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a(0) = 3, a(1) = 8, a(n+2) = 4*a(n+1) - a(n) - 2.
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%I #19 Nov 03 2016 15:20:35

%S 3,8,27,98,363,1352,5043,18818,70227,262088,978123,3650402,13623483,

%T 50843528,189750627,708158978,2642885283,9863382152,36810643323,

%U 137379191138,512706121227,1913445293768,7141075053843,26650854921602,99462344632563,371198523608648

%N a(0) = 3, a(1) = 8, a(n+2) = 4*a(n+1) - a(n) - 2.

%H Colin Barker, <a href="/A102206/b102206.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 (5,-5,1).

%F G.f.: (2x-1)(x-3)/((1-x)(x^2-4x+1)).

%F a(n) = A092184(n+1) + 2; a(n+1) - a(n) = A001834(n+1) (see comment).

%F a(0)=3, a(1)=8, a(2)=27, a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3). - _Harvey P. Dale_, Jul 25 2012

%F a(n) = (2+(2-sqrt(3))^(1+n)+(2+sqrt(3))^(1+n))/2. - _Colin Barker_, Nov 03 2016

%t a[0] = 3; a[1] = 8; a[n_] := a[n] = 4a[n - 1] - a[n - 2] - 2; Table[a[n], {n, 0, 23}] (* Or *)

%t CoefficientList[ Series[(2x - 1)(x - 3)/((1 - x)(x^2 - 4x + 1)), {x, 0, 22}], x] (* _Robert G. Wilson v_, Jan 12 2005 *)

%t LinearRecurrence[{5,-5,1},{3,8,27},30] (* _Harvey P. Dale_, Jul 25 2012 *)

%o (PARI) Vec((2*x-1)*(x-3)/((1-x)*(x^2-4*x+1)) + O(x^30)) \\ _Colin Barker_, Nov 03 2016

%Y Cf. A092184, A001834, A001353, A102207.

%K nonn,easy

%O 0,1

%A _Creighton Dement_, Dec 30 2004

%E More terms from _Robert G. Wilson v_, Jan 12 2005

%E Recurrence in the definition corrected by _R. J. Mathar_, Aug 07 2008