Now that I know what to look for, it seems some of the pieces are starting to fall into place. [**Part 1**][1] was about Levels $n=1,2,8,9$ of [Ramanujan-Sato series][2]. Part 2 below involves Levels $n=4,6,10$. (But I'm missing the odd prime levels $n=3,5,7,11$.)

>**I. Question**

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

$$\frac{1}{C^{3/2}}\sum_{k=0}^\infty s_1(k)\frac{A\,k+B}{C^{k}}=\\\frac{1}{(C+4r)^{3/2}}\sum_{k=0}^\infty s_2(k) \frac{A\,k+B-Dr}{(C+4r)^k}$$

given the sequences,

$$s_1(k) = \tbinom{2k}{k}^{2-p}\sum_{j=0}^k\tbinom{k}{j}^{2+p}\\
\color{red}{s_2(k)} = \tbinom{2k}{k}\sum_{j=0}^k r^{k-j}\tbinom{k}{j} \tbinom{2j}{j}^{1-p}\sum_{m=0}^j\tbinom{j}{m}^{2+p}$$

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

(Again, if the LHS is a formula for a well-known constant, then the RHS guarantees infinitely many such formulas.)

>**II. Ramanujan-Sato level 4**

If $p=0$, then $s_1(k) = \tbinom{2k}{k}^{2}\sum_{j=0}^k\tbinom{k}{j}^{2}=\tbinom{2k}{k}^3$. For example, starting with,

$$\frac1{\pi} =\frac{-2^{12}\sqrt{-1}}{(-2^9)^{3/2}} \sum_{k=0}^\infty\tbinom{2k}{k}^3\,\frac{6k+1}{(-2^9)^k}$$

Since $D = \frac{2A-4B}C = -1/64$, then,

$$\frac1{\pi} =\frac{-2^{12}\sqrt{-1}}{(-2^9+4r)^{3/2}} \sum_{k=0}^\infty s_2(k)\frac{6k+1+r/64}{(-2^9+4r)^k}$$

with the sequence $s_2(k)$ as defined in the question.

>**III. Ramanujan-Sato level 6**

If $p=1$, then $s_1(k) = \tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{j}^{3}$ which is OEIS sequence [A181418][3]. For example, from,

$$\frac1{\pi}=\frac{192\sqrt3}{(2\cdot140^2)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}k\sum_{j=0}^k\tbinom{k}{j}^3\,\frac{140(561k+53)}{(2\cdot140^2)^k}$$

then,

$$\frac1{\pi}=\frac{192\sqrt3}{(2\cdot140^2+4r)^{3/2}}\sum_{k=0}^\infty s_2(k)\frac{140(561k+53)-13r/4}{(2\cdot140^2+4r)^k}$$

>**IV. Ramanujan-Sato level 10**

If $p=2$, then $s_1(k) =\sum_{j=0}^k\tbinom{k}{j}^4$. For example, from

$$\frac1{\pi}=\frac{16\sqrt{\tfrac5{19}}}{(76^2)^{3/2}}\sum_{k=0}^\infty \sum_{j=0}^k\tbinom{k}{j}^4\,\frac{19^2(408k+47)}{(76^2)^k}$$

then,
$$\frac1{\pi}=\frac{16\sqrt{\tfrac5{19}}}{(76^2+4r)^{3/2}}\sum_{k=0}^\infty s_2(k)\frac{19^2(408k+47)-157r/4}{(76^2+4r)^k}$$

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So, like in [Part 1][1], **is the answer to the question a "yes"?**


  [1]: https://mathoverflow.net/questions/338826/generalizing-ramanujans-and-the-chudnovskys-1-pi-formula
  [2]: https://en.wikipedia.org/wiki/Ramanujan%E2%80%93Sato_series
  [3]: https://oeis.org/A181418