cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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.

Original entry on oeis.org

1, 4, 48, 652, 9344, 138004, 2077968, 31712412, 488793088, 7591462564, 118615816048, 1862444310060, 29361743698304, 464472032918196, 7368841675386960, 117200150284494652, 1868129273410953216, 29834667873867329348, 477283242733227391152, 7647021589988643092428
Offset: 0

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Author

Peter Bala, Mar 24 2020

Keywords

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

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
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