My aim is to understand all three coefficients arising in the Chudnovsky-Formula (see also Question [300385][1]). Two of them are easily computed, but I failed with the third: It is known that for all $\tau$ with $Im(\tau)>1.25$ we have \begin{align*} \frac{1}{2\pi Im(\tau)}\sqrt{\frac{J(\tau)}{J(\tau)-1}} &= \sum_{n=0}^\infty \left( \frac{1-s_2(\tau)}{6} + n \right)\cdot \frac{(6n)!}{(3n)!(n!)^3}\cdot \frac{1}{\left(1728J(\tau)\right)^n}\\ \text{with }s_2(\tau) &:= \frac{E_4(\tau)}{E_6(\tau)}\left(E_2(\tau)-\frac{3}{\pi Im(\tau)}\right) \end{align*} Then for $\tau=\frac{1+i\sqrt{163}}{2}$ it is known that $1728J(\tau)=-640320^3$. This gives us \begin{align*} \frac{1}{\pi} &= \sum_{n=0}^\infty \frac{(-1)^n(6n)!}{(3n)!(n!)^3}\cdot \frac{A + B\cdot n}{640320^{3n+3/2}} \end{align*} with $B = \sqrt{163\cdot(1728+640320^3)} = 12\cdot 545140134$ $$\text{and } A = 12\cdot 545140134\cdot\left( \frac{1-s_2(\tau)}{6} \right)$$ It can be easily computed that $A$ is *approximately* $12\cdot13591409$ **Question:** How can I prove that the value of $A$ is *exactly* this number? *Edit*: Thanks to the answer of @HenriCohen, the only thing left to prove is [this][2]. Who can help? [1]: https://mathoverflow.net/questions/300385/why-does-this-quasi-modular-function-have-integral-values [2]: https://mathoverflow.net/questions/301476/why-are-values-of-eisenstein-e-2-algebraic-integers