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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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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 –  Ian Agol Jul 28 '13 at 22:51
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In your formula for $A^{(3)}_n$ you use trinomials with $l+k>n$. They are not covered by your definition. –  Uwe Stroinski Jul 29 '13 at 15:50
    
Thanks! I have corrected the typo. –  Zurab Silagadze Jul 29 '13 at 16:50
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Congruence (2) is valid for $A^{(3)}_{10}$. –  Uwe Stroinski Jul 30 '13 at 12:26
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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 $162$. –  Uwe Stroinski Jul 30 '13 at 14:49
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