Is it true that $$A_n\equiv (1)^n\;\;(\mathrm{mod}\;3)\;\;?$$ Here $A_n$ is the Apery number: $$A_n=\sum\limits_{k=0}^n\binom{n}{k}^2\binom{n+k}{k}^2.$$ What is known about congruence properties modulo 3 for another set of Apery numbers $$B_n=\sum\limits_{k=0}^n\binom{n}{k}^2\binom{n+k}{k}\;?$$
Yes, the first congruence was conjectured by Chowla and Cowles in $1980$ and proved by Y. Mimura in $1983$, see http://www.sciencedirect.com/science/article/pii/0022314X83900380. For $B_n$ the sequence begins $(1,3,19,147,1251,11253,104959...)$, which is $(1,0,1,0,0,0,1...)$ modulo $3$. It seems we have $B_{2^k}\equiv 0 \mod 3$ and $B_n\equiv 1 \mod 3$ otherwise. Perhaps the recurrence $$ (n + 1)^2B_{n + 1} = (11n^2 + 11n + 3)B_n + n^2B_{n1}. $$ is useful.
Edit: I found a reference here: http://arxiv.org/abs/math.CO/0407326. See Theorem $5.8$.

1$\begingroup$ In fact Theorem 5.8 proves a surprising fact that $B_n$ Apery numbers modulo 3 constitute a Cantor set $$1,0,10,0,01,0,10,0,0,0,0,0,0,0,01,0,10,0,01,0,1\ldots$$ $\endgroup$ – Zurab Silagadze Aug 1 '13 at 9:57