TITLE: Numerology with Ramanujan's pi formula
QUESTION [14 upvotes]: Given Ramanujan's famous $\frac1{\pi}$ formula $$\frac 1\pi=\frac {2\sqrt2}{99^2}\sum_{k=0}^\infty\frac {(4k)!}{k!^4}\frac {26390k+1103}{396^{4k}}$$ 
which is a level 2 Ramanujan-Sato series. It can also be expressed as
$$\frac{1}{\pi} =\frac{192 \sqrt 2}{(396^2)^{3/2}} \sum_{k=0}^\infty \tbinom{2k}{k}\tbinom{2k}{k}\tbinom{4k}{2k}\frac{2\cdot58\cdot15015k+72798}{(396^4)^k}$$
where $\binom{n}{k}$ is the binomial coefficient. In this form, its affinity is clear to the following level 8 Ramanujan-Sato series,
$$\begin{aligned}
\frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+4)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_2(k)\,\frac{58\cdot15015k+(72798-37/4)}{(396^2+4)^k}\\
\frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+8)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_3(k)\,\frac{58\cdot15015k+(72798-37/2)}{(396^2+8)^k}\\
\frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+16)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_4(k)\,\frac{58\cdot15015k+(72798-37)}{(396^2+16)^k}\\
\frac{1}{\pi}&\overset{\color{red}?}=\frac{192\sqrt{2}}{(396^2+32)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_5(k)\,\frac{58\cdot15015k+(72798-2\cdot37)}{(396^2+32)^k}\\
\frac{1}{\pi}&\overset{\color{red}?}=\frac{192\sqrt{2}}{(396^2+64)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_6(k)\,\frac{58\cdot15015k+(72798-4\cdot37)}{(396^2+64)^k}\end{aligned}$$
and integer sequences $s_n(k)$ starting with $k=0$,
$$\begin{aligned}
s_2(k)&=\sum_{j=0}^k\tbinom{k}{2j}\tbinom{2j}{j}\tbinom{2j}{j}=1, 1, 5, 13, 61, 221,\dots\\
s_3(k)&=\sum_{j=0}^k\tbinom{k}{2j}\tbinom{2k-4j}{k-2j}\tbinom{2j}{j}=1, 2, 8, 32, 148, 712,\dots\\
s_4(k)&=\sum_{j=0}^k\tbinom{k}{j}\tbinom{2k-2j}{k-j}\tbinom{2j}{j}=1, 4, 20, 112, 676, 4304,\dots\\
s_5(k)&=1, 8, 68, 608, 5668, 54688, 542864,\dots\\
s_6(k)&=1, 16, 260, 4288, 71716, 1215296, 20848016,\dots
\end{aligned}$$

Q: Are all the terms of $s_5(k)$ and $s_6(k)$ integers as well, and does it have a closed-form?

P.S. I have already checked the OEIS.

REPLY [3 votes]: (Per S. Cooper's request.) 

I. Table relating level $1$ with level $9$. 

The general form apparently is,
$$\frac{1}{\pi}=\frac{12}{(C)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k} \tbinom{3k}{k}\tbinom{6k}{3k} \frac{\color{red}3A\,k+B}{(-C^3)^k}$$
and,
$$\frac{1}{\pi}=\frac{12}{(C-4\color{blue}\alpha)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k \color{blue}\alpha^{k-3j} \tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j} \frac{A\,k+B+N\color{blue}\alpha/3}{(-C+4\color{blue}\alpha)^k}$$
for general real $\color{blue}\alpha$ and where,
$$\begin{array}{|c|c|c|c|c|}
\hline
d&A&B&C&N\\
\hline
11&154/9&5&32&4/3\\
19&114&25&96&4\\
43 &5418 &789 &960 &24\\
67 &87234 &10177 &5280 &76\\
163 &181713378 &13591409 &640320 &1448\\
\hline
\end{array}$$
The variables $A,B,C$ are known to have closed-form expressions in terms of $d$. Presumably $N$ should have as well.
P.S. Note also the equivalent forms, 
$$s_k(\color{blue}\alpha)=\sum_{j=0}^k \color{blue}\alpha^{k-3j} \tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j}=\sum_{j=0}^k \color{blue}\alpha^{k-3j} \tbinom{k}{j}\tbinom{k-j}{j}\tbinom{k-2j}{j}$$
where the latter form is used in H. Chan and S. Cooper's paper "Rational analogues of Ramanujan's series for 1/π". The case $\alpha=-3$,
$$s_k(-3) = 1, -3, 9, -21, 9, 297, -2421$$
is one of the six sporadic sequences studied by Zagier and Cooper.

II. Table relating level $2$ with level $8$. 

The general form apparently is,
$$\frac{1}{\pi}=\frac{192\sqrt{2}}{(C)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k} \tbinom{2k}{k}\tbinom{4k}{2k} \frac{\color{red}2A\,k+B}{(C^2)^k}$$
and,
$$\frac{1}{\pi}=\frac{192\sqrt{2}}{(C+4\color{blue}\alpha)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k \color{blue}\alpha^{k-2j} \tbinom{k}{2j}\tbinom{2j}{j}\tbinom{2j}{j} \frac{A\,k+B-M\color{blue}\alpha/4}{(C+4\color{blue}\alpha)^k}$$
for general real $\color{blue}\alpha$ and where,
$$\begin{array}{|c|c|c|c|c|}
\hline
d&A&B&C&M\\
\hline
6&\sqrt2&\sqrt2/4&(4\sqrt3)^2&\sqrt2/12\\
10&10&2&12^2&1/3\\
18 &70\sqrt6 &21\sqrt6/2 &28^2 &\sqrt6/2\\
22 &385\sqrt2 &209\sqrt2/4 &(12\sqrt{11})^2 &17\sqrt2/12\\
58 &870870&72798&396^2&37\\
\hline
\end{array}$$