Entry 109

Let \(q = e^{2\pi i \tau}\) and Dedekind eta function \(\eta(\tau)\). Define the following $$\quad\text{K}_{12}(\tau) = \text{K}_{12}(q) = q\,\prod_{n=1}^\infty\frac{(1-q^{12n-1})(1-q^{12n-11})}{(1-q^{12n-5})(1-q^{12n-7})}\quad $$ It is connected to the mod 6 version, or the cubic continued fraction $$\; \text{K}_{6}(\tau) = \text{K}_{6}(q) = q^{1/3}\prod_{n=1}^\infty\frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})}\quad $$

by the quadratic relation 

$$\frac1{\text{K}_{12}(\tau)}+\text{K}_{12}(\tau) = \frac1{\text{K}_6(\tau)\,\text{K}_6(2\tau)} = \frac{\eta^3(3\tau)\,\eta(4\tau)}{\eta(\tau)\,\eta^3(12\tau)} = \left(\frac{\eta^2(4\tau)\,\eta(6\tau)}{\eta(2\tau)\,\eta^2(12\tau)}\right)^2+1$$ To find exact values of \(\text{K}_{12}(\tau)\), we use discriminants \(d = 12m\) with class number \(h(-d) = 4\) for even \(m=10,14,26,34\) to get the orderly $$\begin{align}\frac1{\text{K}_{12}\Big(\tfrac{\sqrt{-10/3}}4\Big)}+\text{K}_{12}\Big(\tfrac{\sqrt{-10/3}}4\Big) &= 1+\sqrt3\left(2+\sqrt{10}+\sqrt{-1+(2+\sqrt{10})^2}\right)\\ \frac1{\text{K}_{12}\Big(\tfrac{\sqrt{-14/3}}4\Big)}+\text{K}_{12}\Big(\tfrac{\sqrt{-14/3}}4\Big) &= 1+\sqrt3\left(4+\sqrt{21}+\sqrt{1+(4+\sqrt{21})^2}\right)\\ \frac1{\text{K}_{12}\Big(\tfrac{\sqrt{-26/3}}4\Big)}+\text{K}_{12}\Big(\tfrac{\sqrt{-26/3}}4\Big) &= 1+\sqrt3\left(15+4\sqrt{13}+\sqrt{1+(15+4\sqrt{13})^2}\right)\\ \frac1{\text{K}_{12}\Big(\tfrac{\sqrt{-34/3}}4\Big)}+\text{K}_{12}\Big(\tfrac{\sqrt{-34/3}}4\Big) &= 1+\sqrt3\left(28+5\sqrt{34}+\sqrt{-1+(28+5\sqrt{34})^2}\right) \end{align}$$ Update: It turns out all the quartic roots can be factored into fundamentals units \(U_n\), for example $$2+\sqrt{10}+\sqrt{-1+(2+\sqrt{10})^2}=(1+\sqrt2)\big(\tfrac{1+\sqrt5}2\big)^3 = U_2\,U_5^3$$ More on Entry 110.