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,
$$\begin{align} \frac1{j_{36A}(\tau)} &= \left(\frac{d_1\,  d_4\,d_6^2\, d_9\, d_{36}}{ d_2\,d_3^2\,d_{12}^2\, d_{18}}\right)^2\\ \frac1{k_{36A}(\tau)} &= \left(\frac{d_{3}^2\,d_{4}\, d_{36}}{d_{1}\, d_{9}\,d_{12}^2}\right)^2\\ \frac1{\sqrt{j_{18A}(2\tau)}} &=\left(\frac{d_4^2\,d_{6}^2\, d_{36}^2}{d_2\,  d_{12}^4\,d_{18}}\right)\qquad\qquad \end{align}$$