On sums of Apéry polynomials and related congruences

The Apéry polynomials are given byAn(x)=∑k=0n(nk)2(n+kk)2xk(n=0,1,2,…). (Those An=An(1)An=An(1) are Apéry numbers.) Let p be an odd prime. We show that∑k=0p−1(−1)kAk(x)≡∑k=0p−1(2kk)316kxk(modp2), and that∑k=0p−1Ak(x)≡(xp)∑k=0p−1(4kk,k,k,k)(256x)k(modp) for any p  -adic integer x≢0(modp). This enables us to determine explicitly ∑k=0p−1(±1)kAkmodp, and ∑k=0p−1(−1)kAkmodp2 in the case p≡2(mod3). Another consequence states that∑k=0p−1(−1)kAk(−2)≡{4x2−2p(modp2)if p=x2+4y2(x,y∈Z),0(modp2)if p≡3(mod4). We also prove that for any prime p>3p>3 we have∑k=0p−1(2k+1)Ak≡p+76p4Bp−3(modp5) where B0,B1,B2,…B0,B1,B2,… are Bernoulli numbers.