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I. Recurrences

In a previous post, it was mentioned how Almkvist-Zudilin did a computer search for solutions to the recurrence relation,

$$(n+1)^3s_{n+1}=(2n+1)(an^2+an+b)s_n+c\,n^3s_{n-1}$$

within a bound and found 6 non-trivial solutions (essentially related to Zagier's 6 sporadic sequences). But if they used $|c|<256$, then they wouldn't find the four recurrences below where it goes as high as $c = 432^2 = 186624$,

\begin{align} (n+1)^3\alpha_{n+1}&=(2n+1)(432n^2+432n+312)\alpha_n-432^2n^3\alpha_{n-1}\\[6pt] (n+1)^3\beta_{n+1}&=(2n+1)(64n^2+64n+40)\beta_n-64^2n^3\beta_{n-1}\\[6pt] (n+1)^3\gamma_{n+1}&=(2n+1)(27n^2+27n+15)\gamma_n-27^2n^3\gamma_{n-1}\\[6pt] (n+1)^3\delta_{n+1}&=(2n+1)(16n^2+16n+8)\delta_n-16^2n^3\delta_{n-1} \end{align}

While the recurrences found by Zagier, Cooper, et al had a modular interpretation for levels 5, 6, 7, etc, these are for levels 1 to 4. The associated sequences are below.


II. Sequences

Given the binomial $\binom{n}{m}$, then,

\begin{align} \alpha(n) &= (-1)^n\sum_{j=0}^n(-432)^{n-j}\binom{n+j}{n-j}\binom{2j}j\binom{3j}j\binom{6j}{3j}\\ &= 1, 312, 114264, 44196288,\dots\\[6pt] \beta(n) &= (-1)^n\sum_{j=0}^n(-64)^{n-j}\binom{n+j}{n-j}\binom{2j}{j}^2\binom{4j}{2j}\\ &=1, 40, 2008, 109120,\dots \\[6pt] \gamma(n) &= (-1)^n\sum_{j=0}^n(-27)^{n-j}\binom{n+j}{n-j}\binom{2j}{j}^2\binom{3j}{j}\\ &=1, 15, 297, 6495,\dots\\[6pt] \delta(n) &= (-1)^n\sum_{j=0}^n(-16)^{n-j}\binom{n+j}{n-j}\binom{2j}{j}^3\\ &=1, 8, 88, 1088,\dots\\ \end{align}

where all $s(0)=1.$ These can easily be derived (using Method 1 in this post) from Ramanujan's original four sequences for his 1/pi formulas so they are "Ramanujan-type". (My thanks to Michael Somos for providing a Mathematica code to find recurrence relations.)

Incidentally, the 2nd and 4th have "simpler" formulations. Do the other two have as well?

\begin{align} \beta(n) &= \sum_{j=0}^n 16^{n-j}\binom{2j}{j}^3\binom{2n-2j}{n-j}\qquad\\ \delta(n) &= \sum_{j=0}^n\binom{2j}{j}^2\binom{2n-2j}{n-j}^2\qquad\\ \end{align}


III. Modular context and pi formulas

Each of these sequences are associated with a McKay-Thompson series of level 1,2,3,4. The first is connected to the $j$-function and the other three are for the eta quotients,

\begin{align} j_{2B}(\tau) &= \left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}\\ j_{3B}(\tau) &= \left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{12}\\ j_{4C}(\tau) &= \left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{8} \end{align}

while Zagier's sporadic $(11,3,1)$, in a 2nd order recurrence, is for,

$$j_{5B}(\tau) = \left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^{6}$$

For example, let $\tau=\frac12\sqrt{-58},\,$ so $\,j_{2B}(\tau)=64\left(\frac{5+\sqrt{29}}{2}\right)^{12}=D.$ Then,

$$\frac{1}{\pi} = 16\sqrt{2}\,\sum_{n=0}^\infty (-1)^n \,\beta(n)\,\frac{-24184+9801\sqrt{29}\,\left(n+\frac12\right)}{D^{n+\frac12}}$$

so the $\beta$ sequence can be used for a Ramanujan-type pi formula, just like all of Zagier's sporadics. For pi formulas using all four $\alpha, \beta, \gamma, \delta$, see Ramanujan-Sato series.


IV. Continued fractions

The cfracs of the sporadics had closed-forms. Using the polynomials above (and up to $m = 12000$), Wolfram Alpha was accurate only to a few decimal places before timing out,

\begin{align} F_1 &= \frac1{312 + \large{\underset{n=1}{\overset{m}{\mathrm K}} ~ \frac{-432^2 n^6}{(2n+1)(432n^2+432n+312)}}} = 0.0045793\dots\\ F_2 &= \frac1{40 + \large{\underset{n=1}{\overset{m}{\mathrm K}} ~ \frac{-64^2 n^6}{(2n+1)(64n^2+64n+40)}}} \;=\; 0.041425\dots\\ F_3 &= \frac1{15 + \large{\underset{n=1}{\overset{m}{\mathrm K}} ~ \frac{-27^2 n^6}{(2n+1)(27n^2+27n+15)}}} = 0.1366\dots\\ F_4 &= \frac1{8 + \large{\underset{n=1}{\overset{m}{\mathrm K}} ~ \frac{-16^2 n^6}{(2n+1)(16n^2+16n+8)}}} = 0.406\dots\\ \end{align}

with the last having the slowest "convergence". What are these numbers?


V. Questions

  1. Starting with $s(-1) = 0, s(0)=1$, do the recurrences really yield integers for all $n$?
  2. Are there simpler formulas for the other two sequences ($\alpha$ and $\gamma$)?
  3. And are there closed-forms for the continued fractions $F_i$?
  • The answer to 1 is 'yes'. Before I explain why, note that the "HolonomicFunctions" Mathematica package package can quickly verify that the integer-valued $\alpha(n),\beta(n),\gamma(n),\delta(n)$ indeed satisfy your recurrences, answering 1 in the positive. – Ofir Gorodetsky May 28 '23 at 18:43
  • Zagier found 4 hypergeometric and 4 Legendrian solutions for his recurrences. In section 4 of "Generalizations of Clausen's formula and algebraic transformations of Calabi-Yau differential equations" by Almkvist, van Straten and Zudilin, the authors describe the hypergeometric and Legendrian solutions to Zagier's equation (see equations (4.1)-(4.7)) and the connection between the hypergeometric and the Legendrian ones. Your constants 16, 27, 64 and 432 appear explicitly in (4.2). – Ofir Gorodetsky May 28 '23 at 18:46
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    In Theorem 4.1 of that paper, a correspondence is established between solutions to Zagier's equation and the Almkvist-Zudilin equation, which explains why your 4 solutions exist. They are also described in equations (4.9)-(4.11). A binomial formula (that follows from these equations) is given in Q4 in Section 5 here: arxiv.org/abs/2102.11839 . This partially answers 2: there is a unified formula for all 4 sequences as $C_{\alpha}^n \sum_{k=0}^{n} \binom{-\alpha}{n-k}^2 \binom{\alpha-1}{k}^2$, $C_{\alpha} = \alpha^{-3}$ or $2\alpha^{-3}$ and $\alpha \in {1/2, 1/3, 1/4, 1/6}$. – Ofir Gorodetsky May 28 '23 at 18:51
  • I do not know about question 3. – Ofir Gorodetsky May 28 '23 at 18:53
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    As requested: $F_1=0.00457936356455642911364499027248614471814419210576351311808$, $F_2=0.0414268220026378096205680227378000709249113690570134530631$, $F_3=0.136756232683283246080196823032970017605464751401410734953$. On the other hand $F_4$ converges like $A-1/\log(n)$, so I am unable to compute $A$, which is approximately $0.51$. – Henri Cohen May 28 '23 at 20:03
  • @OfirGorodetsky Can you convert your comments as an answer so i can upvote it? You’ve essentially solved 2/3 of the questions. – Tito Piezas III May 28 '23 at 23:17
  • @TitoPiezasIII Sure, will do this later this week due to busyness... – Ofir Gorodetsky May 29 '23 at 12:10
  • @OfirGorodetsky: I just read the Almkvist-Zudilin paper and in Section 4, they mentioned that Zagier found 4 Legendrian solutions in 2009. I looked at Zagier's 2009 paper and in his table of 36 solutions, he explicitly says, and I quote, "three triples", namely \begin{align} &\text{No.}19,; (2\times 16,; 16^2,; 12)\ &\text{No.}25,; (2\times 27,; 27^2,; 21)\ &\text{No.}26,; (2\times 16,; 16^2,; 28)\end{align} I believe he didn't find $64^2$ and $432^2$ as those were beyond the search range. When I saw those numbers, I immediately checked $64^2$ and $432^2$, and they worked. – Tito Piezas III May 29 '23 at 13:36

1 Answers1

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Stupid of me. As O. Gorodetsky mentions, these are classical: $$F_1=(91\zeta(3)-2\pi^3\sqrt{3})/432$$ $$F_2=(28\zeta(3)-\pi^3)/64$$ $$F_3=(117\zeta(3)-2\pi^3\sqrt{3})/243$$

In addition, note that there are almost identical cfracs for the same linear combinations where the $-$ sign is replaced by a $+$ sign: replace in $F_1$ the $312$ by $600$, in $F_2$ the $40$ by $104$, and in $F_3$ the $15$ by $51$.

Added: we have $F_4=(7/16)\zeta(3)=0.525899...$. This is due to Y. Yang and is referred to in my arXiv paper mentioned in the post.

Henri Cohen
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  • What in this answer is stupid of you? – LSpice May 28 '23 at 21:50
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    The fact that I have known this for some time. – Henri Cohen May 28 '23 at 22:16
  • @HenriCohen. Wonderful! Based on the $C_3(a,b,c)$ cfracs, I suspected these would involve $\zeta(3)$ and/or $\pi^3$. – Tito Piezas III May 28 '23 at 23:25
  • @HenriCohen For $F_3$, I changed your "25" to "15". I assume it was just a typo. I noticed if we do these changes also to the recurrence relations, they yield integer sequences as well. I wonder if these new ones can be used for Ramanujan-type pi formulas also? – Tito Piezas III May 29 '23 at 06:14