A156094 5 F(2n) (F(2n) - 1) + 1 where F(n) denotes the n-th Fibonacci number.

1, 1, 31, 281, 2101, 14851, 102961, 708761, 4865911, 33372361, 228792301, 1568309051, 10749725281, 73680695281, 505017569551, 3461448647801, 23725139605861, 162614572159411, 1114576979567761, 7639424583421961, 52361395886149351

Offset

0,3

Comments

Natural bilateral extension (brackets mark index 0): ..., 15401, 2311, 361, 61, 11, [1], 1, 31, 281, 2101, 14851, ... This is A156095-reversed followed by A156094, without repeating the central 1. That is, A156094(-n) = A156095(n).

Links

Table of n, a(n) for n=0..20.

Formula

Let F(n) be the Fibonacci number A000045(n) and let L(n) be the Lucas number A000032(n).

Alternate formula: a(n) = L(4n) - 5 F(2n) - 1.

Recurrence: a(n) - 10 a(n-1) + 23 a(n-2) - 10 a(n-3) + a(n-4) = -5.

Recurrence: a(n) - 11 a(n-1) + 33 a(n-2) - 33 a(n-3) + 11 a(n-4) - a(n-5) = 0.

G.f.: A(x) = (1 - 10 x + 53 x^2 - 60 x^3 + 11 x^4)/(1 - 11 x + 33 x^2 - 33 x^3 + 11 x^4 - x^5) = (1 - 10 x + 53 x^2 - 60 x^3 + 11 x^4)/((1 - x) (1 - 7 x + x^2) (1 - 3 x + x^2)).

a(n)=A056854(n)-5*A001906(n)-1. - R. J. Mathar, Feb 23 2009

a(n)=((2*sqrt(5))/2)*(((3-sqrt(5))/2)^n-((3+sqrt(5))/2)^n)+((7+3*sqrt(5))/2)^n+((7-3*sqrt(5))/2)^n-1. - Tim Monahan, Aug 15 2011

Mathematica

a[n_Integer] := 5 Fibonacci[2n] (Fibonacci[2n] - 1) + 1
5(#*(#-1))&/@Fibonacci[Range[0, 40, 2]]+1 (* Harvey P. Dale, Jan 06 2013 *)

Crossrefs

Cf. A124296, A124297, A001603, A001604, A156095

Sequence in context: A008386 A161558 A336279 * A221430 A115151 A327561

Adjacent sequences: A156091 A156092 A156093 * A156095 A156096 A156097

Keyword

nonn,easy

Author

Stuart Clary, Feb 04 2009

Status

approved