Entry 60

(Under construction.)

Entry 59

(Under construction.)

Entry 58

(Under construction.)

Entry 57

(Summary: Under construction)

Entry 56

For Level 72, these are no longer monster functions. 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 new eta quotient, $\color{blue}{f_p} =\dfrac{\eta(p\tau)}{\eta(2p\tau)}$. Then,

$$\begin{align} \left(\frac{f_{4}^2\, f_{6}^2\, f_{36}^2}{f_{2}\,f_{12}^4\, f_{18}}\right) + 1 &= \left(\frac{f_{3}^2\,f_{4}\, f_{36}}{f_{1}\, f_{9}\,f_{12}^2}\right)f_6^2 \\  \left(\frac{f_{4}^2\, f_{6}^2\, f_{36}^2}{f_{2}\,f_{12}^4\, f_{18}}\right) - 1 &=  \left(\frac{f_{1}\, f_{4}\,f_{6}^2\, f_{9}\, f_{36}}{f_{2}\, f_{3}^2\, f_{12}^2\,f_{18}}\right)f_6^2 \end{align}$$ Or more simply $$a+1=b \qquad \\ a-1=c \qquad$$ From the above, we find $(a,b,c)$. Define, $$\begin{align}\frac{ab}c &= \left(\frac{f_{3}^2\,f_{4}\, f_{36}}{f_{1}\, f_{9}\,f_{12}^2}\right)^2 \\ \frac{bc}a &= \,\big(f_6\big)^4 \\ \frac{ac}b &= \left(\frac{f_{1}\, f_{4}\,f_{6}^2\, f_{9}\, f_{36}}{f_{2}\, f_{3}^2\, f_{12}^2\,f_{18}}\right)^2 \end{align}$$ Then we have the analogous relation,

$$\left(\frac{4a}{bc}+\frac{bc}a\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$

Entry 55

Define $d_k = \eta(k\tau)$ with Dedekind eta function $\eta(\tau)$. Then for Level 60,

$$\begin{align}\frac{d_2\,d_3\,d_5\,d_{12}\,d_{20}\,d_{30}}{d_1\,d_4\,d_6\,d_{10}\,d_{15}\,d_{60}}+1 &= \frac{(d_2\,d_3\,d_5\,d_{12}\,d_{20}\,d_{30})(d_2\, d_{30})^3}{(d_1\,d_4\,d_6\,d_{10}\,d_{15}\,d_{60})^2}\\ \frac{d_2\,d_3\,d_5\,d_{12}\,d_{20}\,d_{30}}{d_1\,d_4\,d_6\,d_{10}\,d_{15}\,d_{60}}-1 &= \frac{(d_6\,d_{10})^3}{d_2\,d_3\,d_5\,d_{12}\,d_{20}\,d_{30}}\end{align}$$

Or more simply $$a+1=b \qquad \\ a-1=c \qquad$$ we find $(a,b,c)$ as the McKay-Thompson series of class 60C for Monster (A145725). They obey,

$$\left(\frac{4a}{bc}+\frac{bc}a\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$

If we express $a=\dfrac{p}{q}$ as

$$\begin{align} p &= d_2\,d_3\,d_5\,d_{12}\,d_{20}\,d_{30}\\ q &= d_1\,d_4\,d_6\,d_{10}\,d_{15}\,d_{60} \end{align}$$

then we have the simple cube, $$\frac{pq+q^2}{p}=\big(d_2\,d_{30})^3$$ where, of course, Level $60 = 2\times30$.

Entry 54

Define $d_k = \eta(k\tau)$ with Dedekind eta function $\eta(\tau)$. Then for Level 36,

$$\begin{align}\left(\frac{d_2\,d_3^2\,d_{12}^2\, d_{18}}{d_1\,  d_4\,d_6^2\, d_9\, d_{36}}\right)^2+1 &= \left(\frac{d_3\,d_{12}}{d_6^2}\right)^2\left(\frac{d_2^2\,d_{18}^2}{d_1\,d_4\,d_9\,d_{36}}\right)^3 \\ \left(\frac{d_2\,d_3^2\,d_{12}^2\, d_{18}}{d_1\,  d_4\,d_6^2\, d_9\, d_{36}}\right)^2-3 &= \left(\frac{d_6^2}{d_3\,d_{12}}\right)^2 \left(\frac{d_1\,d_4\,d_6^{12}\,d_9\,d_{36}}{\big(d_2\,d_3\,d_{12}\,d_{18}\big)^4}\right)\end{align}$$

Or more simply $$a+1=b \qquad \\ a-3=c \qquad$$ and we find $(a,b,c)$ as the McKay-Thompson series of class 36A for Monster (A227585). They obey,

$$a-\frac3{a}-2=\frac{bc}{a}=\left(\frac{d_6^2}{d_3\,d_{12}}\right)^8\quad$$

$$\left(\frac{16a}{bc}+\frac{bc}a\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}\;$$

but this is not one of the 9 linear dependencies by Conway et al.

Entry 53

Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$. Then for Level 30

$$\begin{align}\frac{d_6\, d_{10}}{d_3\, d_5}\left(\frac{d_1\, d_{15}}{d_2\, d_{30}}\right)^2+1 &= \frac{d_1\, d_{15}}{d_3\, d_5}\left(\frac{d_6\, d_{10}}{d_2\, d_{30}}\right)^2\\ \frac{d_6\, d_{10}}{d_3\, d_5}\left(\frac{d_1\, d_{15}}{d_2\, d_{30}}\right)^2+2 &= \left(\frac{d_3\, d_5}{d_2\, d_{30}}\right)\end{align}$$

Expressed as the triple of eta quotients $(a,b,c)$ such that $$a+1 =b\\ a+2=c$$ then $(m,n)=(1,2)$ and
$$\begin{align} \frac{ab}c &= \left(\frac{d_1\, d_6\, d_{10}\, d_{15}}{d_2\,  d_{3}\, d_{5}\, d_{30}}\right)^3\\ \frac{bc}a &= \left(\frac{d_3\, d_5\, d_{6}\, d_{10}}{d_1\,  d_{2}\, d_{15}\, d_{30}}\right)\\ \frac{ac}b &= \left(\frac{d_1\, d_3\, d_{5}\, d_{15}}{d_2\,  d_{6}\, d_{10}\, d_{30}}\right)\end{align}$$ where $(a,b,c)$ are the McKay-Thompson series of class 30G (A133098). They obey $$\left(\frac{a}{bc}+\frac{bc}a\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$ which is one of the $9$ dependencies found by Conway, Norton, and Atkin such that the moonshine functions span a linear space of $172-9=163$ dimensions. Similar identities involving only moonshine functions exist for levels $(6, 10, 12, 18, 30)$. So this is the highest level.

Entry 52

Level 24, Part 3. To summarize the two parts of level 24, we found two eta triples $(a,b,c)$ and $(d,e,f)$. Let $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$. Define $$a=\left(\frac{d_6\, d_{12} }{d_2\, d_ 4}\right) \left(\frac{d_1\, d_8 }{d_3\, d_{24} }\right)^2,\quad b=\left(\frac{d_1\, d_8 }{d_2\, d_4 }\right) \left(\frac{d_6\, d_ {12}}{d_3\, d_{24} }\right)^3,\quad c = \frac{\big( d_2\, d_4\big)^2 }{d_1\, d_3\, d_8\, d_{24} }\quad \\ d=\left(\frac{d_2\, d_{12} }{d_4\, d_ 6}\right) \left(\frac{d_3\, d_8 }{d_1\, d_{24} }\right)^2,\quad e=\frac{d_2^5\, d_3\, d_8\, d_{12}^5 }{\big(d_1\, d_4\, d_6\, d_{24}\big)^3},\qquad f=\frac{\big(d_4\, d_6\big)^4 }{d_1\, d_2^2\, d_3\, d_8\, d_{12}^2\, d_{24}}$$

then they obey a lot of relationships, $$\frac{a}{b c}= \frac{d}{e f}$$

$$a+1 = b,\quad a+3 = c\\ d+1=e,\quad d-1=f$$

$$\frac{ab}c+ \frac{bc}a+ \frac{ac}{b} - \left(\frac{de}f+ \frac{ef}d+ \frac{df}{e}\right) = 2(a-d) = 2(b-e) = 2(c-f-4)$$ as well as

$$\left(\frac{4a}{bc}+\frac{bc}{a}\right)+2a  = \color{red}{\frac{ab}c}+ \frac{bc}a+ \frac{ac}{b}\\ \left(\frac{4d}{ef}+\frac{ef}{d}\right)+2d  = \color{red}{\frac{de}f}+ \frac{ef}d+ \frac{df}{e}\\ \;\quad\left(\frac{4a}{bc}+\frac{bc}{a}\right)+\frac{2cd}{e} - 2 = \color{red}{\frac{ab}c}+\color{red}{\frac{de}f}+\left(\frac{cd}{e}-\frac{3e}{cd}\right)$$ and so on. Combining the last three relations by getting rid of the red non-moonshine terms yields the most complicated of the $9$ dependencies found by Conway, Norton, and Atkin such that the monster functions span a linear space of $172-9=163$ dimensions. 

Entry 51

Level 24, Part 2. As pointed out in Part 1, Level 24 is unusual since two triples $(a,b,c)$ have the special property $$\frac{a_1}{b_1\, c_1} = \frac{a_2}{b_2\, c_2}$$ These also form linear dependencies but involve non-monster functions. But if we combine the two, plus some level 12 functions, then there is a relationship with only monster functions. Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$. Then $$\quad\left(\frac{d_2\, d_{12} }{d_4\, d_ 6}\right) \left(\frac{d_3\, d_8 }{d_1\, d_{24} }\right)^2+1=\frac{d_2^5\, d_3\, d_8\, d_{12}^5 }{\big(d_1\, d_4\, d_6\, d_{24}\big)^3}\\ \qquad \left(\frac{d_6\, d_{12} }{d_2\, d_ 4}\right) \left(\frac{d_3\, d_8 }{d_1\, d_{24} }\right)^2-1=\frac{\big(d_4\, d_6\big)^4 }{d_1\, d_2^2\, d_3\, d_8\, d_{12}^2\, d_{24} }$$  Expressed as the triple of eta quotients $(a,b,c)$ such that $$a+1 =b\\ a-1=c$$ so $(m,n)=(1,-1)$. Then $$\begin{align}\frac{ab}{c} &= \left(\frac{d_2^2\,d_3\,d_8\,d_{12}^2}{d_1\,d_4^2\,d_6^2\,d_{24}}\right)^4\\ \color{blue}{\frac{bc}{a}} &= \left(\frac{d_2\, d_4\, d_6\, d_{12}}{d_1\, d_3\, d_8\, d_{24}}\right)^2 = T_{24B}\\ \frac{ac}{b} &=  \left(\frac{d_4\, d_6}{d_2\,d_{12}}\right)^6 \end{align}$$  where $(a,b,c)$ are the McKay-Thompson series of class 24I (A138688). They obey $$\frac{bc}a = a-\frac1a$$ $$\left(\frac{4a}{bc}+\frac{bc}{a}\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$ This is not one of the 9 linear dependencies by Conway et al since one of the terms, again $\left(\frac{ab}c\right)$, is not a moonshine function. But just like in Part 1, the $\left(\frac{bc}a\right)$ term is the McKay-Thompson series of class 24B (A212771).

Entry 50

Level 24, Part 1. We can find multiple triples $(a,b,c)$, but Level 24 is unusual since two have the special property $$\frac{a_1}{b_1\, c_1} = \frac{a_2}{b_2\, c_2}$$ Naturally, these also form linear dependencies but involve non-monster functions. However, if we combine the two, plus some level 12 functions, then there is a relationship with only monster functions. Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$. Then $$\qquad \left(\frac{d_6\, d_{12} }{d_2\, d_ 4}\right) \left(\frac{d_1\, d_8 }{d_3\, d_{24} }\right)^2+1=\left(\frac{d_1\, d_8 }{d_2\, d_4 }\right) \left(\frac{d_6\, d_ {12}}{d_3\, d_{24} }\right)^3\\ \left(\frac{d_6\, d_{12} }{d_2\, d_ 4}\right) \left(\frac{d_1\, d_8 }{d_3\, d_{24} }\right)^2+3= \frac{\big( d_2\, d_4\big)^2 }{d_1\, d_3\, d_8\, d_{24} }\qquad$$  Expressed as the triple of eta quotients $(a,b,c)$ such that $$a+1 =b\\ a+3=c$$ so $(m,n)=(1,3)$. Then $$\begin{align}\frac{ab}{c} &= \left(\frac{d_1\,d_6\,d_8\,d_{12}}{d_2\,d_3\,d_4\,d_{24}}\right)^4\\ \color{blue}{\frac{bc}{a}} &= \left(\frac{d_2\, d_4\, d_6\, d_{12}}{d_1\, d_3\, d_8\, d_{24}}\right)^2 = T_{24B}\\ \frac{ac}{b} &=  \left(\frac{d_2\, d_4}{d_6\,d_{12}}\right)^2 \end{align}$$ where $(a,b,c)$ are the McKay-Thompson series of class 24C (A206298). They obey $$\frac{bc}a = a+\frac3a+4$$ $$\left(\frac{4a}{bc}+\frac{bc}{a}\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$ However, this is not one of the 9 linear dependencies by Conway et al since one of the terms, namely $\left(\frac{ab}c\right)$, is not a moonshine function. But the $\left(\frac{bc}a\right)$ term is the McKay-Thompson series of class 24B (A212771) and also appears in Part 2.

Entry 49

Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$. Then for Level 20

$$\left(\frac{d_1\, d_4\, d_{10}}{d_2\,  d_5\, d_{20}}\right)^2 +1 =  \left(\frac{d_1\,d_4\,d_{10}^{10}}{d_2^2\,d_5^5\,d_{20}^5}\right)\quad\\ \left(\frac{d_1\, d_4\, d_{10}}{d_2\,  d_5\, d_{20}}\right)^2 + 5 =  \left(\frac{d_2^{8}}{d_1^3\,d_4^3\,d_5\,d_{20}}\right)$$

This level 20 triple can be derived from a level 10. Expressed as the triple of eta quotients $(a,b,c)$ such that $$a+1 =b\\ a+5=c$$ then $(m,n)=(1,5)$ and $$\begin{align}\frac{ab}c &= \left(\frac{d_1\,d_4\,d_{10}^2}{d_2^2\,d_5\,d_{20}}\right)^6\\ \frac{bc}a &= \left(\frac{d_2^2\, d_{10}^2}{d_1\,  d_4\, d_5\, d_{20}}\right)^4\\ \frac{ac}b &= \left(\frac{d_2^4\,d_5\,d_{20}}{d_1\,d_4\,d_{10}^4}\right)^2\end{align}$$ where $(a,b,c)$ are the McKay-Thompson series of class 20C (A145740). They obey $$\left(\frac{16a}{bc}+\frac{bc}a\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$ but is not one of the $9$ dependencies found by Conway et al since some of the terms are not moonshine functions.

Entry 48

Level 18, Part 2. Level 18 has a second triple but it does not include purely moonshine functions. Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$. Then 

$$\begin{align}\left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2-1 &=  \left(\frac{d_1^3\,d_6^2\, d_9^3}{d_2^3\,d_3^2 d_{18}^3}\right)\quad\\ \left(\frac{d_1\,d_6^4\,d_9}{d_2^2\,d_3^2\,d_{18}^2}\right)^2+3 &=\left(\frac{d_3^6}{d_1\,d_2\,d_6^2\,d_9\,d_{18}}\right)\end{align}$$ Expressed as the triple of eta quotients $(a,b,c)$ such that $$a-1 =b\\ a+3=c$$ so $(m,n)=(-1,3)$. Then 

$$\begin{align}\frac{ab}{c} &= \left(\frac{d_1\, d_6^2\, d_9}{d_2\, d_3^2\, d_{18}}\right)^6\\ \frac{bc}{a} &= \left(\frac{d_3}{d_6}\right)^8\\ \frac{ac}{b} &= \left(\frac{d_3^2\, d_6^2}{d_1\, d_2\, d_9\, d_{18}}\right)^{2}\end{align}$$ where $(a,b,c)$ are the McKay-Thompson series of class 18C (A215412). They obey $$\left(\frac{16a}{bc}+\frac{bc}a\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$ but $\dfrac{ab}{c}$ is a non-monster function with expansion (A128512). 

Entry 47

Level 18, Part 1. Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$. Then 
$$\left(\frac{d_2^2\,d_9}{d_1\,d_{18}^2}\right)-1 = \left(\frac{d_6\,d_9^3}{d_3\,d_{18}^3}\right)\quad \\ \left(\frac{d_2^2\,d_9}{d_1\,d_{18}^2}\right)-3 =\left(\frac{d_1^2\,d_6\,d_9}{d_2\,d_3\,d_{18}^2}\right)$$ Expressed as the triple of eta quotients $(a,b,c)$ such that $$a-1 =b\\ a-3=c$$ then $(m,n) = (-1,-3)$ and

$$\begin{align}\qquad\frac{ab}{c} &= \left(\frac{d_2\, d_9}{d_1\,  d_{18}}\right)^{3}\\ \frac{bc}{a} & =  \left(\frac{d_1^3\,d_6^2\, d_9^3}{d_2^3\,d_3^2 d_{18}^3}\right) = \left(\frac{d_1\,d_6^4\, d_9}{d_2^2\,d_3^2 d_{18}^2}\right)^2 -1\\ \frac{ac}{b} &= \left(\frac{d_1\, d_2}{d_9\,  d_{18}}\right) \\ d\, &= \,\left(\frac{d_3^2\, d_6^2}{d_1\, d_2\, d_9\, d_{18}}\right)^{2}\end{align}$$ where $(a,b,c)$ are the McKay-Thompson series of class 18D (A143840, A193261) and $d$ is the McKay-Thompson series of class 18B (A215407). They obey $$\left(\frac{4a}{bc}+\frac{bc}a\right)+5=d$$ $$\left(\frac{4a}{bc}+\frac{bc}a\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$ and the latter is one of the $9$ dependencies found by Conway, Norton, and Atkin such that the moonshine functions span a linear space of $172-9=163$ dimensions. Similar identities involving only moonshine functions exist for levels $(6, 10, 12, 18, 30)$, but level $18$ is special since the first term $\left(\frac{4a}{bc}+\frac{bc}a\right)$ plus an integer is an eta quotient.

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.

Entry 45

Level 12 (Part 2).  For Part 2, one term of the linear dependency is not a moonshine function. Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$. Then $$\left(\frac{d_4^2\,d_6}{d_2\,d_{12}^2}\right)^2-1=\left(\frac{d_1\,d_4^2\,d_6^9}{d_2^3\,d_3^3\,d_{12}^6}\right)\\ \quad\left(\frac{d_4^2\,d_6}{d_2\,d_{12}^2}\right)^2+3=\left(\frac{d_2^7\,d_3}{d_1^3\,d_4^2\,d_6\,d_{12}^2}\right)$$ Expressed as the triple of eta quotients $(a,b,c)$ such that $$a-1 =b\\ a+3=c$$ so $(m,n)=(-1,3)$. Then  $$\begin{align}\frac{ab}{c} &= \left(\frac{d_1\, d_4^2\,d_6^3}{d_2^3\,d_3\, d_{12}^2}\right)^{4}\\ \frac{bc}{a} &= \left(\frac{d_2^3\, d_6^3}{d_1\,d_3\,d_4^2  d_{12}^2}\right)^2\\ \frac{ac}{b} &=  \left(\frac{d_2^2\, d_3}{d_1\,d_6^2}\right)^4\\ d \, &= \, \left(\frac{d_1\, d_3}{d_2\,d_6}\right)^6 \end{align}$$ where $(a,b,c)$ are still the McKay-Thompson series of class 12I (A187144) but $d$ is the McKay-Thompson series of class 6C (A121666). They obey $$\left(\sqrt{\frac{16a}{bc}}\color{red}-\sqrt{\frac{bc}{a}}\right)^2=d$$ $$d+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}\color{red}-8$$ in contrast to Part 1 which used the positive sign. Equivalently, $$\left(\frac{16a}{bc}+\frac{bc}{a}\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}\quad$$ However, this is not one of the 9 linear dependencies by Conway et al since the $\left(\frac{ab}c\right)$ term (which is A193522) is not a moonshine function.

Entry 44

Level 12 (Part 1). We found one triple $(a,b,c)$ each for Levels 6, 8, 10. But for Levels 12, 18, 24, we can find more than one triple, which complicates things. Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$. Then $$\left(\frac{d_4^2\,d_6}{d_2\,d_{12}^2}\right)^2+1=\left(\frac{d_3^3\,d_4}{d_1\,d_{12}^3}\right)\\ \quad\left(\frac{d_4^2\,d_6}{d_2\,d_{12}^2}\right)^2-3=\left(\frac{d_1^3\,d_4\, d_6^2}{d_2^2\,d_3\,d_{12}^3}\right)$$ Expressed as the triple of eta quotients $(a,b,c)$ such that $$a+1 =b\\ a-3=c$$ so $(m,n)=(1,-3)$. Then  $$\frac{ab}{c} = \left(\frac{d_3\, d_4}{d_1\,  d_{12}}\right)^{4}\\ \frac{bc}{a} = \left(\frac{d_1\, d_3}{d_4\,  d_{12}}\right)^{2}\\ \quad\frac{ac}{b} =  \left(\frac{d_1\, d_4\,d_6}{d_2\,d_3\,  d_{12}}\right)^4\\ \qquad d = \left(\frac{d_2^2\, d_6^2}{d_1\,d_3\,d_4\,d_{12}}\right)^6$$ where $(a,b,c)$ are the McKay-Thompson series of class 12I (A187144, A187130) and $d$ is the McKay-Thompson series of class 12A (A112147). They obey $$\left(\sqrt{\frac{16a}{bc}}+\sqrt{\frac{bc}{a}}\right)^2=d$$ $$d+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}+8$$ Equivalently, $$\left(\frac{16a}{bc}+\frac{bc}{a}\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$ which is one of the $9$ dependencies found by Conway, Norton, and Atkin such that the moonshine functions span a linear space of $172-9=163$ dimensions. Similar identities involving only moonshine functions exist for levels $(6, 10, 12, 18, 30)$. However, for levels $(12, 18, 24)$ we can use a fourth eta quotient $d$ to simplify the linear dependency.

Entry 43

Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$. Then for Level 10  $$\left(\frac{d_2^2\,d_5}{d_1\,d_{10}^2}\right)^2-1=\left(\frac{d_2\,d_5^5}{d_1\,d_{10}^5}\right)\\ \left(\frac{d_2^2\,d_5}{d_1\,d_{10}^2}\right)^2-5=\left(\frac{d_1^3\,d_5}{d_2\,d_{10}^3}\right)$$ Expressed as the triple of eta quotients $(a,b,c)$ such that $$a-1 =b\\ a-5=c$$  so $(m,n)=(-1,-5)$. Then  $$\frac{ab}{c} = \left(\frac{d_2\, d_5}{d_1\,  d_{10}}\right)^{6}\\ \frac{bc}{a} = \left(\frac{d_1\, d_5}{d_2\,  d_{10}}\right)^{4}\\ \frac{ac}{b} = \left(\frac{d_1\, d_2}{d_5\,  d_{10}}\right)^{2}$$ where $(a,b,c)$ are the McKay-Thompson series of class 10E (A138516). They obey $$\left(\frac{16a}{bc}+\frac{bc}a\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$ which is one of the $9$ dependencies found by Conway, Norton, and Atkin such that the moonshine functions span a linear space of $172-9=163$ dimensions. Similar identities involving only moonshine functions exist for levels $(6, 10, 12, 18, 30)$.

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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Entry 41

Define $d_k=\eta(k\tau)$ with the Dedekind eta function $\eta(\tau)$. Then for Level 6 $$\left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4-1=\left(\frac{d_2\,d_3^3}{d_1\,d_6^3}\right)^3\\ \left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4-9=\left(\frac{d_1^5\,d_3}{d_2\,d_6^5}\right)$$ Expressed as the triple of eta quotients $(a,b,c)$ such that $$a-1 =b\\ a-9=c$$ then $(m,n)=(-1,-9)$ and  $$\frac{ab}{c} =\left(\frac{d_2\,d_3}{d_1\,d_6}\right)^{12}\\ \frac{bc}{a} = \left(\frac{d_1\,d_3}{d_2\,d_6}\right)^6\\ \frac{ac}{b} = \left(\frac{d_1\,d_2}{d_3\,d_6}\right)^4$$ where $(a,b,c)$ are the McKay-Thompson series of class 6E (A105559, A128633). They obey $$\left(\frac{64a}{bc}+\frac{bc}a\right)+2a  = \frac{ab}c+ \frac{bc}a+ \frac{ac}{b}$$ which is one of the $9$ dependencies found by Conway, Norton, and Atkin such that the moonshine functions span a linear space of $172-9=163$ dimensions. Similar identities involving only moonshine functions exist for levels $(6, 10, 12, 18, 30)$.

Entry 40

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, $$\;\alpha =\frac{16}{\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^8+16} = \left(\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\right)^8  = \left(\frac{\vartheta_2(q)}{\vartheta_3(q)}\right)^4$$ $$\beta = \frac{27}{\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{12}+27} = \left(\frac{3}{\left(\frac{\eta(\tau/3)}{\eta(3\tau)}\right)^3+3}\right)^3 = \left(\frac{c(q)}{a(q)}\right)^3$$ $$\gamma = \frac{64}{\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}+64} = \left(\frac{8}{\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+8}\right)^2 = \left(\frac{f(q)}{d(q)}\right)^2$$ where the functions of $\color{red}{q = e^{2\pi i\tau}}$ are the Jacobi and Borwein theta functions discussed in Entry 38. Also, $\alpha= \lambda(2\tau)$ and $\lambda(\tau)$ is the modular lambda function.

Entry 39

The functions from Entry 38 obey beautiful relations of form $x^n+y^n=1$, $$\vartheta_3^4(q)=\vartheta_4^4(q)+\vartheta_2^4(q)$$ $$a^3(q)=b^3(q)+c^3(q)$$ $$d^2(q)=e^2(q)+f^2(q)$$ 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 case $n=2$ of the three $x^n+y^n=1$ identities 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 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, $$\big(\vartheta_3(q)\big)^2 = 1+4\sum_{n=0}^\infty\left(\frac{q^{4n+1}}{1-q^{4n+1}}-\frac{q^{4n+3}}{1-q^{4n+3}}\right)$$ $$\quad\quad a(q) = 1+6\sum_{n=0}^\infty\left(\frac{q^{3n+1}}{1-q^{3n+1}}-\frac{q^{3n+2}}{1-q^{3n+2}}\right)$$ though I yet haven't found a nice analogy for $d(q)$.

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,
$$\quad \vartheta_3(q) = \sum_{m=-\infty}^{\infty}q^{n^2} = \frac{\eta^5(2\tau)}{\eta^2(\tau)\eta^2(4\tau)}$$ $$\vartheta_4(q) = \sum_{m=-\infty}^{\infty}(-1)^n\,q^{n^2} = \frac{\eta^2(\tau)}{\eta(2\tau)}$$ $$\vartheta_2(q) = \sum_{m=-\infty}^{\infty}q^{(n+1/2)^2} = \frac{2\,\eta^2(4\tau)}{\eta(2\tau)}$$
II. The Borwein cubic theta functions are,
$$a(q)=\sum_{m,n=-\infty}^{\infty}q^{m^2+mn+n^2} = \frac{\eta^3(\tau)}{\eta(3\tau)} +\frac{9\,\eta^3(9\tau)}{\eta(3\tau)}$$ $$b(q)=\sum_{m,n=-\infty}^{\infty}\zeta^{m-n}q^{m^2+mn+n^2} = \frac{\eta^3(\tau)}{\eta(3\tau)}\quad\quad$$ $$\quad c(q)=\sum_{m,n=-\infty}^{\infty}q^{(m+1/3)^2+(m+1/3)(n+1/3)+(n+1/3)^2} = \frac{3\,\eta^3(3\tau)}{\eta(\tau)}$$
with a cube root of unity $\zeta =e^{2\pi i/3}$. 

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 beautiful relations discussed in Entry 39.