A022088 Fibonacci sequence beginning 0, 5.

0, 5, 5, 10, 15, 25, 40, 65, 105, 170, 275, 445, 720, 1165, 1885, 3050, 4935, 7985, 12920, 20905, 33825, 54730, 88555, 143285, 231840, 375125, 606965, 982090, 1589055, 2571145, 4160200, 6731345, 10891545, 17622890, 28514435, 46137325, 74651760, 120789085

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

References

• A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, pp. 15, 34, 52.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Tanya Khovanova, Recursive Sequences
• Kristina Lund, Steven Schlicker and Patrick Sigmon, Fibonacci sequences and the space of compact sets, Involve, 1:2 (2008), pp. 159-165.
• Index entries for linear recurrences with constant coefficients, signature (1,1).

Crossrefs

Cf. A000032, A000045, A000108, A014217, A119457, A267633.

Cf. sequences of the form m*Fibonacci listed in A022086.

Programs

Magma

[5*Fibonacci(n): n in [1..40]]; // Vincenzo Librandi, Sep 03 2015

Mathematica

LinearRecurrence[{1,1},{0,5},40] (* Harvey P. Dale, Jan 13 2012 *)
5*Fibonacci[Range[0, 50]] (* G. C. Greubel, Feb 10 2023 *)

PARI

a(n) = 5*fibonacci(n); \\ Michel Marcus, Sep 03 2015

SageMath

[5*fibonacci(n) for n in range(51)] # G. C. Greubel, Feb 10 2023

Formula

a(n) = round( (2*phi-1)*phi^n ) for n>3. - Thomas Baruchel, Sep 08 2004

a(n) = 5*Fibonacci(n).

a(n) = A119457(n+3,n-1) for n>1. - Reinhard Zumkeller, May 20 2006

G.f.: 5*x/(1-x-x^2). - Philippe Deléham, Nov 20 2008

a(n+2) = A014217(n+4) - A014217(n). - Paul Curtz, Dec 22 2008

a(n) = sqrt(5*(A000032(n)^2 - 4*(-1)^n)). - Alexander Samokrutov, Sep 02 2015

From Tom Copeland, Jan 25 2016: (Start)

The o.g.f. for the shifted series b(0)=0 and b(n) = a(n+1) is G(x) = 5*x*(1+x)/(1-x*(1+x)) = 5 L(-Cinv(-x)), where L(x) = x/(1-x) with inverse Linv(x) = x/(1+x) and Cinv(x) = x*(1-x), the inverse of the o.g.f. for the shifted Catalan numbers of A000108, C(x) = (1-sqrt(1-4*x))/2. Then Ginv(x) = -C(-Linv(x/5)) = (-1 + sqrt(1+4*x/(5+x)))/2.

a(n+1) = 5*Sum_{k=0..n} binomial(n-k,k) = 5 * A000045(n+1), from A267633, with the convention for zeros of the binomial assumed there. (End)

For n > 0, 1/a(n) = Sum_{k>=1} F(n*k)/(L(n+1)^(k+1)), where F(n) = A000045(n) and L(n) = A000032(n). - Diego Rattaggi, Oct 26 2022