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,

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