Level 10

I. Moonshine functions: Define, $$a(\tau) =\left(\frac{d_2^2\,d_5}{d_1\,d_{10}^2}\right)^2,\quad b(\tau) = \left(\frac{d_2\,d_5^5}{d_1\,d_{10}^5}\right),\quad c(\tau) =\left(\frac{d_1^3\,d_5}{d_2\,d_{10}^3}\right)$$ then $$\left(\frac{d_2^2\,d_5}{d_1\,d_{10}^2}\right)^2-1=\left(\frac{d_2\,d_5^5}{d_1\,d_{10}^5}\right)\\ \left(\frac{d_2^2\,d_5}{d_1\,d_{10}^2}\right)^2-5=\left(\frac{d_1^3\,d_5}{d_2\,d_{10}^3}\right)$$ so \(a-1 = b,\,a-5 = c\) and we get the moonshine functions $$\begin{align}
j_{10A}(\tau) &= j_{10B}(\tau)+\frac{4^2}{j_{10B}(\tau)}+8\\
j_{10B}(\tau) &= \left(\frac{d_1\, d_5}{d_2\,  d_{10}}\right)^{4}=\frac{bc}{a}\\
j_{10C}(\tau) &= \left(\frac{d_1\, d_2}{d_5\,  d_{10}}\right)^{2} =\frac{ac}{b}\\
j_{10D}(\tau) &= \left(\frac{d_2\, d_5}{d_1\,  d_{10}}\right)^{6} = \frac{ab}{c}\\
j_{10E}(\tau) &= \left(\frac{d_2^2\,d_5}{d_1\,d_{10}^2}\right)^2 \;=\; a \\
\end{align}$$ Example: \(\quad j_{10A}\left(\sqrt{-19/10}\right) = 76^2\)

II. Relations: Let \(a,b,c\) as defined above and \(m=-1,\,n=-5,\) then the system S1 in the introduction explains all four trinomial identities of level 10 (denoted t10) found by Somos. For S2, let \(a=\color{red}{j_{10E}}\) then, $$\begin{align}
j_{10B} &= a+\frac{5}{a}-6\\
j_{10C} &= a-1-\frac{4}{a-1}-3\\
j_{10D} &= a-5+\frac{20}{a-5}+9
\end{align}$$ as well as S3, $$j_{10A} =  j_{10B}+\frac{4^2}{j_{10B}}+8 =  j_{10C}+\frac{5^2}{j_{10C}}+6 = j_{10D}+\frac{1}{j_{10D}}-2$$ which implies the linear relation between 5 moonshine functions,
$$j_{10A}+2\color{red}{j_{10E}} = j_{10B}+j_{10C}+j_{10D}+8$$
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 \(N = 10\),
$$\begin{align}
a'(\tau) &= -a(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_{10}}{d_2\,d_5\,d_{20}}\right)^2 = j_{20C}(\tau)\\
b'(\tau) &= -b(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_{10}^{10}}{d_2^2\,d_5^5\,d_{20}^5}\right)\\
c'(\tau) &= -c(\tfrac12+\tau) = \left(\frac{d_2^{8}}{d_1^3\,d_4^3\,d_5\,d_{20}}\right)
\end{align}$$ such that \(a'+1=b',\; a'+5=c'\).