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.

A007440 Reversion of g.f. for Fibonacci numbers 1, 1, 2, 3, 5, ....

Original entry on oeis.org

1, -1, 0, 2, -3, -1, 11, -15, -13, 77, -86, -144, 595, -495, -1520, 4810, -2485, -15675, 39560, -6290, -159105, 324805, 87075, -1592843, 2616757, 2136539, -15726114, 20247800, 32296693, -152909577, 145139491, 417959049, -1460704685, 885536173, 4997618808, -13658704994
Offset: 1

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Author

N. J. A. Sloane, May 24 1994

Keywords

Comments

Binomial transform of A104565 (reversion of Pell numbers). - Paul Barry, Mar 15 2005
From Paul Barry, Nov 03 2008: (Start)
Hankel transform of a(n) (starting 0,1,-1,..) is F(n)*(-1)^C(n+1,2).
Hankel transform of a(n+1) is (-1)^C(n+1,2).
Hankel transform of a(n+2) is F(n+2)*(-1)^C(n+2,2).
(End)
The sequence 1,1,-1,0,2,... given by 0^n + Sum_{k=0..floor((n-1)/2)} binomial(n-1,2k)*A000108(k)*(-1)^(n-k-1) has Hankel transform F(n+2)*(-1)^binomial(n+1,2). - Paul Barry, Jan 13 2009
Apart from signs, essentially the same as A343773. For odd terms, a(n) = A343773(n-1), while a(n) = -A343773(n-1) if n is even. - Gennady Eremin, May 19 2021

Examples

			G.f. = x - x^2 + 2*x^4 - 3*x^5 - x^6 + 11*x^7 - 15*x^8 - 13*x^9 + 77*x^10 - 86*x^11 - 144*x^12 + ...
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    A007440 := n -> (-1)^(n+1)*hypergeom([1 - n/2, 1/2 -n/2], [2], -4):
    seq(simplify(A007440(n)), n=1..35); # Peter Luschny, Mar 19 2018, adapted to offset Jul 21 2023
    # Using function CompInv from A357588.
    CompInv(25, n -> combinat:-fibonacci(n)); # Peter Luschny, Oct 07 2022
  • Mathematica
    a[1] = 1; a[2] = -1; a[n_] := a[n] = (-5*(n-2)*a[n-2] + (1-2*n)*a[n-1])/(n+1); Array[a, 36] (* Jean-François Alcover, Apr 18 2014 *)
    Rest[CoefficientList[Series[(-1-x+Sqrt[1+2*x+5*x^2])/(2*x),{x,0,20}],x]] (* Vaclav Kotesovec, Apr 25 2015 *)
  • PARI
    a(n)=polcoeff((-1-x+sqrt(1+2*x+5*x^2+x^2*O(x^n)))/(2*x),n)
    
  • PARI
    Vec(serreverse(x/(1-x-x^2)+O(x^66))) /* Joerg Arndt, Aug 19 2012 */
    
  • Python
    A007440 = [0, 1, -1]
    for n in range(3, 801):
        A007440.append( (-(2*n-1)*A007440[-1]
          - 5*(n-2)*A007440[-2])//(n+1) )
    for n in range(1, 801):
        print(n, A007440[n])  # Gennady Eremin, May 10 2021
  • Sage
    def A007440_list(len):
        T = [0]*(len+1); T[1] = 1; R = [1]
        for n in (1..len-1):
            a,b,c = 1,0,0
            for k in range(n,-1,-1):
                r = a - b - c
                if k < n : T[k+2] = u;
                a,b,c = T[k-1],a,b
                u = r
            T[1] = u; R.append(u)
        return R
    A007440_list(36) # Peter Luschny, Nov 01 2012
    

Formula

D-finite with recurrence (n+3)*a(n+2) = -(2*n + 3)*a(n+1) - 5*n*a(n), a(1) = 1, a(2) = -1.
G.f.: A(x) = (-1 - x + sqrt(1 + 2*x + 5*x^2))/(2*x).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k)*C(k)*(-1)^(n-k), where C(n) is A000108(n). - Paul Barry, May 16 2005
a(n) = (5^((n+1)/2)*LegendreP(n-1,-1/sqrt(5)) + 5^(n/2)*LegendreP(n,-1/sqrt(5)))/(2*n+2). - Mark van Hoeij, Jul 02 2010
a(n) = 2^(-n-1)*Sum_{k=floor((n-1)/2)..n} binomial(k+1,n-k)*5^(n-k)*(-1)^k*C(k), n > 0, where C(k) is A000108. - Vladimir Kruchinin, Sep 21 2010
G.f.: (G(0)-x-1)/(x^2) = 1/G(0) where G(k) = 1 + x + x^2/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 25 2011
From Peter Bala, Jun 23 2015: (Start)
Lucas(n) = [x^n] (x/A(x))^n for n >= 1.
-1/A(-x) = 1/x - 1 + x + x^2 - 2*x^4 - 3*x^5 + x^6 + 11*x^7 + 15*x^8 - 13*x^9 + ... is the Laurent series generating function for A214649. (End)
a(n) = (-1)^n*hypergeom([1/2 - n/2, -n/2], [2], -4). - Peter Luschny, Mar 19 2018
From Gennady Eremin, May 09 2021: (Start)
a(n) = -(-1)^n * A343773(n-1), n > 0.
G.f.: A(x) = x*B(-x), where B(x) is the g.f. of A343773.
Limit_{n->infinity} a(n)/A001006(n) = 0. (End)
G.f. A(x) satisfies A(x) + 1 + x^-1 = 1/A(x). - Gennady Eremin, May 29 2021

Extensions

Extended and signs added by Olivier Gérard
Second formula adapted to offset by Vaclav Kotesovec, Apr 25 2015