**I. Level 6**

This is a long shot, but I am curious where it leads. Given the *Dedekind eta function* $\eta(\tau),$ define,

$$\begin{aligned} j_{6A}(\tau) &= \Big(\sqrt{j_{6B}(\tau)} - \frac{1}{\sqrt{j_{6B}(\tau)}}\Big)^2 \\ j_{6B}(\tau) &= \Big(\tfrac{\eta(2\tau)\,\eta(3\tau)}{\eta(\tau)\,\eta(6\tau)}\Big)^{12}\end{aligned}$$ then, $$\sum_{k=0}^\infty \tbinom{2k}{k}\sum_{j=0}^k\tbinom{k}{j}^3\,\frac1{\big(j_{6A}(\tau)\big)^{k+1/2}}=\sum_{k=0}^\infty \color{blue}{\sum_{j=0}^k\tbinom{k}{j}^2\tbinom{k+j}{j}^2}\,\frac1{\big(j_{6B}(\tau)\big)^{k+1/2}}\tag1$$

where the blue integer sequence $\alpha_k=1,5,73,1445,\dots$ are the * Apery numbers*. These numbers have the known $3$-term recurrence relation,
$$0=k^3\alpha_k-(2k-1)(17k^2-17k+5)\alpha_{k-1}+(k-1)^3\alpha_{k-2}$$
We can use its cubic polynomial coefficients to generate other integer sequences,
$$u_k = k^3 + (k-1)^3 = 1, 9, 35, 91, 189, 341, 559,\dots$$
$$v_k = (2k-1)(17k^2-17k+5)= 5,117,535,1463,\dots$$
both sequences appear in

*cfracs of $\zeta(3)$,*

**two**$$\zeta(3)=\cfrac{1}{1 - \cfrac{1^6}{9 - \cfrac{2^6}{ 35- \cfrac{3^6}{91-\ddots } }}}$$

as well as,

$$\frac{\zeta(3)}6=\cfrac{1}{5 - \cfrac{1^6}{117 - \cfrac{2^6}{ 535- \cfrac{3^6}{1463-\ddots } }}}$$ the latter employed (with other means) by Apery to prove the irrationality of $\zeta(3)$.

**II. Level 10**

Similarly, define, $$\begin{aligned} j_{10A}(\tau) &= \Big(\sqrt{j_{10D}(\tau)} - \frac{1}{\sqrt{j_{10D}(\tau)}}\Big)^2\\ j_{10D}(\tau) &= \Big(\tfrac{\eta(2\tau)\,\eta(5\tau)}{\eta(\tau)\,\eta(10\tau)}\Big)^{6}\end{aligned}$$ then,

$$\sum_{k=0}^\infty \color{red}{\alpha_k}\, \frac1{\big(j_{10A}(\tau)\big)^{k+1/2}}=\sum_{k=0}^\infty \color{blue}{\beta_k}\,\frac1{\big(j_{10D}(\tau)\big)^{k+1/2}}\tag2$$ where,

$\small \alpha_k =\sum_{j=0}^k\tbinom{k}{j}^4 = 1, 2, 18, 164, 1810, 21252, 263844, 3395016\dots$,

$\small \beta_k = 1, 3, 25, 267, 3249, 42795, 594145, 8563035, 126905185, 1921833075, 29609682273, 462653241939, 7313942412825, 116770179560211, 1880087947627377, 30492738838690395,\dots$

Note: Unfortunately, I don't have a closed-form for $\beta_k$ but one can find arbitrarily many terms. The sequence $\alpha_k$ satisfies a 3-term recurrence relation while G. Edgar found that $\beta_k$ has a 5-term recurrence relation.

(* Update:* May 10, 2023. The proposed closed-form of $\beta_k$ in terms of binomial coefficients is given in this MO post.)

**III. Questions:**

- What is the recurrence relation for $\beta_k$?
- This is a long shot: Does its polynomial coefficients somehow appear in the cfrac of $\zeta(5)$?

`https://en.wikipedia.org/wiki/Ramanujan–Sato_series`

on Ramanujan-Sato, under "Level 10", mentions your sequence 1,3,25,... and says closed form is not yet known. If it has a polynomial-cefficient recurrence, it is likely of order higher than 2. $\endgroup$maybe$\zeta(5)$. What about level 8, a 4-term recurrence and $\zeta(4)$? Or level 4, a 2-term recurrence, and $\zeta(2)$? $\endgroup$have 3-term recurrences. For level 10, the identity $(2)$ has an integer sequence on the LHS that has a 3-term recurrence, but the RHS (as you found out) has a 5-term one. And similar disparity for level 4. In this sense, level 6 and its relation to $\zeta(3)$ may be unique. :( $\endgroup$both1more comment