A333481 a(n) = [x^n] S(x)^(2*n), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the large Schröder numbers A006318.

1, 4, 48, 652, 9344, 138004, 2077968, 31712412, 488793088, 7591462564, 118615816048, 1862444310060, 29361743698304, 464472032918196, 7368841675386960, 117200150284494652, 1868129273410953216, 29834667873867329348, 477283242733227391152, 7647021589988643092428

Offset: 0

Author

Peter Bala, Mar 24 2020

Keywords

• nonn
• easy

Comments

Compare with the sequence A103885(n) = [x^n] S(x)^n, where S(x) is the o.g.f. of the large Schröder numbers A006318. See also A333482.

The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for prime p and positive integers n and k.

We conjecture that the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) hold for prime p >= 5 and positive integers n and k.

More generally, we conjecture that for any positive integer a and any integer b the sequence u(a,b;n) := [x^(a*n)] S(x)^(b*n) also satisfies the above congruences.

Examples

			Examples of congruences:
a(17) - a(1) = 29834667873867329348 - 4 = (2^6)*(17^3)*401*236619262717 == 0 ( mod 17^3 ).
a(2*7) - a(2) = 7368841675386960 - 48 = (2^5)*3*(7^4)*577*3229*17159 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 132585158051380247023537388004 - 138004 = (2^4)*(5^6)* 39461*13439614612035199009 == 0 ( mod 5^6 )
		

Crossrefs

Cf. A006318, A103885, A333482.

Programs

Maple

[1, seq((2/3)*add(binomial(3*n,k)*binomial(4*n-k-1,3*n-1), k = 0..n), n = 1..25)];
# alternative program
S := x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
G := (x, n) -> series(S(x)^(2*n), x, 76):
seq(coeff(G(x, n), x, n), n = 0..25);

Mathematica

Join[{1}, Table[2*Binomial[4*n-1, 3*n-1] * Hypergeometric2F1[-3*n, -n, 1 - 4*n, -1]/3, {n,1,20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

Formula

a(n) = (2/3) * Sum_{k = 0..n} C(3*n,k)*C(4*n-k-1,3*n-1) for n >= 1.

P-recursive: P(n)*a(n + 1) = 4*(7805*n^6 - 7132*n^4 + 1559*n^2 - 72)*a(n) -

P(-n)*a(n - 1), where P(n) =(2*n - 1)*(3*n + 1)*(3*n + 2)*(3*n + 3)*(35*n^2 - 35*n + 6).

a(n) ~ 2^(1/4) * (223 + 70*sqrt(10))^n / (5^(1/4) * sqrt(Pi*n) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020

a(n) = (1/2) * [x^n] 1/S(-x)^(4*n) for n >= 1. - Peter Bala, Sep 16 2024