Formulae and integrality for two order four linear recurrences with cubic polynomial coefficients

Question

Why does the following two linear recurrences with cubic polynomial coefficient produce only integers ? What are the explicit formulae for the $A_k$ ?

Sequence 1: $A_0=1,A_1=120,A_2=-9000,A_3=1133760$ $$(k+4)^3A_{k+4}+8(2343+2550k+900k^2+104k^3)A_{k+3}+12288(107+187k+108k^2+20k^3)A_{k+2}$$ $$+2097152(1+k)(8+19k+14k^2)A_{k+1}+1073741824k^3A_{k}=0$$

Sequence 2 :$A_0=1,A_1=12,A_2=-36,A_3=192$ $$(k+4)^3A_{k+4}+2(570+633k+225k^2+26k^3)A_{k+3}+192(26+46k+27k^2+5k^3)A_{k+2}$$ $$+512(1+k)(8+19k+14k^2)A_{k+1}+16384k^3A_k=0$$

Added Context

The recurrences comes from expanding some weight two forms in terms of modular functions at 10 special levels. Let $\ell=1,2,3,5,6,7,11,14,15,23$ which are the values where $\sigma_1(\ell):=\sum_{d| \ell} d$ divides $24$. These are the square-free order of the Mathieu group $M_{23}$ which is the one point stabilizer of $M_{24}$ acting on the Leech lattice via coordinate permutation.

For these 10 special levels and a conjugacy class with cycle pattern $\prod_{d|\ell}d^{24/\sigma_1(\ell})$ there is a level $\ell$ analogue of $\Delta(z)=\Delta_1(z)$ defined by

$\Delta_\ell(z)=\prod_{d | \ell} \eta(dz)^{24/\sigma_1(\ell)},$

and extremal even $\ell$ modular Euclidean lattice of minimal norm $4$ and $2$, $\Lambda_\ell, \Lambda^0_\ell$ which are analogue of the Leech and $E_8$ lattice , so that the space of holomorphic modular form on $\Gamma_0(\ell)+$ are just $\mathbb{C}[\theta_{\Lambda^0_\ell}(z),\Delta_\ell(z)]$ which generalize the fact that $SL_2(\mathbb{Z})$ modular form are generated by $E_4(z)=\theta_{E_8}(z)$ and $\Delta(z)$.

The $A_k$ are needed in 20 formulae for $1/\pi$ in Theorem 3.1 in [1]

https://www.emis.de/journals/EM/expmath/volumes/14/14.3/Chua.pdf

one for each of the lattice $\Lambda_\ell, \Lambda^0_\ell$ defined in Table 2 and also their dimension $D_\ell, d_\ell$.

The $A_k$ are the coefficients of the powers of level $\ell$ theta series of $\Lambda$ scaled to have weight two, $Z_{\Lambda}(q)=\theta_{\Lambda}(q)^{4/dim(\Lambda)}$ in term of the modular invariants $X_\ell(z)=\frac{\Delta_\ell(2z)}{\Delta_\ell(z)}$. The modularity implies they must satisfies some third order ODE (5.3) which is equivalent to recurrence of the $A_k$ (5.7). This is explained in the paragraph after equation (3.13)

I was (and still) unable to find explicit formulae for the $A_k$ so the recurrences are given in Table 5 and 6 for $\Lambda^0_\ell, \Lambda_\ell$ with the initial values in Table 7. So there are twenty recurrences. Sequences 1,2 are the simplest recurrences for $E_8=\Lambda^0_1$ and $A_2=\Lambda^0_3$. 16 of the 20 recurrences (including sequence 1,2) are proven to generate only integers. This requires the dimension to divide 8. For example for $\Lambda_7=A_6^{(2)}$, the Craig lattice of dimension 6, the $A_k$ are not all integral.The coefficients satisfies the congruence $p|A_{p-1}$ similar to Apery's. What are the other initial values which still give rise to integral $A_k$ ?

It seems interesting that one can prove such sequences (including Sequence 1,2) to generate only integer via modular forms. Are there other proofs for example using cluster algebra ? If there is explicit formulae, the Table 5, 6, 7 can be removed. The sequences are not hypergeometric.

Answer 1

Sequence 2

I'll denote the $A_k$ from sequence 2 as $B_k$, such that for $k=1,2,\ldots$ \begin{align} B_k = 12,-36, 192, -1380, 11952, -116928, 1242{}624, -14006{}628, \ldots . \end{align} Observing that \begin{align} \frac{(-1)^{k-1}B_k}{6(k+1)} = 1, 2, 8, 46, 332, 2784, 25888, 259382, \ldots \end{align} also yields an integer sequence, we ask OEIS and find https://oeis.org/A145844, with closed form (thanks Mathematica) \begin{align} A145844(n) &= \sum_{j=0}^{n} \frac{\binom{n}{j}^2 \binom{2 j}{j} \binom{2 (n-j)}{n-j}}{(n-j+1)(j+1)} \label{eq:1a}\tag{1a}\\ &= \frac{1}{n+1} \binom{2n}{n} \,{}_4F_3\left[\begin{matrix}\frac 1 2,\,-n,\,-n,\,-n-1 \\ 1,\,2,\,\frac 1 2{-}n\end{matrix};1\right]\label{eq:1b}\tag{1b}\\ &= 1,2,8,46,\ldots, \end{align} such that for $k>0$ \begin{align} B_k = 6 (-1)^{k-1}\frac{k+1}{k} \binom{2k-2}{k-1} \,{}_4F_3\left[\begin{matrix}\frac 1 2,\,1{-}k,\,1{-}k,\,-k \\ 1,\,2,\,\frac 3 2{-}k\end{matrix};1\right] \label{eq:2}\tag{2}. \end{align} As $A145844(n)$ is combinatorial and therefore an integer sequence, $B_k$ is also an integer sequence.

Maybe a similar expression can be found for sequence 1.

Answer 2

(Too long for a comment.)

I. Pi formulas

I have done some work on Ramanujan-Sato pi formulas up to class $11$ (per the McKay-Thompson series) and this is the first time I've seen,

$$\frac1{\pi}=\sqrt{\frac87}\sum_{k=0}^\infty(\alpha+\beta k) A_k\left(\frac{-3-\sqrt2+\sqrt{11+8\sqrt2}}{8}\right)^k$$

According to your 2005 paper, this uses McKay-Thompson class $14$ so was beyond my scope back then. We are familiar with Ramanujan's $j_{2A}\left(\frac{\sqrt{-58}}2\right) = 396^4$ for $\tau = \frac{\sqrt{-58}}2$. In Table 1, you give similar algebraic numbers $Y_0$ for ten levels $\ell$, the first being $Y_0 = \frac{-7+5\sqrt2}{512} = \left(\frac{-1+\sqrt2}8\right)^3$. While the McKay-Thompson series $T_\ell$ was included, $\tau$ was not.

---

II. Dedekind eta

After a little experimentation, it was implicit in the table after all. For those curious as well, given the Dedekind eta function $\eta(\tau)$, define following Zagier's notation,

\begin{align} j_{2B}(\tau) &= \left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}\\ j_{4C}(\tau) &= \left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{8}\\[4pt] j_{6C}(\tau) &= \left(\frac{\eta(\tau)\,\eta(3\tau)}{\eta(2\tau)\, \eta(6\tau)}\right)^{6}\\[4pt] j_{10B}(\tau) &= \left(\frac{\eta(\tau)\,\eta(5\tau)}{\eta(2\tau)\, \eta(10\tau)}\right)^{4}\\[4pt] j_{14B}(\tau) &= \left(\sqrt{F}-\frac2{\sqrt{F}}\right)^{2}\qquad \end{align}

where $F = \dfrac{\eta^3(2\tau)}{\eta(\tau)\,\eta^2(4\tau)}\times\dfrac{\eta^3(14\tau)}{\eta(7\tau)\,\eta^2(28\tau)}=\left(\dfrac{2^8}{\lambda(2\tau)\,\lambda(14\tau)}\right)^{1/8}\\\\$ and $\lambda(\tau)$ is the modular lambda function. Then for levels $\ell = 1,2,3,5,7,$ we recover Chan's $Y_0$,

\begin{align} \qquad\frac1{j_{2B}\big(\sqrt{-2/1}\big)} &= \frac{-7+5\sqrt2}{512} = \left(\frac{-1+\sqrt2}8\right)^3\\[4pt] \frac1{j_{4C}\big(\sqrt{-2/2}\big)} &= \frac{-4+3\sqrt2}{128} = \left(\frac{-1+\sqrt2}8\right)^2\frac1{\sqrt2}\\[4pt] \frac1{j_{6C}\big(\sqrt{-2/3}\big)} &= \frac{-5+3\sqrt3}{32}\\[4pt] \frac1{j_{10B}\big(\sqrt{-2/5}\big)} &= \frac{-3+\sqrt{10}}8\\[4pt] \frac1{j_{14B}\big(\sqrt{-2/7}\big)} &= \frac{-3-\sqrt2+\sqrt{11+8\sqrt2}}8=\frac1{\left(1+\sqrt{1+2\sqrt2}\right)^3} \end{align}

and so on for five other levels. Quite nice.

---

III. Quintic

Recall $j_{10B}(\tau)$ and define its neighbor $j_{10C}(\tau)$,

\begin{align} j_{10B}(\tau) &= \left(\frac{\eta(\tau)\,\eta(5\tau)}{\eta(2\tau)\, \eta(10\tau)}\right)^{4}\\ j_{10C}(\tau) &= \left(\frac{\eta(\tau)\,\eta(2\tau)}{\eta(5\tau)\, \eta(10\tau)}\right)^{2}\quad \end{align}

Then, suppressing $\tau$ for brevity,

$$\left(\sqrt{j_{10B}}+\frac4{\sqrt{j_{10B}}}\right)^2 = \left(\sqrt{j_{10C}}+\frac5{\sqrt{j_{10C}}}\right)^2-4$$

So both $j_{10B}$ and $j_{10C}$ can appear in the solution of the Bring quintic as in this recent MO post which uses $j_{10C}$.

---

IV. Recurrences

For your question on linear recurrences with cubic polynomial coefficients, I've also made an MO post on that (based on Coopers's 2012 paper) which he used to find more Ramanujan-Sato pi formulas. For example,

$$(n+1)^3 u_{n+1} = (2n+1)(13n^2+13n+4)u_n + 3n(9n^2-1)u_{n-1}$$

which he labels $s_7$ and is used for the McKay-Thompson series of class $7$. This yields $1, 4, 48, 760,\dots$ and is A183204. However my post was in the context of continued fractions with closed-forms. Whether your Sequence 1 yields only integers I cannot yet say, but I wonder if your recurrences can be applied to a continued fraction with a close-form as well? (Though they look more complicated than Cooper's.)