OFFSET
0,2
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).
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) = 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
MATHEMATICA
LinearRecurrence[{1, 1}, {0, 5}, 40] (* Harvey P. Dale, Jan 13 2012 *)
5*Fibonacci[Range[0, 50]] (* G. C. Greubel, Feb 10 2023 *)
PROG
(Magma) [5*Fibonacci(n): n in [1..40]]; // Vincenzo Librandi, Sep 03 2015
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved