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.)

Level 72

These are not monster functions (with the exception of the first). But I'm including them since this seems to be the highest level such that the requirement of the triple relation works, namely that $a,\,a+m,\,a+n$ are all eta quotients. For convenience, first define the eta quotient, $\color{blue}{f_p} = \frac{d_p}{d_{2p}} =\frac{\eta(p\tau)}{\eta(2p\tau)}$. Let,
$$\begin{align} a(\tau) &=  \left(\frac{f_{4}^2\, f_{6}^2\, f_{36}^2}{f_{2}\,f_{12}^4\, f_{18}}\right)\\ b(\tau) &= \left(\frac{f_{3}^2\,f_{4}\, f_{36}}{f_{1}\, f_{9}\,f_{12}^2}\right)f_6^2 \\ c(\tau) &=  \left(\frac{f_{1}\, f_{4}\,f_{6}^2\, f_{9}\, f_{36}}{f_{2}\, f_{3}^2\, f_{12}^2\,f_{18}}\right)f_6^2 \end{align}$$  such that $a+1 = b,\,a-1 = c$. Then define, $$\begin{align} j_{12F}(\tau) &= f_6^4= \left(\frac{d_6}{d_{12}}\right)^4 \;=\; \frac{bc}a\\ k_{72A}(\tau) &= f_6^4+\frac{4}{f_6^4}\\ k_{72B}(\tau) &= \left(\frac{f_{1}\, f_{4}\,f_{6}^2\, f_{9}\, f_{36}}{f_{2}\, f_{3}^2\, f_{12}^2\,f_{18}}\right)^2 = \frac{ac}b\\ k_{72C}(\tau) &= \left(\frac{f_{3}^2\,f_{4}\, f_{36}}{f_{1}\, f_{9}\,f_{12}^2}\right)^2 \;=\; \frac{ac}b\\ \color{red}{k_{72D}}(\tau) &= \left(\frac{f_{4}^2\, f_{6}^2\, f_{36}^2}{f_{2}\,f_{12}^4\, f_{18}}\right)\; = \; a \end{align}$$  Then we have an analogous linear relation,
$$k_{72A}+2\color{red}{k_{72D}} =k_{72B}+k_{72C}+ j_{12F}+2$$   Curiously, notice that the form of the last three $k_{72}$ appear in other functions,
$$\begin{align} \frac1{j_{36A}(\tau)} &= \left(\frac{d_1\,  d_4\,d_6^2\, d_9\, d_{36}}{ d_2\,d_3^2\,d_{12}^2\, d_{18}}\right)^2\\ \frac1{k_{36A}(\tau)} &= \left(\frac{d_{3}^2\,d_{4}\, d_{36}}{d_{1}\, d_{9}\,d_{12}^2}\right)^2\\ \frac1{\sqrt{j_{18A}(2\tau)}} &=\left(\frac{d_4^2\,d_{6}^2\, d_{36}^2}{d_2\,  d_{12}^4\,d_{18}}\right)\qquad\qquad \end{align}$$

Level 60

$$\begin{align}j_{60A}(\tau) & = j_{60E}(\tau)+\frac{1}{ j_{60E}(\tau)} = \sqrt{j_{30B}(2\tau)} \\ j_{60B}(\tau) &= \frac{(d_2\, d_{6}\, d_{10}\, d_{30})^2}{d_1\,  d_{3}\, d_{4}\, d_{5}\,d_{12}\,  d_{15}\, d_{20}\, d_{60}}\\ j_{60C}(\tau) &= \frac{a}{b} = \frac1{a}(d_6\,d_{10})^3+1 = \frac{a}{\,b^2}(d_2\,d_{30})^3-1\\ j_{60D}(\tau) &= \left(\frac{d_1\, d_{12}\, d_{15}\, d_{20}}{d_3\,  d_{4}\, d_{5}\, d_{60}}\right)\\ &=\left(\frac{d_2\, d_6\, d_{10}\, d_{30}}{d_3\,  d_{4}\, d_{5}\, d_{60}}\right)-1\\ j_{60E}(\tau) &= \frac{d_4\, d_6\, d_{20}\, d_{30} } {d_2\,  d_{10}\, d_{12}\, d_{60}} =\sqrt{j_{30D}(2\tau)}\\ j_{60F}(\tau) &= \left(\frac{d_{12}\, d_{30}}{d_6\,  d_{60}}\right) \end{align}$$ where
$$\begin{align} a &= d_2\,d_3\,d_5\,d_{12}\,d_{20}\,d_{30}\\ b &= d_1\,d_4\,d_6\,d_{10}\,d_{15}\,d_{60} \end{align}$$
There are no linear relations between these functions.

Level 48

(Under construction)

Level 40

$$\begin{align} j_{40A}(\tau) &= \sqrt{j_{20B}(\tau)} = \sqrt[4]{j_{10A}(2\tau)}\\ j_{40B}(\tau) &= \sqrt{j_{20A}(2\tau)}= \left(\frac{d_4^2\, d_{20}^2}{d_2\,  d_8\, d_{10}\, d_{40}}\right)^2\\ &= j_{40C}(\tau)+\frac1{j_{40C}(\tau)}\\ j_{40C}(\tau) &= \sqrt{j_{20F}(2\tau)}= \left(\frac{d_8\, d_{10}}{d_2\,  d_{40}}\right) \end{align}$$
There is no linear relation between these functions. $j_{40A}$ is one of the few that involves a 4th root.

Level 36, part 3

(Under construction)

Level 36, part 2

(Under construction)

Level 36, part 1

$$\begin{align} j_{36A}(\tau) &= \left(\frac{d_2\,d_3^2\,d_{12}^2\, d_{18}}{d_1\,  d_4\,d_6^2\, d_9\, d_{36}}\right)^2\\ &=j_{36B}(\tau)+\frac{3}{j_{36B}(\tau)}+3\\ j_{36B}(\tau) &=\left(\frac{d_1\, d_4\, d_{18}}{d_2\,  d_9\, d_{36}}\right)\\ j_{36C}(\tau) &=  \left(\frac{d_6^2\, d_{12}^2}{d_2\,  d_4\,d_{18}\, d_{36}}\right) = \sqrt{j_{18B}(2\tau)}\\ j_{36D}(\tau) &= \left(\frac{d_4\, d_{9}}{d_1\,  d_{36}}\right) \end{align}$$
There is no linear relation between these functions.

Level 30

$$\begin{align} j_{30A}(\tau) &= \left(\frac{d_1\, d_6\, d_{10}\, d_{15}}{d_2\,  d_{3}\, d_{5}\, d_{30}}\right)^3\\ j_{30B}(\tau) &= j_{30D}(\tau)+\frac{1}{ j_{30D}(\tau)}+2\\ j_{30C}(\tau) &= \left(\frac{d_1\, d_3\, d_{5}\, d_{15}}{d_2\,  d_{6}\, d_{10}\, d_{30}}\right)\\ j_{30D}(\tau) &= \left(\frac{d_2\, d_3\, d_{10}\, d_{15}}{d_1\,  d_{5}\, d_{6}\, d_{30}}\right)^2\\ j_{30E}(\tau) &= \left(\frac{d_{6}\, d_{15}}{d_{3}\, d_{30}}\right)^2\\ j_{30F}(\tau) &= \left(\frac{d_3\, d_5\, d_{6}\, d_{10}}{d_1\,  d_{2}\, d_{15}\, d_{30}}\right)\\ j_{30G}(\tau) &= \left(\frac{d_1^2\, d_6\, d_{10}\, d_{15}^2}{d_2^2\,  d_{3}\, d_{5}\, d_{30}^2}\right)\end{align}$$
These functions obey the triple relations. Let $a=\color{red}{j_{30G}},\;m=1,\;n=2$, then,
$$\begin{align} j_{30F} &= a+\frac{mn}{a}+(m+n)\\ j_{30C} &= a+m+\frac{m(m-n)}{a+m}-(2m-n)\\ j_{30A} &= a+n+\frac{n(m-n)}{a+n}+(m-2n) \end{align}$$ as well as,
$$j_{30B} =  j_{30F}+\frac{(m-n)^2}{j_{30F}}+3 =  j_{30C}+\frac{n^2}{j_{30C}}+5 =  j_{30A}+\frac{m^2}{j_{30A}}+7$$ which implies the linear relation between 5 monster functions,
$$j_{30B}+2\color{red}{j_{30G}} = j_{30A}+j_{30C}+j_{30F}+3$$ This is the highest level with linear relations between 5 monster functions.

Level 24, part 3

(Under construction)

Level 24, part 2

(Under construction)

Level 24, part 1

Level 24
$$\begin{align} j_{24A}(\tau) &= \left(\frac{d_4^2\, d_{12}^2}{d_2\, d_6\, d_8\, d_{24}}\right)^3=\sqrt{j_{12A}(2\tau)}\\ &=j_{24D}(\tau)+\frac{4}{j_{24D}(\tau)}\\ &=j_{24H}(\tau)+\frac{1}{j_{24H}(\tau)}\\ \color{blue}{j_{24B}(\tau)} &= \left(\frac{d_2\, d_4\, d_6\, d_{12}}{d_1\, d_3\, d_8\, d_{24}}\right)^2\\ \color{red}{j_{24C}(\tau)} &= \left(\frac{d_1^2\, d_6\, d_8^2\, d_{12}}{d_2\, d_3^2\, d_4\, d_{24}^2}\right)\\ j_{24D}(\tau) &= \left(\frac{d_2\, d_6}{d_8\, d_{24}}\right)=\sqrt{j_{12E}(2\tau)}\\ j_{24E}(\tau) &= \left(\frac{d_{12}^2}{d_6\, d_{24}}\right)^4=\sqrt{j_{12D}(2\tau)}\\ j_{24F}(\tau) &= \left(\frac{d_8\, d_{12}}{d_4\, d_{24}}\right)^3 = \sqrt{j_{12F}(2\tau)}\\ j_{24G}(\tau) &= \left(\frac{d_4\, d_8}{d_{12}\, d_{24}}\right) = \sqrt{j_{12G}(2\tau)}\\ j_{24H}(\tau) &= \left(\frac{d_6\, d_8}{d_2\, d_{24}}\right)^2=\sqrt{j_{12H}(2\tau)}\\ \color{red}{j_{24I}(\tau)} &= \left(\frac{d_2\, d_3^2\, d_8^2\, d_{12}}{d_1^2\, d_4\, d_6\, d_{24}^2}\right)\\ j_{24J}(\tau) &= \;\left(\frac{d_{12}}{d_{24}}\right)^2\;=\;\sqrt{j_{12J}(2\tau)}\\ \end{align}$$ The "important" functions (that aren't just square roots) are $j_{24C}$ and $j_{24I}$. First define the auxiliary non-monster functions, $$\begin{align} U &=\left(\frac{d_1\,d_6\,d_8\,d_{12}}{d_2\,d_3\,d_4\,d_{24}}\right)^4\\ V &=\left(\frac{d_2^2\,d_3\,d_8\,d_{12}^2}{d_1\,d_4^2\,d_6^2\,d_{24}}\right)^4\quad\quad \end{align}$$ I. Let $a_1=\color{red}{j_{24C}},\;m = 1,\;n=3,$ then,
$$\begin{align} \color{blue}{j_{24B}} &= a_1+\frac{mn}{a_1}+(m+n)\\ j_{12G} &= a_1+m+\frac{m(m-n)}{a_1+m}-(2m-n)\\ U &= a_1+n+\frac{n(m-n)}{a_1+n}+(m-2n) \end{align}$$ as well as, $$j_{12C} =  j_{24B}+\frac{(m-n)^2}{j_{24B}}-2 =  j_{12G}+\frac{n^2}{j_{12G}} =  U+\frac{m^2}{U}+4$$
II. Let $a_2=\color{red}{j_{24I}},\;m = 1,\;n=-1,$ then,
$$\begin{align} \color{blue}{j_{24B}} &= a_2+\frac{mn}{a_2}+(m+n)\\ j_{12F} &= a_2+m+\frac{m(m-n)}{a_2+m}-(2m-n)\\ V &= a_2+n+\frac{n(m-n)}{a_2+n}+(m-2n) \end{align}$$ as well as, $$j_{12C} =  j_{24B}+\frac{(m-n)^2}{j_{24B}}-2 =  j_{12F}+\frac{n^2}{j_{12F}} =  V+\frac{m^2}{V}-4$$ These two families imply the two linear relations, $$j_{12C}+2\color{red}{j_{24C}} = j_{24B}+j_{12G}+U-2\\j_{12C}+2\color{red}{j_{24I}} = j_{24B}+j_{12F}+V-2$$  
However, there is one linear relation using only monster functions,
$$j_{12C}-j_{12F}-j_{12G}+j_{12E}-2j_{12I} = 2\big(\color{blue}{j_{24B}}-\color{red}{j_{24C}}-\color{red}{j_{24I}}-3\big)$$ Note that,
$$j_{12C}=j_{12F}+\frac{1}{j_{12F}} = j_{12G}+\frac{9}{j_{12G}}\\ j_{12E} = j_{12I}-\frac{3}{j_{12I}}-2\quad$$

Level 20

I. Monster functions: Let,
$$a(\tau) =  \left(\frac{d_1\, d_4\, d_{10}}{d_2\,  d_5\, d_{20}}\right)^{2},\quad b(\tau) =  \frac{d_1\,d_4\,d_{10}^{10}}{d_2^2\,d_5^5\,d_{20}^5}, \quad c(\tau) =  \frac{d_2^{8}}{d_1^3\,d_4^3\,d_5\,d_{20}}$$
such that $a+1=b,\; a+5=c$. This level 20 triple can be derived from a level 10 triple. Then define,
$$\begin{align} j_{20A}(\tau) &= \left(\frac{d_2^2\, d_{10}^2}{d_1\,  d_4\, d_5\, d_{20}}\right)^4=\frac{bc}{a} \\ &= \left( \sqrt{j_{20F}}+\frac1{\sqrt{ j_{20F}}}\right)^2\\ j_{20B}(\tau) &=\sqrt{j_{10A}(2\tau)}\\ &= j_{20D}(\tau)+\frac4{j_{20D}(\tau)}\\ &= j_{20E}(\tau)-\frac1{j_{20E}(\tau)} \\ \color{red}{j_{20C}}(\tau) &= \left(\frac{d_1\, d_4\, d_{10}}{d_2\,  d_5\, d_{20}}\right)^{2}\;\;=\;\;a\\ j_{20D}(\tau) &= \left(\frac{d_2\, d_{10}}{d_4\,  d_{20}}\right)^{2}=\sqrt{j_{10B}(2\tau)}\\ j_{20E}(\tau) &= \left(\frac{d_4\, d_{10}}{d_2\,  d_{20}}\right)^{3}=\sqrt{j_{10D}(2\tau)}\\ j_{20F}(\tau) &= \left(\frac{d_4\, d_5}{d_1\,  d_{20}}\right)^{2} \end{align}$$
Examples: $\; j_{10A}\left(\tfrac12\sqrt{-19/10}\right) = 76^2,\;\; j_{20B}\left(\tfrac14\sqrt{-19/10}\right) = 76$

II. Non-monster functions. Also define,
$$\begin{align} u(\tau) &= \left(\frac{d_2^4\,d_5\,d_{20}}{d_1\,d_4\,d_{10}^4}\right)^2 = \frac{ac}b\\ v(\tau) &= \left(\frac{d_1\,d_4\,d_{10}^2}{d_2^2\,d_5\,d_{20}}\right)^6 =\frac{ab}c \end{align}$$
III. Relations. Like $j_{30B}$, the function $j_{20A}$ is expressible by others in 4 ways. There is no linear relation between purely monster functions of level 20.

Level 18

This level is also special since we can find two $a(\tau)$ namely, $\color{red}{j_{18D}}$ and $\color{blue}{j_{18C}}$. Using S1 of the Intro, these two explain $2\times4 = 8$ of the 14 trinomial identities of level 18 found by Somos.

I. First triple. Define,
$$a(\tau) =\left(\frac{d_2^2\,d_9}{d_1\,d_{18}^2}\right),\quad b(\tau) = \left(\frac{d_6\,d_9^3}{d_3\,d_{18}^3}\right),\quad  c(\tau) =\left(\frac{d_1^2\,d_6\,d_9}{d_2\,d_3\,d_{18}^2}\right)$$ then,
$$\left(\frac{d_2^2\,d_9}{d_1\,d_{18}^2}\right)-1 = \left(\frac{d_6\,d_9^3}{d_3\,d_{18}^3}\right)\quad \\ \left(\frac{d_2^2\,d_9}{d_1\,d_{18}^2}\right)-3 =\left(\frac{d_1^2\,d_6\,d_9}{d_2\,d_3\,d_{18}^2}\right)$$ so $a-1 = b,\,a-3 = c,$ and we get the moonshine functions,
$$\begin{align} j_{18A}(\tau) &= \left(\frac{d_1\, d_2}{d_9\,  d_{18}}\right) = \frac{ac}{b}\\ j_{18B}(\tau) &= \left(\frac{d_3^2\, d_6^2}{d_1\, d_2\, d_9\, d_{18}}\right)^{2}\\ \color{blue}{j_{18C}}(\tau) &= \left(\frac{d_1\,d_6^4\, d_9}{d_2^2\,d_3^2 d_{18}^2}\right)^2  =\frac{bc}{a}+1\\ \color{red}{j_{18D}}(\tau) &=\left(\frac{d_2^2\,d_9}{d_1\,d_{18}^2}\right)\;=\; a \\ j_{18E}(\tau) &= \left(\frac{d_2\, d_9}{d_1\,  d_{18}}\right)^{3}= \frac{ab}{c}\\ \end{align}$$ Naturally, these also satisfy S2 and S3. We also have
$$j_{18B} =  j_{18A}+\frac{3^2}{j_{18A}}+3 =  j_{18E}+\frac{1}{j_{18E}}-1$$ and the linear relation between these 5 moonshine functions,
$$j_{18B}+2\color{red}{j_{18D}} = j_{18A}+j_{18C}+j_{18E}+4$$
II. Second triple. Define,
$$a(\tau) =\left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2,\quad b(\tau) =  \left(\frac{d_1^3\,d_6^2\, d_9^3}{d_2^3\,d_3^2 d_{18}^3}\right),\quad  c(\tau) =\left(\frac{d_3^6}{d_1\,d_2\,d_6^2\,d_9\,d_{18}}\right)$$ then, $$\left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2-1 =  \left(\frac{d_1^3\,d_6^2\, d_9^3}{d_2^3\,d_3^2 d_{18}^3}\right)\quad\\ \left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2+3 =\left(\frac{d_3^6}{d_1\,d_2\,d_6^2\,d_9\,d_{18}}\right)$$ so $a-1 = b,\,a+3 = c,$ and we get,
$$\begin{align} j_{6d} &= j_{6F}+\frac{16}{j_{6F}}\\ j_{18B}(\tau) &= \left(\frac{d_3^2\, d_6^2}{d_1\, d_2\, d_9\, d_{18}}\right)^{2} =\frac{ac}{b}\\ \color{blue}{j_{18C}}(\tau) &= \left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2 \;\;=\;\;a\\ \beta(\tau) &=\left(\frac{d_1\, d_6^2\, d_9}{d_2\, d_3^2\, d_{18}}\right)^6 =\frac{ab}{c} \\ j_{6F}(\tau) &=\; \left(\frac{d_3}{d_6}\right)^8 \;\;= \;\;\frac{bc}{a}\\ \end{align}$$ where $j_{6d}$ and $\beta$ are non-monster functions with expansions (A007263) and (A128512), respectively. Naturally, these also satisfy S1,2,3. And by adding these additional relations, we get a second linear equation for level 18 of similar form to the first,
$$j_{6d}+2\color{blue}{j_{18C}}=j_{18B}+j_{6F}+\beta-2$$

Level 16

I. Moonshine functions: Define, $$a(\tau)= \left(\frac{d_8^3}{d_4\,  d_{16}^2}\right)^{2}, \quad b(\tau) = \left(\frac{d_1^2\,d_8}{d_2\,d_{16}^2}\right), \quad c(\tau) = \left(\frac{d_2^5\,d_8}{d_1^2\,d_4^2\,d_{16}^2}\right)$$  then, $$\left(\frac{d_8^3}{d_4\,  d_{16}^2}\right)^{2}-2= \left(\frac{d_1^2\,d_8}{d_2\,d_{16}^2}\right)\quad\\ \left(\frac{d_8^3}{d_4\,  d_{16}^2}\right)^{2}+2 = \left(\frac{d_2^5\,d_8}{d_1^2\,d_4^2\,d_{16}^2}\right)$$ so $a-2 = b,\, a+2 = c$. This $a,b,c$ also belong to the class where $a(\tau) = -a\big(\tfrac12+\tau\big)$ and $b(\tau) = -c\big(\tfrac12+\tau\big)$. And the three level 16 moonshine functions, $$\begin{align} j_{16A}(\tau) &= \left(\frac{d_4\, d_8}{d_2\,  d_{16}}\right)^{4}=\sqrt{j_{8A}(2\tau)}\\ \color{red}{j_{16B}}(\tau) &= \left(\frac{d_8^3}{d_4\,  d_{16}^2}\right)^{2}\; = \; a\\ j_{16C}(\tau) &= \left(\frac{d_2^3\,d_8^3}{d_1^2\,  d_4^2\, d_{16}^2}\right)^{2} = \frac{ac}{b} \end{align}$$
II. Relations: Using S1, this $a,b,c$ explains 4 of the 6 trinomial identities of level 16 (prefixed with t16) by Somos. They also obey,
$$\left(\frac{bc}{a}+16\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad$$ In general, $$\left(\frac{bc}{a}+(b-c)^2\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad$$  where $b-c=4$. Or in terms of named functions of level 8 and 16, $$\quad\quad j_{8B}\;+\;2\color{red}{j_{16B}} = j_{8D}+j_{16C}+j_{16C'}$$ where, similar to level 8,  $$j_{16C'}(\tau) =-j_{16C}\big(\tfrac12+\tau\big)= \;\left(\frac{d_1^2\,d_8^3}{d_2^3\, d_{16}^2}\right)^{2}$$ The last 2 of the 6 trinomial identities involves the pair $j_{16C}$ and $j_{16C'}$, hence is just the same function using different arguments, $$\begin{align}j_{16C}(\tau)-2 &=\left(\frac{d_2^3\,d_8^3}{d_1^2\,  d_4^2\, d_{16}^2}\right)^{2}-2 =\frac{d_4^{10}}{d_1^2\,d_2^3\,d_8^3\,d_{16}^2}\\ j_{16C}\big(\tfrac12+\tau\big)-2 &= \; -\left(\frac{d_1^2\,d_8^3}{d_2^3\, d_{16}^2}\right)^{2}-2 \;= \; -\frac{d_1^2\,d_4^{12}}{d_2^9\,d_8^3\,d_{16}^2}\end{align}$$
III. Inter-level quadratic relations: We also have,
$$\begin{align} j_{4C} &= j_{8E}+\frac{16}{j_{8E}}\\ j_{4D} &= j_{8E}-\frac{16}{j_{8E}} \end{align}$$ $$\begin{align}j_{8D} &= j_{16B}-\frac{4}{j_{16B}}\\ j_{8E} &= j_{16B}+\frac{4}{j_{16B}}\quad\end{align}$$ Adding each pair leaves functions with the same argument,
$$\begin{align}  j_{4C}+j_{4D} &= 2\color{red}{j_{8E}}\\  j_{8D}+j_{8E} &= 2\color{red}{j_{16B}} \end{align}$$ which are 2 of the 9 linear relations found by Conway, Norton, and Atkins.

Level 12, part 3

(Under construction)

Level 12, part 2

(Under construction)

Level 12, part 1

Level 12 is very special because, given an eta quotient $a$, we can find $a+m_i$ as an eta quotient for four different integer values $m_i$. However, that also makes the situation more complicated than in other levels. Let,
$$a(\tau) = \left(\frac{d_4^2\,d_6}{d_2\,d_{12}^2}\right)^2,\quad b(\tau) = \frac{d_3^3\,d_4}{d_1\,d_{12}^3}, \quad c(\tau) =\frac{d_1^3\,d_4\, d_6^2}{d_2^2\,d_3\,d_{12}^3}$$ such that $a+1 = b,\,a-3 = c.$ And the monster functions, $$\begin{align} j_{12A}(\tau) &= \left(\frac{d_2^2\, d_6^2}{d_1\,d_3\,d_4\, d_{12}}\right)^{6}\\ \color{red}{j_{12B}}(\tau) &= \left(\frac{d_1\, d_4\,d_6}{d_2\,d_3\,  d_{12}}\right)^4\,=\,\frac{ac}b\\ j_{12C}(\tau) &= \sqrt{j_{6A}(2\tau)}\\ &=j_{12F}(\tau)+\frac{1}{j_{12F}(\tau)}\\ &= j_{12G}(\tau)+\frac{9}{j_{12G}(\tau)}\\ j_{12D}(\tau) &= \left(\frac{d_6^2}{d_3\,  d_{12}}\right)^{8}=\sqrt[3]{j_{4A}(3\tau)}\\ \color{blue}{j_{12E}}(\tau) &= \left(\frac{d_1\, d_3}{d_4\,  d_{12}}\right)^{2}\;=\;\;\frac{bc}a\\ j_{12F}(\tau) &= \left(\frac{d_4\, d_6}{d_2\,  d_{12}}\right)^{6}=\sqrt{j_{6B}(2\tau)}\\ j_{12G}(\tau) &= \left(\frac{d_2\, d_4}{d_6\,  d_{12}}\right)^{2}=\sqrt{j_{6D}(2\tau)}\\ \color{blue}{j_{12H}}(\tau) &= \left(\frac{d_3\, d_4}{d_1\,  d_{12}}\right)^{4}\;=\;\frac{ab}c\\ \color{red}{j_{12I}}(\tau) &= \left(\frac{d_4^2\, d_6}{d_2\,  d_{12}^2}\right)^{2}\;=\; a\\ j_{12J}(\tau) &= \;\left(\frac{d_6}{d_{12}}\right)^{4}\; = \;\sqrt{j_{6F}(2\tau)} \end{align}$$
Example: $\quad j_{12C}\left(\tfrac12\sqrt{-17/6}\right) = 198$

Let $a=\color{red}{j_{12I}},\;m = 1,\;n=-3,$ then,
$$\begin{align} j_{12E} &= a-\frac{3}{a}-2\\ j_{12B} &= a+1+\frac{4}{a+1}-5\\ j_{12H} &= a-3+\frac{12}{a-3}+7 \end{align}$$ as well as,
$$j_{12A} =  j_{12E}+\frac{4^2}{j_{12E}}+8 =  j_{12B}+\frac{3^2}{j_{12B}}+10 =  j_{12H}+\frac{1}{j_{12H}}+2$$ which implies the linear relation between 5 monster functions,
$$j_{12A}+2\color{red}{j_{12I}} =\color{red}{ j_{12B}}+\color{blue}{j_{12E}}+\color{blue}{j_{12H}}+8$$ The functions $j_{12I}$ and  $j_{12B}$ can be used for level 12, while versions of $j_{12E}$ and  $j_{12H}$ will be useful in level 24.

Level 10

I. Moonshine functions: Define, $$a(\tau) =\left(\frac{d_2^2\,d_5}{d_1\,d_{10}^2}\right)^2,\quad b(\tau) = \left(\frac{d_2\,d_5^5}{d_1\,d_{10}^5}\right),\quad c(\tau) =\left(\frac{d_1^3\,d_5}{d_2\,d_{10}^3}\right)$$ then $$\left(\frac{d_2^2\,d_5}{d_1\,d_{10}^2}\right)^2-1=\left(\frac{d_2\,d_5^5}{d_1\,d_{10}^5}\right)\\ \left(\frac{d_2^2\,d_5}{d_1\,d_{10}^2}\right)^2-5=\left(\frac{d_1^3\,d_5}{d_2\,d_{10}^3}\right)$$ so $a-1 = b,\,a-5 = c$ and we get the moonshine functions $$\begin{align} j_{10A}(\tau) &= j_{10B}(\tau)+\frac{4^2}{j_{10B}(\tau)}+8\\ j_{10B}(\tau) &= \left(\frac{d_1\, d_5}{d_2\,  d_{10}}\right)^{4}=\frac{bc}{a}\\ j_{10C}(\tau) &= \left(\frac{d_1\, d_2}{d_5\,  d_{10}}\right)^{2} =\frac{ac}{b}\\ j_{10D}(\tau) &= \left(\frac{d_2\, d_5}{d_1\,  d_{10}}\right)^{6} = \frac{ab}{c}\\ j_{10E}(\tau) &= \left(\frac{d_2^2\,d_5}{d_1\,d_{10}^2}\right)^2 \;=\; a \\ \end{align}$$ Example: $\quad j_{10A}\left(\sqrt{-19/10}\right) = 76^2$

II. Relations: Let $a,b,c$ as defined above and $m=-1,\,n=-5,$ then the system S1 in the introduction explains all four trinomial identities of level 10 (denoted t10) found by Somos. For S2, let $a=\color{red}{j_{10E}}$ then, $$\begin{align} j_{10B} &= a+\frac{5}{a}-6\\ j_{10C} &= a-1-\frac{4}{a-1}-3\\ j_{10D} &= a-5+\frac{20}{a-5}+9 \end{align}$$ as well as S3, $$j_{10A} =  j_{10B}+\frac{4^2}{j_{10B}}+8 =  j_{10C}+\frac{5^2}{j_{10C}}+6 = j_{10D}+\frac{1}{j_{10D}}-2$$ which implies the linear relation between 5 moonshine functions,
$$j_{10A}+2\color{red}{j_{10E}} = j_{10B}+j_{10C}+j_{10D}+8$$
III. Special property. As mentioned in the Intro, there are special triples {$a',b',c'$} of level $N = 6, 10, 12, 18, 30$ that, using a common formula, can generate another triple {$a',b',c'$} of level $2N$. For $N = 10$,
$$\begin{align} a'(\tau) &= -a(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_{10}}{d_2\,d_5\,d_{20}}\right)^2 = j_{20C}(\tau)\\ b'(\tau) &= -b(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_{10}^{10}}{d_2^2\,d_5^5\,d_{20}^5}\right)\\ c'(\tau) &= -c(\tfrac12+\tau) = \left(\frac{d_2^{8}}{d_1^3\,d_4^3\,d_5\,d_{20}}\right) \end{align}$$  such that $a'+1=b',\; a'+5=c'$.

Level 8

I. Moonshine functions: Define, $$a(\tau)=\left(\frac{d_4^3}{d_2\,  d_8^2}\right)^{4},\quad b(\tau) = \left(\frac{d_1^2\,d_4}{d_2\,d_8^2}\right)^2, \quad c(\tau) = \left(\frac{d_2^5}{d_1^2\,d_4\,d_8^2}\right)^2$$   then, $$\left(\frac{d_4^3}{d_2\,  d_8^2}\right)^{4}-4 = \left(\frac{d_1^2\,d_4}{d_2\,d_8^2}\right)^2\quad\\ \left(\frac{d_4^3}{d_2\,  d_8^2}\right)^{4}+4=\left(\frac{d_2^5}{d_1^2\,d_4\,d_8^2}\right)^2$$ so $a-4 = b,\, a+4 = c$. This $a,b,c$ also belong to the class where $a(\tau) = -a\big(\tfrac12+\tau\big)$ and $b(\tau) = -c\big(\tfrac12+\tau\big)$. We then have the six level 8 moonshine functions,
$$\qquad\begin{align} j_{8A}(\tau) &= \left(\frac{d_2\, d_4}{d_1\,  d_8}\right)^{8}\;=\;\frac{ac}b\\ j_{8B}(\tau) &= \left(\frac{d_4^2}{d_2\,  d_8}\right)^{12}=\sqrt{j_{4A}(2\tau)}\\ j_{8C}(\tau) &= \sqrt{j_{4B}(\tau)}\,=\,\sqrt[4]{j_{2A}(2\tau)}\\ &= \sqrt{j_{8A}(\tau)}-\frac{4}{\sqrt{j_{8A}(\tau)}}\\ j_{8D}(\tau) &= \left(\frac{d_2}{d_8}\right)^{4}\\ \color{red}{j_{8E}}(\tau) &= \left(\frac{d_4^3}{d_2\,  d_8^2}\right)^{4}\;=\; a\\ j_{8F}(\tau) &= \left(\frac{d_4}{d_8}\right)^{6} \end{align}$$ Example: $\quad j_{8C}\big(\tfrac14\sqrt{-58}\big) = 396$

II. Relations: Using S1, this $a,b,c$ explains all of the four trinomial identities of level 8 (prefixed with t8) by Somos. They also obey the relation,
$$\left(\frac{bc}{a}+64\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad$$ In general, $$\left(\frac{bc}{a}+(b-c)^2\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad$$  where $b-c=8$. Or in terms of named functions of level 4 and 8,
$$\quad\quad j_{4B}\;+\;2\color{red}{j_{8E}} = j_{4D}+j_{8A}+j_{8A'}$$ where, $$j_{8A'}(\tau) = -j_{8A}\big(\tfrac12+\tau\big) = \left(\frac{d_1\, d_4^2}{d_2^2\,  d_8}\right)^{8}$$

Level 6

I. Moonshine functions:  Let, $$a(\tau)=\left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4,\quad b(\tau)= \left(\frac{d_2\,d_3^3}{d_1\,d_6^3}\right)^3,\quad c(\tau)=\left(\frac{d_1^5\,d_3}{d_2\,d_6^5}\right)$$ then, $$\left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4-1=\left(\frac{d_2\,d_3^3}{d_1\,d_6^3}\right)^3\\ \left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4-9=\left(\frac{d_1^5\,d_3}{d_2\,d_6^5}\right)$$  So $m = -1,\,n=-9,$ and we get the moonshine functions, $$\begin{align} j_{6A}(\tau) &= j_{6D}(\tau)+\frac{9^2}{j_{6D}(\tau)}+18\\ j_{6B}(\tau) &= \left(\frac{d_2\,d_3}{d_1\,d_6}\right)^{12} = \frac{ab}{c} \\ j_{6C}(\tau) &= \left(\frac{d_1\,d_3}{d_2\,d_6}\right)^6 = \frac{bc}{a} \\ j_{6D}(\tau) &= \left(\frac{d_1\,d_2}{d_3\,d_6}\right)^4 = \frac{ac}{b} \\ j_{6E}(\tau) &= \left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4 = a\\ j_{6F}(\tau) &=\;\left(\frac{d_3}{d_6}\right)^{8}\\ \end{align}$$ Example: $\; j_{6A}\left(\sqrt{-17/6}\right) = 198^2$

II. Relations: Let $a,b,c$ as defined above and $m=-1,\,n=-9,$ then the system S1 given in the Introduction,
$$\begin{align} a+m &= b\\ a+n &= c\\ a(m-n)+bn &=cm\\ b-c &= (m-n)\end{align}$$ explains all four trinomial identities of level 6 (with prefix t6) found by Somos. For S2, let $a=\color{red}{j_{6E}}$ then,
$$\begin{align} j_{6B}(\tau)  &= a-9+\frac{72}{a-9}+17\\ j_{6C}(\tau)  &= a+\frac{9}{a}-10\\ j_{6D}(\tau) &= a-1-\frac{8}{a-1}-7\\ \end{align}$$ as well as S3, $$j_{6A} =  j_{6B}+\frac{1}{j_{6B}}+2 = j_{6C}+\frac{8^2}{j_{6C}}+20 = j_{6D}+\frac{9^2}{j_{6D}}+18$$ which implies the linear relation between 5 moonshine functions,
$$j_{6A}+2\color{red}{j_{6E}} = j_{6B}+j_{6C}+j_{6D}+20$$ These are general phenomena but, in levels 6, 10, 12, 18, 30, the five functions are ALL moonshine.

III. Special property. As mentioned in the Intro, there are special triples {$a',b',c'$} of level $N = 6, 10, 12, 18, 30$ that using a common formula can generate another triple {$a',b',c'$} of level $2N$. For $N = 6$,
$$\begin{align} a'(\tau) &= -a(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_6}{d_2\,d_3\,d_{12}}\right)^4 = j_{12B}(\tau)\\ b'(\tau) &= -b(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_6^6}{d_2^2\,d_3^3\,d_{12}^3}\right)^3\\ c'(\tau) &= -c(\tfrac12+\tau) = \left(\frac{d_2^{14}}{d_1^5\,d_3\,d_4^5\,d_6^2\,d_{12}}\right) \end{align}$$  such that $a'+1=b',\; a'+9=c'$.

Level 4

I. Moonshine function Define, $$a(\tau) = \left(\frac{d_2^3}{d_1\,d_4^2}\right)^8,\quad b(\tau) = \left(\frac{d_1}{d_4}\right)^8+8$$ then $$\left(\frac{d_2^3}{d_1\,d_4^2}\right)^8-8=\left(\frac{d_1}{d_4}\right)^8+8$$ Equivalently, using the reciprocal of $a(\tau)$, $$\left(\frac{\sqrt2\,d_1\,d_4^2}{d_2^3}\right)^8+\left(\frac{d_1^2\,d_4}{d_2^3}\right)^8 = 1$$ which are just versions of the single trinomial identity for level 4. We also have the four level 4 moonshine functions, $$\qquad\qquad\begin{align} j_{4A}(\tau) &= \left(\left(\frac{d_1}{d_4}\right)^4+16\left(\frac{d_4}{d_1}\right)^4\right)^2=\left(\frac{d_2^2}{d_1\,d_4}\right)^{24}\\ j_{4B}(\tau) &=\left(\left(\frac{d_2\,d_4}{d_{1}\;d_8}\right)^4-4\left(\frac{d_{1}\;d_8}{d_2\,d_4}\right)^4\right)^2 = \sqrt{j_{2A}(2\tau)} \\ &=j_{4D}(\tau)+\frac{2^6}{j_{4D}(\tau)} \\ \color{red}{j_{4C}}(\tau) &= \left(\frac{d_1}{d_4}\right)^8+8\; = \;b(\tau)\\ j_{4D}(\tau) &= \left(\frac{d_2}{d_4}\right)^{12} \end{align}$$ Note that the function $j_{4C}(\tau) = \frac{16}{\lambda(2\tau)}-8=\left(\frac{d_1}{d_4}\right)^8+8$ with modular lambda function $\lambda(\tau)$ is a normalized Hauptmodul.

II. Comparisons: Notice that, $$j_{4C} =\left(\frac{d_2^3}{d_1\,d_4^2}\right)^8-8,\quad  j_{8E} =\left(\frac{d_4^3}{d_2\,d_8^2}\right)^4,\quad  j_{16B} =\left(\frac{d_8^3}{d_4\,d_{16}^2}\right)^2$$ The last two will play analogous roles for levels 8 and 16. Also, $j_{4A}$ is remarkably a 24th power. Versions appear in levels that are 4m divisors of 24, $$j_{4A}=\left(\frac{d_2^2}{d_1\,d_4}\right)^{24},\quad j_{8B}=\left(\frac{d_4^2}{d_2\,d_8}\right)^{12},\quad j_{12D}=\left(\frac{d_6^2}{d_3\,d_{12}}\right)^{8},\quad j_{24E}=\left(\frac{d_{12}^2}{d_6\,d_{24}}\right)^{4}$$ Similarly, versions of $j_{2B}$ appear in levels that are 2m divisors of 24, $$j_{2B}(\tau) = \left(\frac{d_1}{d_2}\right)^{24},\quad j_{4D}(\tau) = \left(\frac{d_2}{d_4}\right)^{12},\quad j_{6F}(\tau) = \left(\frac{d_3}{d_6}\right)^{8},\\ j_{8F}(\tau) = \left(\frac{d_4}{d_8}\right)^{6},\quad j_{12J}(\tau) = \left(\frac{d_6}{d_{12}}\right)^{4},\quad j_{24J}(\tau) = \left(\frac{d_{12}}{d_{24}}\right)^{2}$$

Entry 39

Let $\color{red}{q = e^{2\pi i\tau}}$. Given $\tau = \sqrt{-n}$ or $\tau = \frac{1+\sqrt{-n}}{2}$, then the solutions $\alpha, \beta,\gamma$ to the following equations,
$$\frac{_2F_1\big(\tfrac12,\tfrac12;1;\,1-\alpha\big)}{_2F_1\big(\tfrac12,\tfrac12;1;\,\alpha\big)}\,i=\color{red}{\sqrt{4}\,\tau}$$ $$\frac{_2F_1\big(\tfrac13,\tfrac23;1;\,1-\beta\big)}{_2F_1\big(\tfrac13,\tfrac23;1;\,\beta\big)}\,i=\color{red}{\sqrt{3}\,\tau}$$ $$\frac{_2F_1\big(\tfrac14,\tfrac34;1;\,1-\gamma\big)}{_2F_1\big(\tfrac14,\tfrac34;1;\,\gamma\big)}\,i=\color{red}{\sqrt{2}\,\tau}$$ are given by, $$\;\alpha =\frac{16}{u_1^8+16} = \left(\frac{\sqrt2 }{u_2}\right)^8  = \left(\frac{\vartheta_2(q)}{\vartheta_3(q)}\right)^4$$ $$\beta = \frac{27}{v_1^{12}+27} = \left(\frac{3}{v_2^3+3}\right)^3 = \left(\frac{c(q)}{a(q)}\right)^3$$ $$\gamma = \frac{64}{w_1^{24}+64} = \left(\frac{8}{w_2^8+8}\right)^2 = \left(\frac{f(q)}{d(q)}\right)^2$$ where, $$\quad u_1 = \frac{\eta(\tau)}{\eta(4\tau)},\quad u_2 = \frac{\eta^3(2\tau)}{\eta(\tau)\,\eta^2(4\tau)}$$ $$v_1 = \frac{\eta(\tau)}{\eta(3\tau)},\quad v_2 = \frac{\eta(\tau/3)}{\eta(3\tau)}$$ $$w_1 = \frac{\eta(\tau)}{\eta(2\tau)},\quad w_2 = \frac{\eta(\tau/2)}{\eta(2\tau)}$$ and where the functions of $\color{red}{q = e^{2\pi i\tau}}$ are the Jacobi and Borwein theta functions discussed in Entry 38. Also, $\alpha= \lambda(2\tau)$ with the modular lambda function $\lambda(\tau)$.

Entry 38

For consistency, let the variable $q$ be the nome's square $q=e^{2\pi i \tau}\,$throughout. 

I. The null Jacobi theta functions (with $z=0$) are,
$$\qquad\qquad\quad \vartheta_3(q) = \sum_{m=-\infty}^{\infty}q^{n^2} = \frac{\eta^2(\tau)}{\eta(2\tau)}+\frac{4\,\eta^2(16\tau)}{\eta(8\tau)}=\frac{\eta^5(2\tau)}{\eta^2(\tau)\eta^2(4\tau)}$$ $$\vartheta_4(q) = \sum_{m=-\infty}^{\infty}(-1)^n\,q^{n^2} = \frac{\eta^2(\tau)}{\eta(2\tau)}$$ $$\vartheta_2(q) = \sum_{m=-\infty}^{\infty}q^{(n+1/2)^2} = \frac{2\,\eta^2(4\tau)}{\eta(2\tau)}$$
II. The Borwein cubic theta functions are,
$$a(q)=\sum_{m,n=-\infty}^{\infty}q^{m^2+mn+n^2} = \frac{\eta^3(\tau)}{\eta(3\tau)} +\frac{9\,\eta^3(9\tau)}{\eta(3\tau)}$$ $$b(q)=\sum_{m,n=-\infty}^{\infty}\zeta^{m-n}q^{m^2+mn+n^2} = \frac{\eta^3(\tau)}{\eta(3\tau)}\quad\quad$$ $$\quad c(q)=\sum_{m,n=-\infty}^{\infty}q^{(m+1/3)^2+(m+1/3)(n+1/3)+(n+1/3)^2} = \frac{3\,\eta^3(3\tau)}{\eta(\tau)}$$
where $\zeta =e^{2\pi i/3}$. 

III. The derived Jacobi theta functions are,
$$\qquad\qquad d(q)= \vartheta_4^4(q)+2\,\vartheta_2^4(q) = \frac{\eta^8(\tau)}{\eta^4(2\tau)} +\frac{32\,\eta^8(4\tau)}{\eta^4(2\tau)}$$ $$\quad e(q)= \vartheta_4^4(q)= \frac{\eta^8(\tau)}{\eta^4(2\tau)}$$ $$\quad f(q)= \tfrac12 \vartheta_2^4(q^{1/2})= \frac{8\,\eta^8(2\tau)}{\eta^4(\tau)}$$ These obey the beautiful relations, $$\vartheta_3^4(q)=\vartheta_4^4(q)+\vartheta_2^4(q)$$ $$a^3(q)=b^3(q)+c^3(q)$$ $$d^2(q)=e^2(q)+f^2(q)$$ As well as, $$\vartheta_3(q)\vartheta_3(q^3)=\vartheta_4(q)\vartheta_4(q^3)+\vartheta_2(q)\vartheta_2(q^3)$$ $$a(q)a(q^2)=b(q)b(q^2)+c(q)c(q^2)$$ $$d(q)d(q)=e(q)e(q)+f(q)f(q)$$ the last of which naturally leads to the third one above. Also, $$\vartheta_3(q^4)=\vartheta_4(q)+\vartheta_2(q^4)$$ $$a(q^3)=b(q)+c(q^3)$$ $$d(q^2)=e(q)+f(q^2)$$

The Jacobi thetas can be expressed in terms of each other,
$$\vartheta_3(q) = \vartheta_4(q)+2\,\vartheta_2(q^4)$$ $$\vartheta_4(q) = 2\,\vartheta_3(q^4)-\vartheta_3(q)$$ $$\vartheta_2(q) = \vartheta_3(q^{1/4})-\vartheta_3(q)$$ similarly for the Borwein thetas, $$a(q) = b(q) + 3\,c(q^3)$$ $$2\,b(q) = 3a(q^3)-a(q)$$ $$2\,c(q) = a(q^{1/3})-a(q)$$ and also similarly for the derived Jacobi thetas, $$d(q) = e(q) + 4\,f(q^2)$$ $$3\,e(q) = 4d(q^2)-d(q)$$ $$3\,f(q) = d(q^{1/2})-d(q)$$ Furthermore, we have the similar, $$\big(\vartheta_3(q)\big)^2 = 1+4\sum_{n=0}^\infty\left(\frac{q^{4n+1}}{1-q^{4n+1}}-\frac{q^{4n+3}}{1-q^{4n+3}}\right)$$ $$\quad\quad a(q) = 1+6\sum_{n=0}^\infty\left(\frac{q^{3n+1}}{1-q^{3n+1}}-\frac{q^{3n+2}}{1-q^{3n+2}}\right)$$