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?
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> 
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**P.S.** I have already checked the OEIS.



Numerology with Ramanujan's pi formula
(*Too long for a comment*.) After staring hard at my question and recalling an old [MSE post](https://math.stackexchange.com/questions/1952204/the-chudnovsky-pi-formula-1-pi-revisited) of mine, I made an inspired guess and found,



> 
> Level 8
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$$\frac{1}{\pi}=\frac{192\sqrt{2}}{(396^2+4\color{blue}\alpha)^{3/2}}\sum\_{k=0}^\infty\tbinom{2k}{k}\sum\_{j=0}^k \color{blue}\alpha^{k-2j} \tbinom{k}{2j}\tbinom{2j}{j}\tbinom{2j}{j} \frac{58\cdot15015k+72798-37\color{blue}\alpha/4}{(396^2+4\color{blue}\alpha)^k}$$


for ***general*** real $\alpha$, so turns out it is unnecessary to restrict it to powers of $2$. (Thus, $s\_5(k)$ and $s\_6(k)$ ***do*** have a closed-form and are integer sequences.)



> 
> Level 9
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Similarly, the Chudnovsky formula (a level 1 Ramanujan-Sato) yields,


$$\frac{1}{\pi}=\frac{12}{(640320-4\color{blue}\alpha)^{3/2}}\sum\_{k=0}^\infty\tbinom{2k}{k}\sum\_{j=0}^k \color{blue}\alpha^{k-3j} \tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j} \frac{163A\,k+B+1448\color{blue}\alpha/3}{(-640320+4\color{blue}\alpha)^k}$$


where $A=1114806, B=13591409$.




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(***Added later.***) Now that I know what to look for, one can find other families.



> 
> Level 6
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For example, starting with H. H. Chan, W. Zudilin, et al's


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


mentioned in H. H. Chan and S. Cooper's *"Rational analogues of Ramanujan's series for 1/π"*, we find,
$$\frac1{\pi}=\frac{192\sqrt3}{(2\cdot140^2+4\color{blue}\alpha)^{3/2}}\sum\_{k=0}^\infty\tbinom{2k}k\sum\_{j=0}^k \color{blue}\alpha^{k-j}\tbinom{k}{j}\sum\_{m=0}^j\tbinom{j}{m}^3\,\frac{140(561k+53)-13\color{blue}\alpha/4}{(2\cdot140^2+4\color{blue}\alpha)^k}$$



> 
> Level 10
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> 


$$\frac1{\pi}=\frac{16\sqrt{5}}{\sqrt{19}(76^2+4\color{blue}\alpha)^{3/2}}\sum\_{k=0}^\infty\tbinom{2k}k\sum\_{j=0}^k \color{blue}\alpha^{k-j}\tbinom{k}{j}\tbinom{2j}{j}^{-1}\sum\_{m=0}^j\tbinom{j}{m}^4\,\frac{19^2(408k+47)-157\color{blue}\alpha/4}{(76^2+4\color{blue}\alpha)^k}$$


**P.S.** However, I do not have a rigorous proof for these families and the relevant literature do not seem to address general $\alpha$.


**Note:** The level $\color{red}m$ is given by an eta quotient $\frac{\eta(\tau)}{\eta(\color{red}m\,\tau)}$. For example, define


$$\lambda(\tau) =-6+\big(\tfrac{\eta^2(3\tau)}{\eta(\tau)\,\eta(\color{red}9\tau)}\big)^6 -27 \big(\tfrac{\eta(\tau)\,\eta(\color{red}9\tau)}{\eta^2(3\tau)}\big)^6$$


so $\lambda\Big(\tfrac{3+\sqrt{-163}}{6}\Big)=-640320$ is level $9$.


