Some years ago, I asked in MSE a question about the Chudnovsky brothers pi formula. Later, I asked in MO a related question. The former was unanswered until a few days ago when L. Miller gave me a clue to find a missing piece of the puzzle. (It seems it may take years for a question to be resolved.)

I. Question

Given $A,B,C$ and binomial $\binom{n}{k}$. If the series below converges, is it true that,

$$\frac{1}{C^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k} \tbinom{pk}{k}\tbinom{2pk}{pk} \frac{\color{blue}pA\,k+B}{C^{\color{blue}p\,k}}=\\ \frac{1}{(C+4r)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k r^{k-pj}\tbinom{k}{pj}\tbinom{2j}{j}\tbinom{pj}{j} \frac{A\,k+B-Dr}{(C+4r)^k}$$

where $D = \frac{2A-4B}C$, and generic $r$ such that $C(C+4r)>0$?

II. Ramanujan

Ramanujan's formula is then the case $p=2$,

$$\frac{1}{\pi} =\frac{192 \sqrt 2}{(396^2)^{3/2}} \sum_{k=0}^\infty \tbinom{2k}{k}\tbinom{2k}{k}\tbinom{4k}{2k}\frac{\color{blue}2Ak+B}{(396^2)^{\color{blue}2k}}$$

yielding

$$\frac{1}{\pi}=\frac{192\sqrt{2}}{(396^2+4r)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k r^{k-2j} \tbinom{k}{2j}\tbinom{2j}{j}\tbinom{2j}{j} \frac{Ak+B-37r/4}{(396^2+4r)^k}$$

where $A=58\cdot15015$ and $B=72798$ and $C=396^2$.

III. Chudnovsky

The Chudnovsky formula is the case $p=3$,

$$\frac{1}{\pi}=\frac{12\sqrt{-1}}{(-640320)^{3/2}} \sum_{k=0}^\infty \tbinom{2k}{k}\tbinom{3k}{k}\tbinom{6k}{3k}\frac{\color{blue}3A\,k+B}{(-640320)^{\color{blue}3k}}$$

yielding

$$\frac{1}{\pi}=\frac{12\sqrt{-1}}{(-640320+4r)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k r^{k-3j} \tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j} \frac{A\,k+B+1448r/3}{(-640320+4r)^k}$$

where $A=163\cdot1114806$ and $B=13591409$ and $C=-640320$.

I am not aware of a formula that uses $p=4$. So is the answer to the question a "yes"?