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

Author

Stuart Clary, Feb 04 2009

Keywords

• nonn
• easy

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).

Crossrefs

Cf. A124296, A124297, A001603, A001604, A156095

Programs

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 *)

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