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a(n) = Lucas(4n+3) - 1, or Lucas(2n+1)*Lucas(2n+2).
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%I #28 Dec 16 2023 17:45:08

%S 3,28,198,1363,9348,64078,439203,3010348,20633238,141422323,969323028,

%T 6643838878,45537549123,312119004988,2139295485798,14662949395603,

%U 100501350283428,688846502588398,4721424167835363,32361122672259148

%N a(n) = Lucas(4n+3) - 1, or Lucas(2n+1)*Lucas(2n+2).

%D Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

%H G. C. Greubel, <a href="/A081019/b081019.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 (8,-8,1).

%F a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).

%F G.f.: (3+4*x-2*x^2)/((1-x)*(1-7*x+x^2)). - _Colin Barker_, Jun 22 2012

%p with(combinat): luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 25 do printf(`%d,`,luc(4*n+3)-1) od: # _James A. Sellers_, Mar 03 2003

%t LucasL[4*Range[0,30] +3] -1 (* _G. C. Greubel_, Jul 14 2019 *)

%t LinearRecurrence[{8,-8,1},{3,28,198},20] (* _Harvey P. Dale_, Nov 17 2020 *)

%o (PARI) Vec((2*x^2-4*x-3)/((x-1)*(x^2-7*x+1)) + O(x^30)) \\ _Michel Marcus_, Dec 23 2014

%o (PARI) vector(30, n, n--; f=fibonacci; f(4*n+4)+f(4*n+2)-1) \\ _G. C. Greubel_, Jul 14 2019

%o (Magma) [Lucas(4*n+3)-1: n in [0..30]]; // _G. C. Greubel_, Jul 14 2019

%o (Sage) [lucas_number2(4*n+3,1,-1)-1 for n in (0..30)] # _G. C. Greubel_, Jul 14 2019

%o (GAP) List([0..30], n-> Lucas(1,-1,4*n+3)[2] -1); # _G. C. Greubel_, Jul 14 2019

%Y Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers).

%K nonn,easy

%O 0,1

%A _R. K. Guy_, Mar 01 2003

%E More terms from _James A. Sellers_, Mar 03 2003