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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.)

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 $\zeta(5)$?
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  • $\begingroup$ The Wikipedia page 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. $\endgroup$
    – Gerald Edgar
    Mar 27, 2017 at 16:03
  • $\begingroup$ @GeraldEdgar: I had assumed that, analogous to the Apery numbers, it would be a three-term relation with quintic polynomials, I tried to solve it using Mathematica but couldn't find it. Maybe it involves more than three terms. :( $\endgroup$
    – Tito Piezas III
    Mar 27, 2017 at 16:12
  • $\begingroup$ Is there a place where I can get more terms for this? The terms you give are enough to show that there are no low degrees low order recurrences where low is quite low! So I could use a few more terms :) $\endgroup$
    – Vladimir Dotsenko
    Mar 27, 2017 at 16:33
  • $\begingroup$ OK ... At level 6 we get a 3-term recurrence and $\zeta(3)$; level 10 we get a 5-term recurrence and maybe $\zeta(5)$. What about level 8, a 4-term recurrence and $\zeta(4)$? Or level 4, a 2-term recurrence, and $\zeta(2)$? $\endgroup$
    – Gerald Edgar
    Mar 28, 2017 at 11:50
  • $\begingroup$ @GeraldEdgar: For level 6, the identity $(1)$ involves an integer sequence on the LHS and RHS that both 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. :( $\endgroup$
    – Tito Piezas III
    Mar 28, 2017 at 12:55

2 Answers 2

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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 $$ j_{10A} = j_{10D} + \frac{1}{j_{10D}} - 2 $$

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  • $\begingroup$ Whew, that's a beast! $\endgroup$
    – user78249
    Mar 27, 2017 at 17:52
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    $\begingroup$ Very nice! Some remarks: $n^4+6n^3+12n^2+10n+3=(n+1)^3(n+3)$, $n^4+14n^3+72n^2+160n+128=(n+2)(n+4)^3$, other three polynomial coefficients appear to be irreducible. Also, retaining the highest terms in $n$ of the recurrence (like Poincaré would do) gives a constant coefficient recurrence with the characteristic polynomial $t^4-20t^3+38t^2-20t+1=(t-1)^2(t^2-18t+1)$ with roots $1$ of multiplicity two and $(2\pm\sqrt{5})^2$ of multiplicity one. I wonder what the double root $1$ may mean, if anything at all. $\endgroup$
    – Vladimir Dotsenko
    Mar 27, 2017 at 18:53
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    $\begingroup$ Thanks for this G. Edgar! So while the Apery numbers were a $3$-term recurrence, these involve a $5$-term. Hm... $\endgroup$
    – Tito Piezas III
    Mar 28, 2017 at 2:05
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    $\begingroup$ @VladimirDotsenko: If we do the same for the Apery numbers $\alpha$, the characteristic polynomial is $t^2-34t+1=0$ with root $\sigma^4$ and silver ratio $\sigma$. The limiting ratio of $\displaystyle\frac{\alpha_{k+1}}{\alpha_k}=\sigma^4$. If we do the same for $\beta$, then $(2+\sqrt{5})^2=\phi^6$ with golden ratio $\phi$. I assume then that $\displaystyle\frac{\beta_{k+1}}{\beta_k}=\phi^6$. $\endgroup$
    – Tito Piezas III
    Mar 28, 2017 at 2:41
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    $\begingroup$ @Marty: There is the known, $$\zeta(5) = \cfrac{1}{w_1-\cfrac{1^{10}}{w_2-\cfrac{2^{10}}{w_3-\cfrac{3^{10}}{w_4-\ddots}}}}$$ where $w_n = (n-1)^5+n^5 = (2n-1)(n^4-2n^3+4n^2-3n+1) = 1, 33, 275, 1267, \dots$ Unfortunately, no one has yet found an accelerated version analogous to what Apery did for $\zeta(3)$. $\endgroup$
    – Tito Piezas III
    Mar 28, 2017 at 17:10
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Since $\alpha_{k}$ is known, according to the below relation \begin{eqnarray} \sum_{k=0}^{\infty}\frac{\beta_{k}}{\left(j_{10D}(\tau)\right)^{k}}&=&{\sqrt{\frac{j_{10D}(\tau)}{j_{10A}(\tau)}}\sum_{k=0}^{\infty}\frac{\alpha_{k}}{\left(j_{10A}(\tau)\right)^{k}}}\\&=&\alpha_{1}+q\left(\alpha_{1}+\alpha_{2}\right)+q^{2}\left(-5\alpha_{1}-3\alpha_{2}+\alpha_{3}\right)\\&&+q^{3}\left(4\alpha_{1}-15\alpha_{2}-7\alpha_{3}+\alpha_{4}\right)+O\left(q^{4}\right)\\&=&QM_{\alpha}\alpha+O\left(q^{4}\right)\\&&\beta_{1}+\beta_{2}q+q^{2}\left(-6\beta_{2}+\beta_{3}\right)\\&&q^{3}\left(15\beta_{2}-12\beta_{3}+\beta_{4}\right)+O\left(q^{4}\right)\\&=&QM_{\beta}\beta+O\left(q^{4}\right)\\Q&=&\left[\begin{array}{c} 1\\ q\\ q^{2}\\ q^{3} \end{array}\right]^{T},\alpha=\left[\begin{array}{c} \alpha_{1}\\ \alpha_{2}\\ \alpha_{3}\\ \alpha_{4} \end{array}\right],\beta=\left[\begin{array}{c} \beta_{1}\\ \beta_{2}\\ \beta_{3}\\ \beta_{4} \end{array}\right]\\M_{\alpha}&=&\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ -5 & -3 & 1 & 0\\ 4 & -15 & -7 & 1 \end{array}\right],M_{\beta}=\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & -6 & 1 & 0\\ 0 & 15 & -12 & 1 \end{array}\right] \end{eqnarray} By comparing the coefficients of q in the two sides, we have \begin{eqnarray} \beta&=&{T}\alpha\\{T}&=&1\\&&1,1\\&&1,3,1\\&&1,6,5,1\\&&1,10,15,7,1\\&&1,15,35,28,9,1\\&&1,21,70,84,45,11,1\\&&1,28,126,210,165,66,13,1\\&&\cdots \end{eqnarray} Notice that T is exactly A085478(https://oeis.org/A085478) and $T_{n,k}=\left(\begin{array}{c}n+k\\2k\end{array}\right)$, therefore \begin{eqnarray} \beta_{n}&=&\sum_{k=0}^{n}T_{n,k}\alpha_{k}\\&=&\sum_{k=0}^{n}\sum_{j=0}^{k}\left(\begin{array}{c} n+k\\ 2k \end{array}\right)\left(\begin{array}{c} k\\ j \end{array}\right)^{4} \end{eqnarray} which is the closed form of $\beta_n$.

All computations are done by GP/PARI: https://pari.math.u-bordeaux.fr/

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  • $\begingroup$ Thanks. This is a nice development. I have actually also found a closed-form for three sequences $(\beta_n, \gamma_n, \delta_n)$ in this post. Can you do a similar analysis for $\gamma_n$ and $\delta_n$ in that post and see if you can find a simpler alternative closed-form? $\endgroup$
    – Tito Piezas III
    Jun 13 at 16:18
  • $\begingroup$ Fine, I would have a try. $\endgroup$
    – xiaoshuchong
    Jun 14 at 8:30
  • $\begingroup$ The result is so messy, I fail to find a simpler closed-form with the similar analysis. $\endgroup$
    – xiaoshuchong
    Jun 14 at 11:04
  • $\begingroup$ Thanks for the effort. However, your approach has given me an idea on how to check for alternative formulas. $\endgroup$
    – Tito Piezas III
    Jun 14 at 11:11
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    $\begingroup$ I'm glad my effort is helpful to you. $\endgroup$
    – xiaoshuchong
    Jun 14 at 11:23

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