A156095 5 F(2n) (F(2n) + 1) + 1 where F(n) denotes the n-th Fibonacci number.

1, 11, 61, 361, 2311, 15401, 104401, 712531, 4875781, 33398201, 228859951, 1568486161, 10750188961, 73681909211, 505020747661, 3461456968201, 23725161388951, 162614629188281, 1114577128871281, 7639424974303651, 52361396909490901

Offset

0,2

Comments

Natural bilateral extension (brackets mark index 0): ..., 14851, 2101, 281, 31, 1, [1], 11, 61, 361, 2311, 15401, ... This is A156095-reversed followed by A156095, without repeating the central 1. That is, A156095(-n) = A156094(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 - 27 x^2 + 20 x^3 + x^4)/(1 - 11 x + 33 x^2 - 33 x^3 + 11 x^4 - x^5) = (1 - 27 x^2 + 20 x^3 + x^4)/((1 - x) (1 - 7 x + x^2) (1 - 3 x + x^2)).

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

Crossrefs

Cf. A124296, A124297, A001603, A001604, A156094.

Sequence in context: A137410 A268863 A199414 * A068846 A092164 A354336

Adjacent sequences: A156092 A156093 A156094 * A156096 A156097 A156098

Keyword

nonn,easy

Author

Stuart Clary, Feb 04 2009

Status

approved