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A357507
a(n) = A005259(n)^5 * (A005259(n-1))^7.
6
3125, 161958718203125, 69598400094777710760545478125, 514885225734532980507136994998009584838203125, 15708056924221066705174364772957342407662356116035885781253125, 1125221282019374727979322420623179115437017599670596496532725068048858642578125
OFFSET
1,1
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
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);
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Oct 01 2022
STATUS
approved