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 these four recurrences where it is 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 et al had a modular interpretation for levels 5, 6, 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,

\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}

Do the other two have as well?

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 while 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}

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 Zagier's. For pi formulas using all four, 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 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)}}} \;=\; 0.041424\dots\\ F_3 &= \frac1{15 + \large{\underset{n=1}{\overset{m}{\mathrm K}} ~ \frac{-27^2 n^6}{(2n+1)(27n^2+27n+15)}}} = 0.13659\dots\\ F_4 &= \frac1{8 + \large{\underset{n=1}{\overset{m}{\mathrm K}} ~ \frac{-16^2 n^6}{(2n+1)(16n^2+16n+8)}}} = 0.4064\dots\\ \end{align}

with the last one 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?
3. And are there closed-forms for the continued fractions $F_i$?