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.

A122868 Expansion of 1/sqrt(1-6x-3x^2).

Original entry on oeis.org

1, 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201, 1254816463923, 7873205412825, 49482344889261, 311457546052659, 1963051327342449, 12387750763156227, 78258731003169435
Offset: 0

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Author

Paul Barry, Sep 16 2006

Keywords

Comments

Binomial transform of A084609. Central coefficients of (1+3x+3x^2)^n.
The number of free (3,3)-Motzkin paths of length n, where free (k,t)-Motzkin paths are the free Motzkin paths with level steps of weight k and down steps of weight t. For example a(2)=15 because there are 9, 3, 3 paths consisting of two level steps, UD's and DU's, respectively. - Carol J. Wang (cerlined7(AT)hotmail.com), Nov 27 2007
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. - Peter Bala, Jan 07 2022

Crossrefs

Top row of array in A232973.

Programs

  • Maple
    b:= proc(x, y) option remember; `if`([x, y]=[0$2], 1,
          `if`(x>0, add(b(x-1, y+j), j=-1..1), 0)+
          `if`(y>0, b(x, y-1), 0)+`if`(y<0, b(x, y+1), 0))
        end:
    a:= n-> b(n, 0):
    seq(a(n), n=0..23);  # Alois P. Heinz, Sep 21 2021
    r := (-3)^(1/2): seq(simplify(r^n*LegendreP(n, -r)), n=0..10); # Mark van Hoeij, Nov 13 2022
  • Mathematica
    CoefficientList[Series[1/Sqrt[1-6*x-3*x^2], {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 19 2012 *)
  • Maxima
    a(n):=coeff(expand((1+3*x+3*x^2)^n),x,n);
    makelist(a(n),n,0,12);
    
  • PARI
    my(x = 'x + O('x^30)); Vec(1/sqrt(1-6*x-3*x^2)) \\ Michel Marcus, Jan 29 2016

Formula

a(n) = Sum_{k=0..floor(n/2)} C(n,2k)*C(2k,k)*3^(n-k).
E.g.f. : exp(3x)*Bessel_I(0,2*sqrt(3)x).
D-finite with recurrence: n*a(n) + 3*(1-2*n)*a(n-1) + 3*(1-n)*a(n-2) = 0. - R. J. Mathar, Nov 14 2011 [proved in Belbachir et al. (see Table 1)]
a(n) ~ (1+sqrt(3))*(3+2*sqrt(3))^n/(2*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 19 2012
a(n) = (-3)^(n/2)*LegendreP(n, -(-3)^(1/2)). - Mark van Hoeij, Nov 13 2022