Entry 159

Given the McKay-Thompson series of Class 4A for the Monster (A097340)

$$j_{4A}(\tau)=\left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)}\right)^{24} = \frac1q+24 + 276q + 2048q^2 +\dots$$ We take the \(6\)th root, scale \(\tau \to \frac{\tau}2\), and use the version, $$(x_d)^4=\left(\frac{\eta^2(\tau)}{\eta(\tfrac{\tau}2)\,\eta(2\tau)}\right)^4=\frac{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)}{\eta^2(\tau)}=\frac{\frac2{\pi}\times K(k_d)}{\eta^2(\tau)}$$ where \(K(k_d)\) is a complete elliptic integral of the first kind. Compared also to Ramanujan's G-function, $$2^{1/4}G_n = \frac{\eta^2(\tau)}{\eta\big(\tfrac{\tau}{2}\big)\eta(2\tau)} = q^{-\frac{1}{24}}\prod_{n>0}(1+q^{2n-1})\qquad\qquad $$ where Ramanujan used odd \(n\) for \(G_n\) and calculated a lot of \(n\). Mathworld also has a partial list of closed-forms for \(K(k_d)\). Knowing \(x_d\) or \(G_n\) immediately leads to a closed-form for \(\eta^2(\tau)\) as well. Examples, 

$$\begin{align}\frac{2}{\pi}\times K(k_7) &= x_7^2\,\frac{\Gamma\big(\tfrac17\big)\,\Gamma\big(\tfrac27\big)\,\Gamma\big(\tfrac47\big)}{7^{1/4}\,(2\pi)^2}\\ \eta^2(\sqrt{-7}) &= \frac1{x_7^2}\frac{\Gamma\big(\tfrac17\big)\,\Gamma(\tfrac27\big)\,\Gamma\big(\tfrac47\big)}{7^{1/4}\,(2\pi)^2}\end{align}$$

where \(x_7 =\sqrt2\). And

$$\begin{align}\frac{2}{\pi}\times K(k_{11}) &= x_{11}^2\,\frac{\Gamma\big(\tfrac1{11}\big)\,\Gamma\big(\tfrac3{11}\big)\,\Gamma\big(\tfrac4{11}\big)\,\Gamma\big(\tfrac5{11}\big)\,\Gamma\big(\tfrac9{11}\big)}{11^{1/4}\,(2\pi)^3}\\ \eta^2(\sqrt{-11}) &= \frac1{x_{11}^2}\frac{\Gamma\big(\tfrac1{11}\big)\,\Gamma\big(\tfrac3{11}\big)\,\Gamma\big(\tfrac4{11}\big)\,\Gamma\big(\tfrac5{11}\big)\,\Gamma\big(\tfrac9{11}\big)}{11^{1/4}\,(2\pi)^3}\end{align}$$

where \(x_{11} =\dfrac1T+1\) and \(T\) is the tribonacci constant, the real root of \(x^3-x^2-x-1=0\). For \(d=11,19,43,67,163\), then \(x_d\) is just the real root of a cubic.