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A357506
a(n) = A005258(n)^3 * A005258(n-1).
6
27, 20577, 60353937, 287798988897, 1782634331587527, 13011500170881726987, 106321024671550496694837, 943479109706472533832704097, 8916177779855571182824077866307, 88547154924474394601268826256953077, 915376390434997094066775480671975209017
OFFSET
1,1
COMMENTS
The Apéry numbers B(n) = A005258(n) satisfy the supercongruences B(p) == 3 (mod p^3) and B(p-1) == 1 (mod p^3) for all primes p >= 5 (see, for example, Straub, Example 3.4). It follows that a(p) == 27 (mod p^3) for all primes p >= 5. We conjecture that, in fact, the stronger congruence a(p) == 27 (mod p^5) holds for all primes p >= 3 (checked up to p = 251). Compare with the congruence B(p) + B(p-1) == 4 (mod p^5) conjectured to hold for all primes p >= 5. See A352655.
Conjecture: for r >= 2, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for all primes p >= 5. - Peter Bala, Oct 13 2022
EXAMPLE
Example of a supercongruence:
a(7) - a(1) = 106321024671550496694837 - 27 = 2*(3^3)*5*(7^5)* 11*18143* 117398731273 == 0 (mod 7^5)
MAPLE
A005258 := n -> add(binomial(n, k)^2*binomial(n+k, k), k = 0..n):
seq(A005258(n)^3*A005258(n-1), n = 1..20);
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Oct 01 2022
STATUS
approved