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

Level 8

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

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

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 $A=1114806, B=13591409$.

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

Level 6

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

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

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 $9$. [1]: https://math.stackexchange.com/questions/1952204/the-chudnovsky-pi-formula-1-pi-revisited

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

Level 8

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

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

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 $A=1114806, B=13591409$.

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

Level 6

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

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

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 $9$.

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

Level 8

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

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

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 $A=1114806, B=13591409$.

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

Level 6

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

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

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 $9$. [1]: http://math.stackexchange.com/questions/1952204/the-chudnovsky-pi-formula-1-pi-revisited

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

Level 8

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

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

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 $A=1114806, B=13591409$.

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

Level 6

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

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

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 $9$. [1]: https://math.stackexchange.com/questions/1952204/the-chudnovsky-pi-formula-1-pi-revisited

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

Level 8

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

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

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 $A=1114806, B=13591409$.

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

Level 6

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

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

given in "New analogues of Clausen’s identities arising from the theory of modular forms", 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

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

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

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

Level 8

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

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

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 $A=1114806, B=13591409$.

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

Level 6

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

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

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 $9$. [1]: http://math.stackexchange.com/questions/1952204/the-chudnovsky-pi-formula-1-pi-revisited