A364299 a(n) = [x^n] 1/(1 + x) * Legendre_P(n, (1 - x)/(1 + x))^(-1) for n >= 0.

1, 1, 19, 721, 49251, 5370751, 859748023, 190320431953, 55743765411043, 20884452115700251, 9745388924112505269, 5543574376457462884111, 3776677001062829977964007, 3036161801705682492174749691, 2844274879825369072829081331519

Offset: 0

Author

Peter Bala, Jul 18 2023

Keywords

• nonn
• easy

Comments

Row 1 of A364298.

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

A005258 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. A005258, A364298, A364300, A364301, A364302.

Programs

Maple

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

Formula

Conjectures:

1) 13*a(p) - 7*a(p-1) == 6 (mod p^5) for all primes p >= 3 (checked up to p = 101).

2) for r >= 2, 13*a(p^r) - 7*a(p^r - 1) == 13*a(p^(r-1)) - 7*a(p^(r-1) - 1) (mod p^(3*r+3)) for all primes p >= 5.

3) a(p)^13 == a(p-1)^7 (mod p^5) for all primes p >= 3 (checked up to p = 101).

4) for r >= 2, a(p^r)^13 * a(p^(r-1) - 1)^7 == a(p^(r-1))^13 * a(p^r - 1)^7 (mod p^(3*r+3)) for all primes p >= 5.