A362408 a(n) = [x^n] (F(x)/F(-x))^n where F(x) = (1 + x)*(1 + x^3).

1, 2, 8, 44, 256, 1502, 8912, 53510, 324352, 1980332, 12160008, 75015162, 464566144, 2886488906, 17985045464, 112333392044, 703119387648, 4409231140086, 27696141476336, 174229516043630, 1097501783152256, 6921721148337452, 43701895245221848

Offset: 0

Author

Peter Bala, Apr 18 2023

Keywords

• nonn
• easy

Comments

Compare with A228960(n) = [x^n] F(x)^n.

Let k and m be positive integers and let f(x) be a finite product of cyclotomic polynomials. Define b(n) = [x^(k*n)] (f(x)/f(-x))^(m*n). Then we conjecture that the supercongruences a(p) == a(1) (mod p^3) and, for n >= 2, a(n*p) == a(n) (mod p^2) hold for all primes p, with a finite number of exceptions depending on f(x).

The present sequence is the case k = m = 1 and f(x) = (1 + x)*(1 + x^3) = C(2,x)^2 * C(6,x), where C(n,x) denotes the n-th cyclotomic polynomial. See A002003 for the case k = m = 1 and f(x) = (1 + x).

Links

• Wikipedia, Cyclotomic polynomial

Crossrefs

Cf. A002003, A228960, A333579.

Programs

Maple

F(x) := (1 + x)*(1 + x^3): G(x) := taylor(F(x)/F(-x),x = 0, 50); seq(coeftayl(G(x)^n, x = 0, n), n = 0..50);

Formula

Conjectures: 1) the supercongruence a(p) == 2 (mod p^3) holds for all primes p >= 5 (checked up to p = 47).

2) for n >= 2, the supercongruence a(n*p) == a(n) (mod p^2) holds for all primes p >= 5.