A104565 Reversion of Pell numbers A000129(n+1).

1, -2, 3, -2, -6, 28, -61, 54, 158, -860, 2062, -2004, -5804, 33720, -84509, 86054, 247862, -1492908, 3838298, -4019452, -11537556, 71101832, -185868978, 198310460, 567902572, -3555617432, 9404104764, -10168382696, -29069700056, 184127171952, -491229517661

Offset: 0

Author

Paul Barry, Mar 15 2005

Keywords

• easy
• sign

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000

Programs

Maple

a:= proc(n) a(n):= `if`(n<2, 1-3*n,
      ((8-8*n)*a(n-2)-(4*n+2)*a(n-1))/(n+2))
    end:
seq (a(n), n=0..40);  # Alois P. Heinz, Nov 09 2012

Mathematica

CoefficientList[Series[(Sqrt[1+4*x+8*x^2]-1-2*x)/(2*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 08 2014 *)
Table[(-2)^n Hypergeometric2F1[1/2-n/2, -n/2, 2, -1], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 07 2015 *)

Sage

def A104565_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 - 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
A104565_list(30)  # Peter Luschny, Nov 01 2012

Formula

G.f.: (sqrt(1+4*x+8*x^2)-1-2*x)/(2*x^2).

a(n) = sum{k=0..floor(n/2), binomial(n, 2k)*C(k)*(-1)^(n-k)2^(n-2k)}, where C(n) is A000108. - Paul Barry, May 16 2005

G.f. 1/G(0) where G(k)= 1 + 2*x + x^2/G(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Aug 10 2012

G.f.: (2/W(0)-1)/x where W(k)= 1 + 1/(1 + 2*x/(1 + 2*x/W(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Sep 21 2012

D-finite with recurrence (n+2)*a(n) +2*(2*n+1)*a(n-1) +8*(n-1)*a(n-2)=0. - R. J. Mathar, Nov 09 2012

G.f.: G(0)/x^2 - 1/x - 1/x^2 where G(k)= 1 + 2*x/(1 + 1/(1 + 2*x/G(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 23 2012

G.f.: 1/(x^2*Q(0)) - 1/(x^2) - 1/x, where Q(k)= 1 - (4*k+1)*x*(1+2*x)/(k+1 - x*(1+2*x)*(2*k+2)*(4*k+3)/(2*x*(1+2*x)*(4*k+3) - (2*k+3)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 15 2013

Lim sup n->infinity |a(n)|^(1/n) = 2*sqrt(2). - Vaclav Kotesovec, Feb 08 2014

a(n) = (-2)^n*hypergeom([1/2-n/2,-n/2], [2], -1). - Vladimir Reshetnikov, Nov 07 2015