A364116 a(n) = [x^n] 1/(1 - x) * Legendre_P(n, (1 + x)/(1 - x))^n for n >= 0.

1, 3, 73, 5623, 908001, 251831261, 106898093065, 64439674636863, 52344140654486017, 55113399257643294769, 73004404532578627776801, 118810038754810358401521065, 233027150139808176596750408337, 542098915811219991386976197616441

Offset: 0

Author

Peter Bala, Jul 08 2023

Keywords

• nonn
• easy

Comments

Main diagonal of A364113.

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, A364113, A364114, A364115, A364117, A364301.

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);

Mathematica

Table[SeriesCoefficient[1/(1 - x) * LegendreP[n, (1 + x)/(1 - x)]^n, {x,0,n}], {n,0,20}] (* Vaclav Kotesovec, Jul 09 2023 *)

Formula

Conjectures:

1) a(p) == 2*p + 1 (mod p^4) for all primes p >= 3 (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 >= 5 (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.

From Vaclav Kotesovec, Jul 10 2023: (Start)

a(n) ~ c * d^n * n^(2*n - 1/2), where d = 2.102423770105721036432437141524634595160013830317976222331887376263238499... (the same as for A033935) and c = 1.325068544739430738025458046917491360304162175529817456184402029433873399...

a(n) ~ A033935(n) * exp(2*n + 1) / (2*Pi*n).

a(n) ~ A033935(n) * exp(1) * n^(2*n) / n!^2. (End)