%I #33 Jul 23 2024 01:00:39
%S 1,4,6,19,46,124,321,844,2206,5779,15126,39604,103681,271444,710646,
%T 1860499,4870846,12752044,33385281,87403804,228826126,599074579,
%U 1568397606,4106118244,10749957121,28143753124,73681302246,192900153619,505019158606,1322157322204
%N a(n) = L(3*n)/L(n), where L(n) = Lucas number.
%C Subsidiary sequences: a(n) = L((2k+1)*n)/L(n) for k = 2,3, etc. This is the sequence for k = 1.
%H Colin Barker, <a href="/A110391/b110391.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F From _R. J. Mathar_, Oct 18 2010: (Start)
%F a(n) = A005248(n) - (-1)^n.
%F a(n) = +2*a(n-1) +2*a(n-2) -a(n-3).
%F G.f.: ( 1+2*x-4*x^2 ) / ( (1+x)*(x^2-3*x+1) ). (End)
%F Exp( Sum_{n >= 1} a(n)*t^n/n ) = 1 + 4*t + 11*t^2 + 29*t^3 + ... is the o.g.f. for A002878. This is the case x = 1 of the general result exp( Sum_{n >= 1} L(3*n,x)/L(n,x)*t^n/n ) = Sum_{n >= 0} L(2*n + 1,x)*t^n, where L(n,x) is the n-th Lucas polynomial of A114525. - _Peter Bala_, Mar 18 2015
%F a(n) = 2^(-n)*(-(-2)^n+(3-sqrt(5))^n+(3+sqrt(5))^n). - _Colin Barker_, Jun 03 2016
%e a(1) = L(3)/L(1) = 4/1 = 4.
%p with(combinat): L:=n->fibonacci(n+2)-fibonacci(n-2): seq(L(3*n)/L(n),n=0..30); # _Emeric Deutsch_, Jul 31 2005
%t Table[LucasL[3 n]/LucasL[n], {n, 0, 27}] (* _Michael De Vlieger_, Mar 18 2015 *)
%t LinearRecurrence[{2,2,-1},{1,4,6},40] (* _Harvey P. Dale_, Aug 20 2020 *)
%o (PARI) Vec((1+2*x-4*x^2)/((1+x)*(x^2-3*x+1)) + O(x^30)) \\ _Colin Barker_, Jun 03 2016
%o (Magma) [Lucas(3*n)/Lucas(n): n in [0..30]]; // _G. C. Greubel_, Dec 17 2017
%o (PARI) for(n=0,30, print1((fibonacci(3*n+1) + fibonacci(3*n-1))/( fibonacci(n+1) + fibonacci(n-1)), ", ")) \\ _G. C. Greubel_, Dec 17 2017
%Y Cf. A000204, A000032, A002878, A114525, A153173, A153175, A153177, A153179, A153180.
%K easy,nonn
%O 0,2
%A _Amarnath Murthy_, Jul 27 2005
%E Corrected and extended by _Emeric Deutsch_ and _Erich Friedman_, Jul 31 2005