A357507 a(n) = A005259(n)^5 * (A005259(n-1))^7.

3125, 161958718203125, 69598400094777710760545478125, 514885225734532980507136994998009584838203125, 15708056924221066705174364772957342407662356116035885781253125, 1125221282019374727979322420623179115437017599670596496532725068048858642578125

Offset: 1

Author

Peter Bala, Oct 01 2022

Keywords

• nonn
• easy

Comments

The Apéry numbers A(n) = A005259(n) satisfy the supercongruences A(p) == 5 (mod p^3) and A(p-1) == 1 (mod p^3) for all primes p >= 5 (see, for example, Straub, Introduction). It follows that a(p) == 3125 (mod p^3) for all primes p >= 5. We conjecture that, in fact, the stronger congruence a(p) == 3125 (mod p^5) holds for all primes p >= 3 (checked up to p = 251). Compare with the congruence A(p) + 7*A(p-1) == 12 (mod p^5) conjectured to hold for all primes p >= 5. See A212334.

Conjecture: a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for r >= 2 and all primes p >= 5. - Peter Bala, Oct 26 2022

Links

• A. Straub, Multivariate Apéry numbers and supercongruences of rational functions, arXiv:1401.0854 [math.NT] (2014).

Crossrefs

Cf. A005259, A212334, A339946, A352655, A357506, A357508, A357509, A357567, A357568, A357569, A357956, A357957, A357958, A357959.

Programs

Maple

A005259 := n -> add(binomial(n,k)^2*binomial(n+k,k)^2, k = 0..n):
seq(A005259(n)^5 * A005259(n-1)^7, n = 1..10);