A005592 a(n) = F(2n+1) + F(2n-1) - 1.

1, 2, 6, 17, 46, 122, 321, 842, 2206, 5777, 15126, 39602, 103681, 271442, 710646, 1860497, 4870846, 12752042, 33385281, 87403802, 228826126, 599074577, 1568397606, 4106118242, 10749957121, 28143753122, 73681302246, 192900153617, 505019158606, 1322157322202

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

Comments

For any m, the maximum element in the continued fraction of F(2n+m)/F(m) is a(n). - Benoit Cloitre, Jan 10 2006

The continued fraction [a(n);1,a(n)-1,1,a(n)-1,...] = phi^(2n), where phi = 1.618... is the golden ratio, A001622. - Thomas Ordowski, Jun 07 2013

a(n) is the number of labeled subgraphs of the n-cycle C_n. For example, a(3)=17. There are 7 subgraphs of the triangle C_3 with 0 edges, 6 with 1 edge, 3 with 2 edges, and 1 with 3 edges (C_3 itself); here 7+6+3+1 = 17. - John P. McSorley, Oct 31 2016

a(n) equals the sum of the n-th row of triangle A277919. - John P. McSorley, Nov 25 2016

Examples

			G.f. = 1 + 2*x + 6*x^2 + 17*x^3 + 46*x^4 + 122*x^5 + 321*x^6 + 842*x^7 + ...
		

References

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

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000 (terms n = 1..200 from Vincenzo Librandi)
• Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 41, 56.
• M. D. McIlroy, The number of states of a dynamic storage system, Computer J., Vol. 25, No. 3 (1982), pp. 388-392.
• M. D. McIlroy, The number of states of a dynamic storage system, Computer J., Vol. 25, No. 3 (1982), pp. 388-392. (Annotated scanned copy)
• Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
• Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
• Jesús Salas and Alan D. Sokal, Transfer Matrices and Partition-Function Zeros for Antiferromagnetic Potts Models. V. Further Results for the Square-Lattice Chromatic Polynomial, J. Stat. Phys., Vol. 135 (2009), pp. 279-373; arXiv preprint, arXiv:0711.1738 [cond-mat.stat-mech], 2007-2009. Mentions this sequence. - N. J. A. Sloane, Mar 14 2014
• Robert S. Seamons, Problem B-89, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 4, No. 2 (1966), p. 190; A Close Approximation, Solution to Problem B-89 by Douglas Lind, ibid., Vol. 5, No. 1 (1967), pp. 108-109.
• Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Crossrefs

Equals A004146+1 and A005248+1.

Bisection of A014217; the other bisection is A002878, which also bisects A000032.

Cf. A000045, A065034.

Programs

Haskell

a005592 n = a005592_list !! (n-1)
a005592_list = map (subtract 1) $
                   tail $ zipWith (+) a001519_list $ tail a001519_list
-- Reinhard Zumkeller, Aug 09 2013

Magma

[Fibonacci(2*n+1)+Fibonacci(2*n-1)-1: n in [1..30]]; // Vincenzo Librandi, Aug 23 2011

Maple

A005592:=-(2-2*z+z**2)/(z-1)/(z**2-3*z+1); # conjectured by Simon Plouffe in his 1992 dissertation
# second Maple program:
F:= n-> (<<0|1>, <1|1>>^n)[1,2]:
a:= n-> F(2*n+1)+F(2*n-1)-1:
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 04 2016

Mathematica

Table[Fibonacci[2n+1]+Fibonacci[2n-1]-1,{n,30}] (* Harvey P. Dale, Aug 22 2011 *)
a[n_] := LucasL[2n]-1; Array[a, 30] (* Jean-François Alcover, Dec 09 2015 *)

PARI

a(n)=fibonacci(2*n+1)+fibonacci(2*n-1)-1 \\ Charles R Greathouse IV, Aug 23 2011

Sage

[lucas_number2(n,3,1)-1 for n in range(1,29)] # Zerinvary Lajos, Jul 06 2008

Formula

a(n) = Lucas(2*n)-1, with Lucas(n)=A000032(n).

a(n) = floor(r^(2*n)), where r = golden ratio = (1+sqrt(5))/2.

a(n) = floor(Fibonacci(5*n)/Fibonacci(3*n)). - Gary Detlefs, Mar 11 2011

a(n) = +4*a(n-1) -4*a(n-2) +1*a(n-3). - Joerg Arndt, Mar 11 2011

a(n) = A001519(2*n-1) + A001519(2*n+1) - 1. - Reinhard Zumkeller, Aug 09 2013

a(n) = 3*a(n) - a(n-1) + 1; a(n) = A004146(n) + 1, n>0. - Richard R. Forberg, Sep 04 2013

a(n) = 2*cosh(2*n*arcsinh(1/2)) - 1. - Ilya Gutkovskiy, Oct 31 2016

a(n) = floor(sqrt(5)*Fibonacci(2*n)), for n > 0 (Seamons, 1966). - Amiram Eldar, Feb 05 2022

Extensions

Formulae and comments by Clark Kimberling, Nov 24 2010

a(0)=1 prepended by Alois P. Heinz, Nov 04 2016