On four Ramanujan-type "Legendrian" sequences with a 3-term recurrence?

Question

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.

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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}

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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.

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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?

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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 $F_i$?

Answer 1

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.

Answer 2

[As Henri Cohen already answered the last question, evaluating the continued fractions in terms of $\zeta(3)$, let me answer the first two questions, which are answered in the literature.]

Zagier, in his original paper, classified (at least conjecturally) the integral solutions $(A,B,\lambda)$ that make $$(\star)\, (n+1)^2 u_{n+1}-(An^2+An+\lambda) u_n+Bn^2 u_{n-1}=0$$ integral for all $n$ (with $u_0=1$). He was mostly interested in sporadic solutions, not fitting in one of the two families he found. Almkvist and Zudilin performed a related search in their paper, looking for integral solutions $(a,b,c)$ that make $$(\star\star)\, (n + 1)^3 u_{n+1} − (2n + 1)(an^2 + an + b)u_n+cn^3 u_{n-1}$$ integral for all $n$. Almkvist, van Straten and Zudilin revisit the aforementioned searches in Section 4 of "Generalizations of Clausen's formula and algebraic transformations of Calabi-Yau differential equations".

They show that $(16,0,4)$, $(27,0,6)$, $(64,0,12)$ and $(432,0,60)$ lead to integral solutions to $(\star)$. (The first 3 were explicitly written down by Zagier - all 4 are part of a family he calls hypergeometric). These 4 tuples appear in equations (4.2)-(4.3) of A-vS-Z's paper, implicitly. To see this, recall that instead of talking about $(u_n)_n$ defined by $(\star)$ one can talk about $\sum_n u_n z^n$ which will be annihilated by the differential operator $$\theta^2 -z(A\theta^2+A\theta+\lambda)+Bz^2(\theta+1)^2,\qquad \theta:=z \frac{d}{dz}.$$ In equations (4.2)-(4.3) it is shown that $$\theta^2 -C_{\alpha} z(\theta+\alpha)(\theta+1-\alpha)$$ annihilates a power series with integral coefficients, for $$\alpha \in \{1/2,1/3,1/4,1/6\}, \qquad C_{\alpha} := \begin{cases}\alpha^{-3} &\text{if }\alpha\in \{1/3,1/4\},\\ 2\alpha^{-3}&\text{if }\alpha\in \{1/2,1/6\}.\end{cases}$$ This corresponds to $(A,B,\lambda)=(C_{\alpha},0,C_{\alpha}\alpha(1-\alpha))$, which are exactly $(16,0,4)$, $(27,0,6)$, $(64,0,12)$ and $(432,0,60)$.

Next, they show that your tuples, $(16,8,16^2)$, $(27,15,27^2)$, $(64,40,64^2)$ and $(432,312,432^2)$ lead to integral solutions to $(\star \star)$. These 4 tuples appear in equations (4.10)-(4.11) of A-vS-Z's paper, implicitly. To see this, recall that instead of talking about $(u_n)_n$ defined by $(\star\star)$ one can talk about $\sum_n u_n z^n$ which will be annihilated by the differential operator $$\theta^3 -z(2\theta+1)(a\theta^2+a\theta+b)+cz^2(\theta+1)^3.$$ In equations (4.10)-(4.11) it is shown that $$\theta^3 -C_{\alpha} z(2\theta+1)(\theta(\theta+1)+\alpha^2+(1-\alpha)^2)+C_{\alpha}^2 z^2(\theta+1)^3$$ annihilates a power series with integral coefficients, for the same $\alpha$ and $C_{\alpha}$ as before. This corresponds to $(a,b,c)=(C_{\alpha},C_{\alpha}(\alpha^2+(1-\alpha)^2),C^2_{\alpha})$, which are your 4 tuples!

Implicit in equation (4.9) is a unified formula for the sequences $u_n$ in $(\star \star)$ defined via $(a,b,c)=(C_{\alpha},C_{\alpha}(\alpha^2+(1-\alpha)^2),C^2_{\alpha})$, namely $$u_n= C_{\alpha}^n \sum_{k=0}^{n} \binom{-\alpha}{n-k}^2\binom{\alpha-1}{k}^2.$$ This agrees with your formulas. See Q4 in Section 5 here for the formula appearing explicitly.

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This should answer your first two questions, but let me mention that the punchline of Section 4 is that a solution $(A,B,\lambda)$ of $(\star)$ leads to a solution $$(a,b,c) = (A,A-2\lambda,A^2-4B)$$ of $(\star\star)$. See Theorem 4.1 for the full statement and a computational proof, and the discussion following it for a geometric proof. In particular, the solutions $(16,0,4)$, $(27,0,6)$, $(64,0,12)$ and $(432,0,60)$ of $(\star)$ lead to your solutions $(16,8,16^2)$, $(27,15,27^2)$, $(64,40,64^2)$ and $(432,312,432^2)$ of $(\star \star)$.