Entry 11

The previous entry featured the Chudnovsky algorithm which uses the negative discriminant $d=-163$ and has class number $h(d)=1$. Just like Ramanujan's well-known formula which uses $d=-4\times58$ and has $h(d)=2$, the algorithm is also connected to Pell equations. Let $n=3\times163=489$. The fundamental solution to $$u^2-489v^2 = 1$$ is then given by the expansion of the fundamental unit $$U_{n} = \left(\tfrac{1}{18}(\color{brown}{640320}-6)\sqrt{3}+4826\sqrt{163}\right)^2=u+v\sqrt{489}$$ Recall that $e^{\pi\sqrt{163}}\approx\color{brown}{640320}^3+743.99999999999925.$ Some experimentation yielded $$3\sqrt{3}\big(U_n^{1/2}-U_n^{-1/2}\big)+18 = \color{brown}{2^2\cdot3^3\cdot7^2\cdot11^2}$$ $$\sqrt{163}\big(U_n^{1/2}+U_n^{-1/2}\big) = \color{brown}{2^2\cdot19\cdot127\cdot163}$$ $$3\sqrt{3}\big(U_n^{1/2}-U_n^{-1/2}\big)+6 = \color{brown}{640320}$$ and we find these integers in the Chudnovsky algorithm $$12\sum_{k=0}^\infty (-1)^k \frac{(6k)!}{k!^3(3k)!}  \frac{\color{brown}{2\cdot3^2\cdot7\cdot11\cdot19\cdot127\cdot163}\,k+13591409}{(\color{brown}{640320}^3)^{k+1/2}} = \frac{1}{\pi}$$ The same thing happens with the other large $d$ with $h(d)=1$, for example $3\times67=201$ $$U_{201} = \left(\tfrac{1}{18}(\color{brown}{5280}-6)\sqrt{3}+62\sqrt{67}\right)^2$$ and $e^{\pi\sqrt{67}}\approx\color{brown}{5280}^3+743.9999986.$ Similar expressions using $U_{201}$ will yield the integers for its corresponding pi formula.