A364301 a(n) = [x^n] 1/(1 + x) * Legendre_P(n, (1 - x)/(1 + x))^(-n) for n >= 0.

1, 1, 73, 10805, 3100001, 1479318759, 1062573281785, 1073267499046525, 1451614640844881665, 2534009926232394596267, 5548110762587726241026801, 14890865228866506199602545427, 48084585660733078332263158771313, 183923731031112887024255817209295155, 822427361894711201025101782425695273529

Offset: 0

Author

Peter Bala, Jul 18 2023

Keywords

• nonn
• easy

Comments

Main diagonal of A364298 (with extra initial term 1). Compare with A364116.

Compare with the two types of Apéry numbers A005258 and A005259, which are related to the Legendre polynomials by A005258(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x)) and A005259(n) = [x^k] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^2.

A005258 is the main diagonal of A108625 and A005259 is the main diagonal of A143007.

Crossrefs

Cf. A005258, A005259, A108625, A143007, A364116, A364298, A364302.

Programs

Maple

a(n) := coeff(series( 1/(1 + x) * LegendreP(n, (1 - x)/(1 + x))^(-n), x, 21), x, n):
seq(a(n), n = 0..20);

Formula

Conjectures:

1) a(p) == 2*p - 1 (mod p^4) for all primes p >= 5 (checked up to p = 101).

More generally, the supercongruence a(p^k) == 2*p^k - 1 (mod p^(3+k)) may hold for all primes p >= 5 and all k >= 1.

2) a(p-1) == 1 (mod p^3) for all primes p except p = 3 (checked up to p = 101).

More generally, the supercongruence a(p^k - p^(k-1)) == 1 (mod p^(2+k)) may hold for all primes p >= 5 and all k >= 1.