$$\begin{align}
j_{30A}(\tau) &= \left(\frac{d_1\, d_6\, d_{10}\, d_{15}}{d_2\, d_{3}\, d_{5}\, d_{30}}\right)^3\\
j_{30B}(\tau) &= j_{30D}(\tau)+\frac{1}{ j_{30D}(\tau)}+2\\
j_{30C}(\tau) &= \left(\frac{d_1\, d_3\, d_{5}\, d_{15}}{d_2\, d_{6}\, d_{10}\, d_{30}}\right)\\
j_{30D}(\tau) &= \left(\frac{d_2\, d_3\, d_{10}\, d_{15}}{d_1\, d_{5}\, d_{6}\, d_{30}}\right)^2\\
j_{30E}(\tau) &= \left(\frac{d_{6}\, d_{15}}{d_{3}\, d_{30}}\right)^2\\
j_{30F}(\tau) &= \left(\frac{d_3\, d_5\, d_{6}\, d_{10}}{d_1\, d_{2}\, d_{15}\, d_{30}}\right)\\
j_{30G}(\tau) &= \left(\frac{d_1^2\, d_6\, d_{10}\, d_{15}^2}{d_2^2\, d_{3}\, d_{5}\, d_{30}^2}\right)\end{align}$$
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.
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