the pair of trinomial identities each for level 126 and level 252, respectively, and the latter seems to be the highest known.

Levels 42 & 84

I. Moonshine functions. The four functions for level 42,
$$\begin{align} j_{42A}(\tau) &= \left(\sqrt{j_{42B}}+\frac1{\sqrt{j_{42B}}}\right)^2\\ j_{42B}(\tau) &= \left(\frac{d_1\,d_6\,d_{14}\,d_{21}}{d_2\,d_3\,d_7\,d_{42}}\right)^2\\ j_{42C}(\tau) &= \sqrt[3]{j_{14B}(3\tau)} = \left(\frac{d_3\,d_{21}}{d_6\,d_{42}} \right)\\ j_{42D}(\tau) &= \left(\frac{d_2\,d_6\,d_7\,d_{21}}{d_1\,d_3\,d_{14}\,d_{42}} \right) \end{align}$$ and the three functions for level 84,
$$\begin{align} j_{84A}(\tau) &= \sqrt{j_{42A}(2\tau)}\\ j_{84B}(\tau) &= \sqrt{j_{42B}(2\tau)}\\ j_{84C}(\tau) &= \sqrt[3]{j_{28B}(3\tau)}= \left(\frac{d_6^2\,d_{42}^2}{d_3\,d_{12}\,d_{21}\,d_{84}}\right)\\ \end{align}$$
II. Non-moonshine functions. Define the two pairs of eta quotients, $$\begin{align} a(\tau) &= \left(\frac{d_1^2\,d_{14}^2}{d_2\,d_3\,d_{7}\,d_{42}}\right), \quad b(\tau) = \left(\frac{d_2^2\,d_{7}^2}{d_1\,d_6\,d_{14}\,d_{21}}\right)\\ c(\tau) &= \left( \frac{d_3^2\,d_{42}^2}{d_1\,d_6\,d_{14}\,d_{21}}\right), \quad d(\tau) = \left(\frac{d_6^2\,d_{21}^2}{d_2\,d_3\,d_{7}\,d_{42}}\right)\end{align}$$ Interestingly, these satisfy $$a(\tau)\, b(\tau)\, c(\tau)\, d(\tau) = 1$$ And in terms of the moonshine functions,
$$\begin{align} \frac{a}{b} &= \frac{j_{42B}}{j_{42D}}\\ \frac{d}{c} &= j_{42B}\,j_{42D}\\ \end{align}$$ They obey, $$\begin{align} a(\tau)+3 &= b(\tau)\\ c(\tau)+1 &= d(\tau)\end{align}$$ $$\begin{align} a\big(\tfrac12+\tau\big)+3 &= b\big(\tfrac12+\tau\big)\\ c\big(\tfrac12+\tau\big)+1 &= d\big(\tfrac12+\tau\big)\end{align}$$ which is the pair of trinomial identities each for level 42 and level 84, respectively.

Levels 14 & 28

For certain even levels divisible by $7$, the situation is now different. We can still find an eta quotient $a$ such that $a+m = b$ is also an eta quotient, but only for one rational $m$.

I. Moonshine functions: Define $$a(\tau) =\left(\frac{d_1\,d_7}{d_4\,d_{28}}\right),\quad b(\tau) = \left(\frac{d_2^3\,d_{14}^3}{d_1\,d_4^2\,d_7\,d_{28}^2}\right)$$ then $a+2 = b$ or, $$\left(\frac{d_1\,d_7}{d_4\,d_{28}}\right)+2 = \left(\frac{d_2^3\,d_{14}^3}{d_1\,d_4^2\,d_7\,d_{28}^2}\right)$$  for the unique $m=2$. This is the single trinomial identity of level 28. First, we have the 3 moonshine functions of level 14,
$$\begin{align} j_{14A}(\tau) &= \left(\sqrt{j_{14C}}+\frac{1}{\sqrt{j_{14C}}}\right)^2\\ j_{14B}(\tau) &= \left(\frac{d_1\,d_7}{d_2\,d_{14}}\right)^3\\ &= b+\frac{4}{b}-4\\ j_{14C}(\tau) &= \left(\frac{d_2\,d_7}{d_1\,d_{14}}\right)^4 \end{align}$$ and the 4 moonshine functions of level 28,
$$\begin{align}\quad j_{28A}(\tau) &= \sqrt{j_{14A}(2\tau)} = j_{28D} (\tau)+\frac{1}{j_{28D} (\tau)}\\ j_{28B}(\tau) &= \left(\frac{d_2^2\,d_{14}^2}{d_1\,d_4\,d_7\,d_{28}}\right)^3\\ &= a+\frac{4}{a}+4\\ \color{red}{j_{28C}}(\tau) &=  \frac{d_1\,d_7}{d_4\,d_{28}}\;=\; a\\ j_{28D}(\tau) &=  \sqrt{j_{14C}(2\tau)} = \left(\frac{d_4\,d_{14}}{d_2\,d_{28}}\right)^2 \end{align}$$ Note that $j_{14B}\big(\tfrac12+\tau\big) = -j_{28B}(\tau)$.

Summary

(Under construction)

Level 80

(Under construction.)