9
$\begingroup$

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 $\zeta(5)$?
$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$ 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. :( $\endgroup$ 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 :) $\endgroup$ 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 $\zeta(2)$? $\endgroup$ 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. :( $\endgroup$ Mar 28, 2017 at 12:55

2 Answers 2

10
$\begingroup$

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

$\endgroup$
9
  • $\begingroup$ Whew, that's a beast! $\endgroup$
    – user78249
    Mar 27, 2017 at 17:52
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 27, 2017 at 18:53
  • 1
    $\begingroup$ Thanks for this G. Edgar! So while the Apery numbers were a $3$-term recurrence, these involve a $5$-term. Hm... $\endgroup$ Mar 28, 2017 at 2:05
  • 1
    $\begingroup$ @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 $\displaystyle\frac{\beta_{k+1}}{\beta_k}=\phi^6$. $\endgroup$ Mar 28, 2017 at 2:41
  • 4
    $\begingroup$ @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 $\zeta(3)$. $\endgroup$ Mar 28, 2017 at 17:10
1
$\begingroup$

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/

$\endgroup$
7
  • $\begingroup$ 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? $\endgroup$ Jun 13 at 16:18
  • $\begingroup$ Fine, I would have a try. $\endgroup$ Jun 14 at 8:30
  • $\begingroup$ The result is so messy, I fail to find a simpler closed-form with the similar analysis. $\endgroup$ 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. $\endgroup$ Jun 14 at 11:11
  • 1
    $\begingroup$ I'm glad my effort is helpful to you. $\endgroup$ Jun 14 at 11:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.