Level 6

I. Moonshine functions:  Let, $$a(\tau)=\left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4,\quad b(\tau)= \left(\frac{d_2\,d_3^3}{d_1\,d_6^3}\right)^3,\quad c(\tau)=\left(\frac{d_1^5\,d_3}{d_2\,d_6^5}\right)$$ then, $$\left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4-1=\left(\frac{d_2\,d_3^3}{d_1\,d_6^3}\right)^3\\ \left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4-9=\left(\frac{d_1^5\,d_3}{d_2\,d_6^5}\right)$$ So \(m = -1,\,n=-9,\) and we get the moonshine functions, $$\begin{align}
j_{6A}(\tau) &= j_{6D}(\tau)+\frac{9^2}{j_{6D}(\tau)}+18\\
j_{6B}(\tau) &= \left(\frac{d_2\,d_3}{d_1\,d_6}\right)^{12} = \frac{ab}{c} \\
j_{6C}(\tau) &= \left(\frac{d_1\,d_3}{d_2\,d_6}\right)^6 = \frac{bc}{a} \\
j_{6D}(\tau) &= \left(\frac{d_1\,d_2}{d_3\,d_6}\right)^4 = \frac{ac}{b} \\
j_{6E}(\tau) &= \left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4 = a\\
j_{6F}(\tau) &=\;\left(\frac{d_3}{d_6}\right)^{8}\\
\end{align}$$ Example: \(\; j_{6A}\left(\sqrt{-17/6}\right) = 198^2\)

II. Relations: Let \(a,b,c\) as defined above and \(m=-1,\,n=-9,\) then the system S1 given in the Introduction,
$$\begin{align}
a+m &= b\\
a+n &= c\\
a(m-n)+bn &=cm\\
b-c &= (m-n)\end{align}$$ explains all four trinomial identities of level 6 (with prefix t6) found by Somos. For S2, let \(a=\color{red}{j_{6E}}\) then,
$$\begin{align}
j_{6B}(\tau)  &= a-9+\frac{72}{a-9}+17\\
j_{6C}(\tau)  &= a+\frac{9}{a}-10\\
j_{6D}(\tau) &= a-1-\frac{8}{a-1}-7\\
\end{align}$$ as well as S3, $$j_{6A} =  j_{6B}+\frac{1}{j_{6B}}+2 = j_{6C}+\frac{8^2}{j_{6C}}+20 = j_{6D}+\frac{9^2}{j_{6D}}+18 $$ which implies the linear relation between 5 moonshine functions,
$$j_{6A}+2\color{red}{j_{6E}} = j_{6B}+j_{6C}+j_{6D}+20$$These are general phenomena but, in levels 6, 10, 12, 18, 30, the five functions are ALL moonshine.

III. Special property. As mentioned in the Intro, there are special triples {\(a',b',c'\)} of level \(N = 6, 10, 12, 18, 30\) that using a common formula can generate another triple {\(a',b',c'\)} of level \(2N\). For \(N = 6\),
$$\begin{align}
a'(\tau) &= -a(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_6}{d_2\,d_3\,d_{12}}\right)^4 = j_{12B}(\tau)\\
b'(\tau) &= -b(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_6^6}{d_2^2\,d_3^3\,d_{12}^3}\right)^3\\
c'(\tau) &= -c(\tfrac12+\tau) = \left(\frac{d_2^{14}}{d_1^5\,d_3\,d_4^5\,d_6^2\,d_{12}}\right)
\end{align}$$ such that \(a'+1=b',\; a'+9=c'\).