(*Too long for a comment*.) After staring hard at my question and recalling an old [MSE post][1] of mine, I made a guess and found,

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

for ***general*** $\alpha$, 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.)

Similarly, the Chudnovsky formula (a level 1 Ramanujan-Sato) generalizes to a level 9,

$$\small\frac{1}{\pi}=\frac{12}{(640320-4\color{blue}\beta)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k \color{blue}\beta^{k-3j} \tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j} \frac{163\cdot1114806k+13591409+1448\color{blue}\beta/3}{(-640320+4\color{blue}\beta)^k}$$

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


  [1]: http://math.stackexchange.com/questions/1952204/the-chudnovsky-pi-formula-1-pi-revisited