Entry 2

Here's another pi formula from Ramanujan,$$\frac{1}{\pi} =\frac{2 \sqrt 2}{\color{blue}{9}} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{5\cdot\color{blue}{2\cdot1}\,k+1}{\color{blue}{(12^4)}^k}$$ Note that $$e^{\pi\sqrt{10}} = 12^4-104.21\dots$$The approximation is not as impressive as in Entry 1, but shows the same pattern and "excess". It also has the same connection to Pell equations. Given the golden ratio (and fundamental unit) \(\phi = \frac{1+\sqrt{5}}{2}\), then $$\phi^3=2+\sqrt{5},\quad \text{thus}\;\;\color{blue}{2}^2-5\cdot\color{blue}{1}^2=-1$$ $$\phi^6=9+4\sqrt{5},\quad \text{thus}\;\;\color{blue}{9}^2-5\cdot4^2=1$$ $$2^6\left(\phi^6+\phi^{-6}\right)^2 =\color{blue}{12^4}$$ and, just like the previous entry, we find these integers all over the formula. However, there are only two negative fundamental discriminants of form \(d = -4(8n+2)\) with class number \(h(d)=2\), namely \(d = -4\times10\) and \(d=-4\times58\) so this nice direct connection to Pell equations, among Ramanujan's many pi formulas, is just limited to these two.