# Conjectured congruence for the Apery numbers

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?

• It follows directly from the recurrence relation satisfied by the Apery numbers. – Lucia May 30 '14 at 6:51

The Apery numbers satisfy the recurrence $$n^3 A_n = (34n^3-51n^2+27n-5)A_{n-1}- (n-1)^3 A_{n-2}.$$ If $n\equiv 1\pmod 5$ this recurrence gives $$A_n \equiv (4\cdot 1-1\cdot 1+2\cdot 1-5)A_{n-1} \equiv 0 \pmod 5.$$ If $n\equiv 3 \pmod 5$ it gives $$2 A_n \equiv (4 \cdot 2-1\cdot 4+2\cdot 3 -5)A_{n-1}-3A_{n-2} \equiv 2 A_{n-2} \equiv 0 \pmod 5,$$ by the result just established.