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A156095 5 F(2n) (F(2n) + 1) + 1 where F(n) denotes the n-th Fibonacci number. 4

%I

%S 1,11,61,361,2311,15401,104401,712531,4875781,33398201,228859951,

%T 1568486161,10750188961,73681909211,505020747661,3461456968201,

%U 23725161388951,162614629188281,1114577128871281,7639424974303651,52361396909490901

%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): ..., 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).

%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 - 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)).

%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

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

%K nonn,easy

%O 0,2

%A _Stuart Clary_, Feb 04 2009

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Last modified May 16 08:07 EDT 2021. Contains 343940 sequences. (Running on oeis4.)