On sums of Apéry polynomials and related congruences

Mathematics – Number Theory

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Scientific paper

The Ap\'ery polynomials are given by $$A_n(x)=\sum_{k=0}^n\binom(n,k)^2\binom(n+k,k)^2x^k.$$ (Those A_n=A_n(1) are Ap\'ery numbers.) Let p be an odd prime. We show that $$\sum_{k=0}^{p-1}(-1)^kA_k(x)=\sum_{k=0}^{p-1}\binom(2k,k)^3x^k/16^k (mod p^2)$$, and that $$\sum_{k=0}^{p-1}A_k(x)=(x/p)\sum_{k=0}^{p-1}((4k)!/(k!)^4)/(256x)^k (mod p^2)$$ for any p-adic integer x not divisible by p. This enables us to determine explicitly $\sum_{k=0}^{p-1}(\pm1)^kA_k$ mod $p$, and $\sum_{k=0}^{p-1}(-1)^kA_k$ mod $p^2$ in the case p=2 (mod 3). Another consequence states that $\sum_{k=0}^{p-1}(-1)^kA_k(-2)$ is congruent to 4x^2-2p (mod p^2) if p=x^2+4y^2, and congruent to 0 (mod p^2) if p=3 (mod 4). We also prove that for any prime $p>3$ we have $$\sum_{k=0}^{p-1}(2k+1)A_k=p+7/6*p^4B_{p-3} (mod p^5)$$ where B_0,B_1,B_2,... are Bernoulli numbers.

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