Numerical evidence for the first hundred Apery numbers $$A_n=\sum\limits_{k=0}^n\binom{n}{k}^2\binom{n+k}{k}^2$$ suggests the following congruence relation $$A_n\equiv 0\; (\mathrm{mod}\; 5),\;\;\mathrm{if}\;\; n\equiv \{1,3\}\;(\mathrm{mod}\; 5).$$ Was this congruence ever proved?

3$\begingroup$ It follows directly from the recurrence relation satisfied by the Apery numbers. $\endgroup$ – Lucia May 30 '14 at 6:51
The Apery numbers satisfy the recurrence $$ n^3 A_n = (34n^351n^2+27n5)A_{n1} (n1)^3 A_{n2}. $$ If $n\equiv 1\pmod 5$ this recurrence gives $$ A_n \equiv (4\cdot 11\cdot 1+2\cdot 15)A_{n1} \equiv 0 \pmod 5. $$ If $n\equiv 3 \pmod 5$ it gives $$ 2 A_n \equiv (4 \cdot 21\cdot 4+2\cdot 3 5)A_{n1}3A_{n2} \equiv 2 A_{n2} \equiv 0 \pmod 5, $$ by the result just established.
P.S. This proof was also given in the paper S. Chowla, J. Cowles, M. Cowles, Congruence properties of Apéry numbers J. Number Theory, 12 (1980), pp. 188–190, as indicated by Pietro Majer.

$\begingroup$ I corrected a typo in the recurrence relation and added a note about S. Chowla et al. paper, indicated by Pietro Majer. $\endgroup$ – Zurab Silagadze May 30 '14 at 7:35
Here yours and other congruences are proven: