Entry 177

From Entry 176, we gave the fundamental unit \(U_d\) $$U_{163}=64080026 + 5019135\sqrt{163}= \left(\frac{\color{blue}{8005} + 627\sqrt{163}}{\sqrt2}\right)^2$$ and from \(e^{\pi\sqrt{163}}\approx 640320^3+744\), observed that $$640320=80\,(\color{blue}{8005}-1)$$ This may be just coincidence, but not when \(U_{3d}\) for \(d = 7,11,19,43,67,163\). This was also observed by H. H. Chan. Given the fundamental units

$$\begin{align}U_{21} &=\left(\tfrac{\sqrt3+\sqrt{7}}2\right)^2\\ U_{33} &=\big(2\sqrt3+\sqrt{11}\big)^2\\ U_{57} &=\big(5\sqrt3+2\sqrt{19}\big)^2\\ U_{129} &=\big(53\sqrt3+14\sqrt{43}\big)^2\\ U_{201} &=\big(293\sqrt3+62\sqrt{67}\big)^2\\ U_{489} &=\big(35573\sqrt3+4826\sqrt{163}\big)^2 \end{align}$$

Define the function

$$F_d = 3\sqrt3\big(\sqrt{U_{3d}}-1/\sqrt{U_{3d}}\big)+6$$

then we get the rather familiar$$\begin{align}F_{7} &= 15\\ F_{11} &= 6(1+\sqrt{33})\\ F_{19} &= 96\\ F_{43} &= 960\\ F_{67} &= 5280\\ F_{163} &= 640320\end{align}$$

which (except for \(d=11\)) are the cube roots of the j-function (negated). 

P.S. I don't know why \(d=11\) does not obey the pattern, but it does yield the integer \(42\) if the positive sign of the second square root \(\pm 1/\sqrt{U_{3d}}\) is used, though the correct value should be \(32\).