Given Ramanujan's famous pi formula $$\frac 1\pi=\frac {2\sqrt2}{99^2}\sum_{k=0}^\infty\frac {(4k)!}{k!^4}\frac {26390k+1103}{396^{4k}}$$

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 Ramanujan-Sato series,

$$\begin{aligned} \frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+4)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_1(k)\,\frac{58\cdot15015k+(72798-37/4)}{(396^2+4)^k}\\ \frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+8)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_2(k)\,\frac{58\cdot15015k+(72798-37/2)}{(396^2+8)^k}\\ \frac{1}{\pi}&=\frac{192\sqrt{2}}{(396^2+16)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_3(k)\,\frac{58\cdot15015k+(72798-37)}{(396^2+16)^k}\\ \frac{1}{\pi}&\overset{\color{red}?}=\frac{192\sqrt{2}}{(396^2+32)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_4(k)\,\frac{58\cdot15015k+(72798-2\cdot37)}{(396^2+32)^k}\\ \frac{1}{\pi}&\overset{\color{red}?}=\frac{192\sqrt{2}}{(396^2+64)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\,s_5(k)\,\frac{58\cdot15015k+(72798-4\cdot37)}{(396^2+64)^k}\end{aligned}$$

and integer sequences $s_n(k)$ starting with $k=0$,

$$\begin{aligned} s_1(k)&=\sum_{j=0}^k\tbinom{k}{2j}\tbinom{2j}{j}\tbinom{2j}{j}=1, 1, 5, 13, 61, 221,\dots\\ s_2(k)&=\sum_{j=0}^k\tbinom{k}{2j}\tbinom{2k-4j}{k-2j}\tbinom{2j}{j}=1, 2, 8, 32, 148, 712,\dots\\ s_3(k)&=\sum_{j=0}^k\tbinom{k}{j}\tbinom{2k-2j}{k-j}\tbinom{2j}{j}=1, 4, 20, 112, 676, 4304,\dots\\ s_4(k)&=1, 8, 68, 608, 5668, 54688, 542864,\dots\\ s_5(k)&=1, 16, 260, 4288, 71716, 1215296, 20848016,\dots \end{aligned}$$

Q: Are all the terms of $s_4(k)$ and $s_5(k)$ integers as well, and does it have a closed-form?

P.S. Of course I already checked the OEIS.