Entry 67

Let \(j=j(\tau)\) be the j-function. Define $$j_{1B}(\tau)=432\frac{\sqrt{j}+\sqrt{j-1728}}{\sqrt{j}-\sqrt{j-1728}}=\frac1q-120+10260q-901120q^2+\dots$$ which is (A299954). Then the following values $$\begin{align}j_{1B}\big(\tfrac{1+\sqrt{-19}}2\big) &= -432\,U_{19a}=-432\left(\sqrt{96^3/12^3}+\sqrt{96^3/12^3+1}\right)^2\\ j_{1B}\big(\tfrac{1+\sqrt{-43}}2\big) &= -432\,U_{43b}=-432\left(\sqrt{960^3/12^3}+\sqrt{960^3/12^3+1}\right)^2\\ j_{1B}\big(\tfrac{1+\sqrt{-67}}2\big) &= -432\,U_{67c}=-432\left(\sqrt{5280^3/12^3}+\sqrt{5280^3/12^3+1}\right)^2\\ j_{1B}\big(\tfrac{1+\sqrt{-163}}2\big) &= -432\,U_{163c}=-432\left(\sqrt{640320^3/12^3}+\sqrt{640320^3/12^3+1}\right)^2\end{align}$$ are fundamental units, solutions to Pell equations \(x^2-ny^2=\pm1\) and where \((a,b,c,d) = (6, 15, 330, 10005)\). The smaller Heegner numbers like \(d=11\) don't yield fundamental units. And the second one \(U_{43b}=U_{645}\) involves a cube (Wolfram computation), $$j_{1B}\big(\tfrac{1+\sqrt{-43}}2\big) = -432\,U_{645}=-432\left(\frac{127+5\sqrt{645}}2\right)^3$$ which seems surprising. Note that $$432\big(\sqrt{-U_{645}}+1/\sqrt{-U_{645}}\big)^2=-960^3$$