Entry 85

The eta quotients \(\left(\frac{\eta(\tau)}{\eta(k\tau)}\right)\) discussed previously are useful for j-function formulas. The easy levels \(k\) are when \(m=24/(k-1)\) is an integer. For prime \(k=(2,3,5,7,13)\) yields \(m=(24,12,6,4,2)\) which are the exponents of \(x\) below. $$j(\tau) =\frac{(x-16)^3}x, \quad\text{with}\;\; x=\left(\frac {\sqrt2\,\eta(2\tau)}{\;\eta(\tau)}\right)^{24}$$

$$\; j(\tau) =\frac{(x+3)^3(x+27)}x, \quad\text{with}\;\; x=\left(\frac {\sqrt3\,\eta(3\tau)}{\;\eta(\tau)}\right)^{12}$$

$$j(\tau) \,=\,\frac{(x^2+10x+5)^3}x, \;\quad\text{with}\;\; x=\left(\frac {\sqrt5\,\eta(5\tau)}{\;\eta(\tau)}\right)^6$$

$$j(\tau) =\frac{(x^2+5x+1)^3(x^2+13x+49)}x, \quad\text{with}\;\; x=\left(\frac {\sqrt7\,\eta(7\tau)}{\;\eta(\tau)}\right)^4$$

$$j(\tau) =\frac{(x^4+7x^3+20x^2+19x+1)^3(x^2+5x+13)}x, \;\text{with}\;\; x=\left(\frac {\sqrt{13}\,\eta(13\tau)}{\;\eta(\tau)}\right)^2$$