A122871 Expansion of (1 - 2*x - sqrt(1 - 4*x - 8*x^2))/(6*x^2).

1, 2, 7, 26, 106, 452, 1999, 9074, 42046, 198044, 945430, 4564100, 22243060, 109285256, 540738943, 2692103714, 13475973238, 67784600108, 342439638418, 1736727343436, 8839203054604, 45132514680248, 231121351433158

Offset: 0

Author

Paul Barry, Sep 16 2006

Keywords

• easy
• nonn

Comments

Series reversion of x/(1+2x+3x^2). Binomial transform is A107264. Counts colored Motzkin paths. Second binomial transform of 1,0,3,0,18,0,... or 3^n*binomial(n) (A005159) with interpolated zeros.

Hankel transform is 3^binomial(n+1,2). - Paul Barry, Oct 01 2009

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi)
• Paul Barry, Characterizations of the Borel triangle and Borel polynomials, arXiv:2001.08799 [math.CO], 2020.
• Serkan Demiriz, Adem Şahin, and Sezer Erdem, Some topological and geometric properties of novel generalized Motzkin sequence spaces, Rendiconti Circ. Mat. Palermo Ser. 2 (2025) Vol. 74, No. 136. See p. 4.
• Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.

Programs

Mathematica

CoefficientList[Series[(1-2*x-Sqrt[1-4*x-8*x^2])/(6*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 19 2012 *)

PARI

x='x+O('x^50); Vec((1 - 2*x - sqrt(1 - 4*x - 8*x^2))/(6*x^2)) \\ G. C. Greubel, Mar 19 2017

Sage

def A122871_list(n):  # n>=1
    T = [0]*(n+1); R = [1]
    for m in (1..n-1):
        a,b,c = 1,0,0
        for k in range(m,-1,-1):
            r = a + 2*b + 3*c
            if k < m : T[k+2] = u;
            a,b,c = T[k-1],a,b
            u = r
        T[1] = u; R.append(u)
    return R
A122871_list(23)  # Peter Luschny, Nov 01 2012

Formula

E.g.f.: exp(2*x)*Bessel_I(1, sqrt(3)*2*x)/(sqrt(3)x).

a(n) = Sum_{k=0..floor(n/2)} binomial(n,2*k)*binomial(k)3^k*2^(n-2k).

G.f.: 1/(1-2x-3x^2/(1-2x-3x^2/(1-2x-3x^2/(1-2x-3x^2/(1-.... (continued fraction). - Paul Barry, Oct 01 2009

D-finite with recurrence: (n+2)*a(n) - 2*(2n+1)*a(n-1) + 8*(1-n)*a(n-2) = 0. - R. J. Mathar, Nov 14 2011

a(n) ~ 2*sqrt(9+5*sqrt(3))*(2+2*sqrt(3))^n/(3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 19 2012