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


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


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](https://mathoverflow.net/q/446654/12905).)




---


**III. Questions:**


1. What is the recurrence relation for $\beta\_k$?
2. This is a long shot: Does its polynomial coefficients somehow appear in the cfrac of $\zeta(5)$?



About a Ramanujan-Sato formula of level 10, a recurrence, and $\zeta(5)$?
Maple found this recurrence:  

`(n^4+6*n^3+12*n^2+10*n+3)*beta(n)+(-20*n^4-152*n^3-420*n^2-508*n-229)*beta(n+1)+(38*n^4+380*n^3+1416*n^2+2330*n+1431)*beta(n+2)+(-20*n^4-248*n^3-1140*n^2-2292*n-1689)*beta(n+3)+(n^4+14*n^3+72*n^2+160*n+128)*beta(n+4), beta(0) = 1, beta(1) = 3, beta(2) = 25, beta(3) = 267`


\begin{align}
 0 = \;&\left( {n}^{4}+6\,{n}^{3}+12\,{n}^{2}+10\,n+3 \right) \beta\_n
\\& + \left( -20\,{n}^{4}-152\,{n}^{3}-420\,{n}^{2}-508
\,n-229 \right) \beta\_{n+1}
\\& + \left( 38\,{n}^{4}+380\,{n}
^{3}+1416\,{n}^{2}+2330\,n+1431 \right) \beta\_{n+2}
\\& +
 \left( -20\,{n}^{4}-248\,{n}^{3}-1140\,{n}^{2}-2292\,n-1689 \right) 
\beta\_{n+3}
\\& + \left( {n}^{4}+14\,{n}^{3}+72\,{n}^{2}+160
\,n+128 \right) \beta\_{n+4},
\end{align}


with $\beta\_0=1,\beta\_1 =3,\beta\_2 =25,\beta\_3 =267$. Equivalently, by shifting indices


\begin{align}0=&\;(k+1)(k-1)^3\,\beta\_{k-2}\\ 
&+ (-20k^4 + 8k^3 + 12k^2 - 12k + 3)\,\beta\_{k-1}\\ 
&+ (38k^4 + 76k^3 + 48k^2 + 10k + 3)\,\beta\_{k}\\ 
&+ (-20k^4 - 88k^3 - 132k^2 - 68k - 1)\,\beta\_{k+1}\\ 
&+ k(k+2)^3\,\beta\_{k+2}\end{align}


P.S. to Vladimir: For more terms that I needed, I used only
$$
j\_{10A} = j\_{10D} + \frac{1}{j\_{10D}} - 2
$$


