A364300 a(n) = [x^n] 1/(1 + x) * Legendre_P(n, (1 - x)/(1 + x))^(-2) for n >= 0.

1, 3, 73, 3747, 329001, 44127003, 8405999785, 2160445363107, 720972846685225, 303256387595475003, 157007652309393485073, 98141188253799911132091, 72882030213423405890701449, 63436168183711463443127520699, 63968150042375034921379294100073, 73985402858435691329113991048739747

Offset: 0

Author

Peter Bala, Jul 18 2023

Keywords

• nonn
• easy

Comments

Row 2 of A364298.

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

A005259 satisfies the supercongruences

1) u (n*p^r) == u(n*p^(r-1)) (mod p^(3*r))

and the shifted supercongruences

2) u (n*p^r - 1) == u(n*p^(r-1) - 1) (mod p^(3*r))

for all primes p >= 5 and positive integers n and r.

We conjecture that the present sequence also satisfies the supercongruences 1) and 2).

Crossrefs

Cf. A005259, A364298, A364299, A364301, A364302.

Programs

Maple

a(n) := coeff(series( 1/(1 + x) * LegendreP(n, (1 - x)/(1 + x))^(-2), x, 21), x, n):
seq(a(n), n = 0..20);

Formula

Conjectures:

1) 17*a(p) - 11*a(p-1) == 40 (mod p^5) for all primes p >= 7 (checked up to p = 101).

2) for r >= 2, 17*a(p^r) - 11*a(p^r - 1) == 17*a(p^(r-1)) - 11*a(p^(r-1) - 1) (mod p^(3*r+3)) for all primes p >= 5.

3) a(p)^(3*17) == a(1)^(3*17) * a(p-1)^11 (mod p^5) for all primes p except p = 5 (checked up to p = 101).

4) for r >= 2, a(p^r)^(3*17) * a(p^(r-1) - 1)^11 == a(p^(r-1))^(3*17) * a(p^r - 1)^11 (mod p^(3*r+3)) for all primes p >= 5.