Entry 86

As discussed in the previous entry, the eta quotients $\left(\frac{\eta(\tau)}{\eta(k\tau)}\right)$ are useful for j-function formulas. The easy levels $k$ are when $m=24/(k-1)$ is an integer. For square $k=(4,9,25)$ yields $m=(8,3,1)$ which can be found in the exponents below. $$j(\tau) =\frac{(x^2-48)^3}{x^2-64}, \quad\text{with}\quad x=\left(\frac {\sqrt4\,\eta(4\tau)}{\eta(\tau)}\right)^8+8$$

$$j(\tau) =\frac{x^3(x^3-24)^3}{x^3-27}, \quad\text{with}\quad x=\left(\frac {\sqrt9\,\eta(9\tau)}{\eta(\tau)}\right)^3+3$$

$$j(\tau) = \frac{-(x^{20}+12 x^{15}+14 x^{10}-12 x^5+1)^3}{x^{25} (x^{10}+11 x^5-1)} , \quad\text{with}\;\; x^{-1}-x=\left(\frac {\sqrt{25}\,\eta(25\tau)}{\eta(\tau)}\right)^1+1$$

Alternatively,

$$j(\tau) =\frac{(x^2+192)^3}{(x^2-64)^2}, \quad\text{with}\quad x=\left(\frac {\eta(\tau/2)}{\eta(2\tau)}\right)^8+8$$

$$j(\tau) =\frac{x^3(x^3+216)^3}{(x^3-27)^3}, \quad\text{with}\quad x=\left(\frac {\eta(\tau/3)}{\eta(3\tau)}\right)^3+3$$

$$j(\tau) = \frac{-(r^{20}-228 r^{15}+494 r^{10}+228 r^5+1)^3}{r^5 (r^{10}+11 r^5-1)^5} , \quad\text{with}\;\; r^{-1}-r=\left(\frac {\eta(\tau/5)}{\eta(5\tau)}\right)^1+1$$ After some manipulation, these eta quotients are related to q-continued fractions with octahedral, tetrahedral, and icosahedral symmetries, with $r=R(q)$ being the well-known Rogers-Ramanujan continued fraction.