Entry 99

Level 6. Using the eta quotients $(a,b)$ from Entry 98 their ratio $\dfrac{a}{b}$ is $$\begin{align}\text{K}_6(q) &= q^{1/3}\prod_{n=1}^\infty \frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})}\\ &=q^{1/3}\prod_{n=1}^\infty \frac{(1-q^{2n-1})}{\; (1-q^{6n-3})^3} = q^{1/3}\prod_{n=1}^\infty \frac{(1+q^{3n})^3}{(1+q^{n})\;}\\  &= \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}\\ & = \cfrac{q^{1/3}}{1+\cfrac{q+q^2}{1+\cfrac{q^2+q^4}{1+\cfrac{q^3+q^6}{1+\ddots}}}}\end{align}$$

this is also known as the cubic continued fraction. For appropriate $\tau$, then $(a, b, \text{K}_6)$ are radicals. The formula for the j-function using $\text{K}_6(q)$ employs polynomial invariants of the tetrahedron and the integer $12$ of $b= q^{-1/12}B(q)$ in Entry 98 reflects the order $12$ of the tetrahedral group. Note that the cube of the reciprocal of $\text{K}_6(q)$ is the McKay-Thompson series of class 6E of the Monster (A105559) $$\left(\frac{\eta(2\tau)\,\eta^3(3\tau)}{\eta(\tau)\,\eta^3(6\tau)}\right)^3 + 1 = \left(\frac{\eta^2(2\tau)\,\eta(3\tau)}{\eta(\tau)\,\eta^2(6\tau)}\right)^4$$