Entry 133

Given fundamental discriminants \(d=4m\) with class number \(h(-d)=2\), there are exactly four even \(m = 6, 10, 22, 58\). The most well-known is \(m=58\) because of the near-integer $$\quad e^{\pi\sqrt{58}} = 396^4-104.00000017\dots$$ and the appearance of \(396^4\) in the denominator of Ramanujan's famous \(1/\pi\) formula. These \(m=2p\) for prime \(p\) have other interesting properties. Recall the modular lambda function \(\lambda(\tau)\) also discussed in Entry 112 $$\lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac12\tau\big)\,\eta^2\big(2\tau\big)}{\eta^3(\tau)}\right)^8$$ We focus on \(m=2p\) for prime \(p = 3\,\text{mod}\,4\) hence $$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-6})}} &= U_3\sqrt{U_6} = (2+\sqrt3)(\sqrt2+\sqrt3)\\ \frac1{\sqrt{\lambda(\sqrt{-22})}} &= U_{11}\sqrt{U_{22}}= (10+3\sqrt{11})(7\sqrt2+3\sqrt{11}) \end{align}$$ with fundamental units \(U_n\). However, special \(d\) with class number \(h(-d)=2^k\) surprisingly can be expressed by nested radicals using only the square root of 2. So,$$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-6})}} &= (1+\sqrt2)^2+\sqrt{1+ (1+\sqrt2)^4}\\ \frac1{\sqrt{\lambda(\sqrt{-22})}} &=  (1+\sqrt2)^6+\sqrt{1+ (1+\sqrt2)^{12}} \end{align}\quad$$ Similar behavior can also be observed for \(2p\) for \(p=7,23,71\) which now have class number 4.