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It is well known (Beukers 1987) that the Apery numbers $$A_n\equiv A_n^{(2)}=\sum\limits_{k=0}^n\binom{n}{k}^2\binom{n+k}{k}^2$$ satisfy the fancy recurrence relation $$n^3A_n=(34n^3-51n^2+27n-5)A_{n-1}-(n-1)^3A_{n-2},\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$$ with $A_0=1$ and $A_1=5$.

Introducing trinomial coefficients $$\binom{n}{k,\;l}=\frac{n!}{k!\;l!\;(n-k-l)!},$$ let's generalize the Apery numbers as follows $$A_n^{(3)}=\sum\limits_{k=0}^n\sum\limits_{l=0}^{n-k}\binom{n}{k,\;l}^2\binom{n+k+l}{k,\; l}^2.$$ Do these numbers satisfy some analog of the recurrence relation (1)?

The first $A_n^{(3)}$ numbers are: $$\begin{array}{l} 9\\ 721\\ 82089\\ 12230001\\ 2120202009\\ 406989480241\\ 84181340789289\\ 18415254766978801\\ 4208936841232398009\\ \end{array}$$ Note that $$\hspace{50mm}A_n^{(3)}\equiv 0 \;(\mathrm{mod}\; 9),\hspace{50mm} (2)$$ if $n=1,3,4,5,7,9$. This sequence belongs to the so called vile numbers http://oeis.org/A003159 Is the congruence (2) true for any vile number $n$? Note also that $A_5^{(3)}\equiv A_1^{(3)}\;(\mathrm{mod}\; 3^3)$, $A_8^{(3)}\equiv A_2^{(3)}\;(\mathrm{mod}\; 3^3)$ and $A_9^{(3)}\equiv A_1^{(3)}\;(\mathrm{mod}\; 5^3)$. What about Beukers-like congruence $$\hspace{50mm}A_{np-1}^{(3)}\equiv A_{n-1}^{(3)} \;(\mathrm{mod}\; p^3),\hspace{50mm} (3)$$ for any prime $p$ and positive integer $n$, is it true?

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    $\begingroup$ You can express the trinomials as a product of binomials. For sums of products of binomials, there are computer packages for finding recurrence relations due to Zeilberger and others. See the book "A=B":math.upenn.edu/~wilf/AeqB.html $\endgroup$
    – Ian Agol
    Jul 28, 2013 at 22:51
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    $\begingroup$ In your formula for $A^{(3)}_n$ you use trinomials with $l+k>n$. They are not covered by your definition. $\endgroup$
    – Uwe Stroinski
    Jul 29, 2013 at 15:50
  • $\begingroup$ You can take the coefficients in front of $A^{(3)}_n, A^{(3)}_{n-1}$ and $A^{(3)}_{n-2}$ as general polynomials of say degree 6 and generate say 20 linear equations by plugging in the $A^{(3)}_i$ for $0\leq i\leq 20$. I did just that and solved the resulting linear system for the coefficients of the polynomial to only get the trivial all zero solution. Thus one of your polynomials has at least degree 7 or the recursion goes beyond $n-2$. $\endgroup$
    – Uwe Stroinski
    Jul 30, 2013 at 11:50
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    $\begingroup$ Congruence (2) is valid for $A^{(3)}_{10}$. $\endgroup$
    – Uwe Stroinski
    Jul 30, 2013 at 12:26
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    $\begingroup$ For $n\leq 175$ we have that $A^{(3)}_n$ is not divisible by $9$ for $n\in\{2,6,8,18,20,24,26,54,56,60,62,72,74,78,80,162,164,168,170\}$. Note the big gap between $80$ and $162$. $\endgroup$
    – Uwe Stroinski
    Jul 30, 2013 at 14:49

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