How to compute Coefficients in Chudnovsky's Formula?

Question

My aim is to understand all three coefficients arising in the Chudnovsky-Formula (see also Question 300385). 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: a reference is needed, why $\sqrt{D}E_2^*(\tau)/\eta^4(\tau)$ is an algebraic integer.

EDIT: Answer:

Thanks to the answers of Henri Cohen (see below) and Michael Griffin (see here), an explicit calculation of the coefficients along with a complete proof of their exactness can now be found in chapter 10 and the appendix of this arXiv-preprint.

Answer 1

Let $\tau$ be any CM point. By basic theorems of complex multiplication, if you choose a suitable period $\omega(\tau)$, $E_4(\tau)/\omega(\tau)^4$, $E_6(\tau)/\omega(\tau)^6$, and $\sqrt{D}E_2^*(\tau)/\omega(\tau)^2$ (with $E_2^*(\tau)=E_2(\tau)-3/(\pi\Im(\tau))$ and $D$ the discriminant of $\tau$) will be algebraic numbers of known degree, and if you choose $\omega(\tau)=\eta(\tau)^2$, they will even be algebraic integers. Incidentally (but this is not needed) the Chowla--Selberg formula gives $\eta(\tau)$ explicitly as a product of gamma values.

In the case $\tau=(1+\sqrt{-163})/2$, we easily find that $E_4/\omega^4=-640320\rho$, $E_6/\omega^6=-40133016\sqrt{-163}$, and $\sqrt{-163}E_2^*/\omega^2=-8688\rho^2$ with $\rho=(-1+\sqrt{-3})/2$ a cube root of unity. This gives $s_2(\tau)=77265280/90856689$ which implies the result.