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%I #31 Sep 08 2022 08:45:30
%S 1,3,7,9,47,123,161,843,2207,2889,15127,39603,51841,271443,710647,
%T 930249,4870847,12752043,16692641,87403803,228826127,299537289,
%U 1568397607,4106118243,5374978561,28143753123,73681302247,96450076809,505019158607,1322157322203
%N a(n) = (F(2*n-1) + F(2*n+1))*(5/6 - cos(2*Pi*n/3)/3), where F(n) = Fibonacci(n).
%C The a(n+1) are the numerators of A178381(4*n+3)/A178381(4*n+2). For the denominators see A179133(n). - _Johannes W. Meijer_, Jul 01 2010
%H Vincenzo Librandi, <a href="/A128052/b128052.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,18,0,0,-1).
%F Lim_{n->infinity} A128052(n+1)/A179133(n) = 1 + cos(Pi/5). - _Johannes W. Meijer_, Jul 01 2010
%F a(n) = Lucas(2*n)*(Fibonacci(n) mod 2 + 1)/2, Lucas(n)=A000032, Fibonacci(n)=A000045. - _Gary Detlefs_, Jan 19 2001
%F From _Colin Barker_, Jun 27 2013: (Start)
%F a(n) = 18*a(n-3) - a(n-6).
%F G.f: -(3*x^5 + 7*x^4 + 9*x^3 - 7*x^2 - 3*x - 1) / ((x^2 - 3*x + 1)*(x^4 + 3*x^3 + 8*x^2 + 3*x + 1)). (End)
%F With L(n) the Lucas number A000032, a(n) = L(2*n)/2 or L(2*n) according as n is, or is not, divisible by 3. - _David Callan_, Jul 17 2019
%p with(combinat): nmax:=25; for n from 0 to nmax do a(n):= (fibonacci(2*n-1)+fibonacci(2*n+1))*(5/6-cos(2*Pi*n/3)/3) od: seq(a(n),n=0..nmax); # _Johannes W. Meijer_, Jul 01 2010
%t LinearRecurrence[{0, 0, 18, 0, 0, -1}, {1, 3, 7, 9, 47, 123}, 40] (* _Vincenzo Librandi_, Jul 17 2019 *)
%o (Magma) I:=[1,3,7,9,47,123]; [n le 6 select I[n] else 18*Self(n-3)-Self(n-6): n in [1..30]]; // _Vincenzo Librandi_, Jul 17 2019
%Y Cf. A128053.
%Y Cf. A179134. Trisection: A023039.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Feb 13 2007
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