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

%I #11 Feb 10 2014 01:31:25

%S 1,1,31,281,2101,14851,102961,708761,4865911,33372361,228792301,

%T 1568309051,10749725281,73680695281,505017569551,3461448647801,

%U 23725139605861,162614572159411,1114576979567761,7639424583421961,52361395886149351

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

%C 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).

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

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

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

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

%F 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)).

%F a(n)=A056854(n)-5*A001906(n)-1. - _R. J. Mathar_, Feb 23 2009

%F 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

%t a[n_Integer] := 5 Fibonacci[2n] (Fibonacci[2n] - 1) + 1

%t 5(#*(#-1))&/@Fibonacci[Range[0,40,2]]+1 (* _Harvey P. Dale_, Jan 06 2013 *)

%Y Cf. A124296, A124297, A001603, A001604, A156095

%K nonn,easy

%O 0,3

%A _Stuart Clary_, Feb 04 2009