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,
$$j_{6d}+2\color{blue}{j_{18C}}=j_{18B}+j_{6F}+\beta-2$$