TITLE: About a Ramanujan-Sato formula of level 10, a recurrence, and $\zeta(5)$?
QUESTION [7 upvotes]: This is a long shot, but I am curious where it leads. First, recall the Dedekind eta function $\eta(\tau)$.

I. Level 6

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 polynomial coefficient to generate another integer sequence,
$$v_k = (2k-1)(17k^2-17k+5)= 5,117,535,1463,\dots$$
which appears in the cfrac of $\zeta(3)$,
$$\zeta(3)=\cfrac{6}{5 - \cfrac{1^6}{117 - \cfrac{2^6}{ 535- \cfrac{3^6}{1463-\ddots } }}}$$
and employed (among 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,
$$\sum_{k=0}^\infty \sum_{j=0}^k\tbinom{k}{j}^4\,\frac1{\big(j_{10A}(\tau)\big)^{k+1/2}}=\sum_{k=0}^\infty \color{blue}{\beta_k}\,\frac1{\big(j_{10D}(\tau)\big)^{k+1/2}}\tag2$$
where,
$\small \beta_k = 1, 3, 25, 267, 3249, 42795, 594145, 8563035, 126905185, 1921833075, 29609682273, 462653241939, 7313942412825, 116770179560211, 1880087947627377, 30492738838690395,\dots$
Unfortunately, I don't have a closed-form for $\beta_k$ but one can find arbitrarily many terms. 

Questions:


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

REPLY [9 votes]: 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
$$
j_{10A} = j_{10D} + \frac{1}{j_{10D}} - 2
$$