A133641 a(n) = 2*L(n) + L(n-1) - n, L(n) = n-th Lucas number A000204(n).

1, 5, 8, 14, 24, 41, 69, 115, 190, 312, 510, 831, 1351, 2193, 3556, 5762, 9332, 15109, 24457, 39583, 64058, 103660, 167738, 271419, 439179, 710621, 1149824, 1860470, 3010320, 4870817, 7881165, 12752011, 20633206, 33385248, 54018486, 87403767, 141422287, 228826089, 370248412

Offset

1,2

Comments

Limit_{n->infinity} a(n)/a(n-1) = phi.

Links

Table of n, a(n) for n=1..39.

W. Kuszmaul, Fast Algorithms for Finding Pattern Avoiders and Counting Pattern Occurrences in Permutations, arXiv preprint arXiv:1509.08216 [cs.DM], 2015-2017.

Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Formula

Given n-th Lucas number A000204(n), a(n) = 2*L(n) + L(n-1) - n.

G.f.: -x*(1-5*x^2+x^3+2*x+2*x^4)/(-1+x+x^2)/(-1+x)^2. - R. J. Mathar, Nov 14 2007

a(n) = A000032(n+2) - n = Fibonacci(n+3) + Fibonacci(n+1) - n, n > 1. - R. J. Mathar, Jul 20 2009, extended by David A. Corneth, Aug 08 2018

Example

a(5) = 24 = 2*L(5) + L(4) - n = 2*11 + 7 - 5.

Mathematica

a[1] = 1; a[n_] := LucasL[n+2] - n;
Array[a, 14] (* Jean-François Alcover, Aug 08 2018, after R. J. Mathar *)

Prog

(PARI) a(n) = {if(n==1, 1, fibonacci(n+3)+fibonacci(n+1)-n)} \\ David A. Corneth, Aug 08 2018

Crossrefs

Cf. A000032, A000204.

Sequence in context: A124011 A101835 A192522 * A164094 A246319 A302649

Adjacent sequences: A133638 A133639 A133640 * A133642 A133643 A133644

Keyword

nonn,easy

Author

Gary W. Adamson, Sep 19 2007

Status

approved