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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.

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.

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. Who can help?

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.

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?

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. Who can help?