Entry 70

 III. Level 3. The McKay-Thompson series of class 3A for the Monster (A030197)

$$\begin{align}j_{3A}(\tau) &= \left(\left(\frac{d_1}{d_3}\right)^6+3^3\left(\frac{d_3}{d_1}\right)^6\right)^2\\ &=  \left(\left(\frac{d_1}{d_3}\right)^2+9\Big(\frac{d_9^3}{d_1\,d_3^2}\Big)\right)^6\end{align}$$ The second form shows they may be \(6\)th powers. Examples: 

$$\begin{align}j_{3A}\Big(\tfrac{1+\sqrt{-5/3}}{2}\Big) &= -(\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-17/3}}{2}\Big) &= -(2\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-41/3}}{2}\Big) &= -(4\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-89/3}}{2}\Big) &= -(10\sqrt3)^6\end{align}$$ Compare to a similar phenomenon for Level 2. These have discriminant \(d=3p\) for prime \(p=5,17,41,89\) and have class number 2. For class number 6,

$$\begin{align}j_{3A}\Big(\tfrac{1+\sqrt{-29/3}}{2}\Big) &= -(x\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-113/3}}{2}\Big) &= -(y\sqrt3)^6\\ j_{3A}\Big(\tfrac{1+\sqrt{-137/3}}{2}\Big) &= -(z\sqrt3)^6\end{align}$$ where \((x,y,z)\) are the real roots of the cubics

$$x^3 - x^2 - 4x - 5 = 0\\ y^3 - 14y^2 - 4y - 16 =0\\ z^3 - 22z^2 + 44z - 32 = 0$$ and so on. For class number 10, $$j_{3A}\Big(\tfrac{1+\sqrt{-53/3}}{2}\Big) = -(u\sqrt3)^6$$ where \(u\) is the real roof of the solvable quintic $$u^5 - 8u^4 + 19u^3 - 26u^2 + 16u - 11 = 0$$ as well as for other \(d\).