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%I #28 Dec 18 2023 10:14:02
%S 1,1,2,5,11,22,43,85,170,341,683,1366,2731,5461,10922,21845,43691,
%T 87382,174763,349525,699050,1398101,2796203,5592406,11184811,22369621,
%U 44739242,89478485,178956971,357913942,715827883,1431655765,2863311530
%N Sequence is identical to its third differences: a(n+3) = 3*a(n+2) - 3*a(n+1) + 2*a(n), with a(0)=a(1)=1, a(2)=2.
%C The inverse binomial transform is 1,0,1,... repeated with period 3, essentially A011655. - _R. J. Mathar_, Aug 28 2023
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,2).
%F 3*a(n) = 2^(n+1) + A087204(n+1).
%F Also first differences of A024494.
%F G.f.: (1-2x+2x^2)/(1-3x+3x^2-2x^3).
%F Binomial transform of [1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, ...]; i.e., ones in positions 2, 5, 8, 11, ... and the rest zeros. [Corrected by _Gary W. Adamson_, Jan 07 2008]
%t a[n_] := a[n] = 3 a[n - 1] - 3 a[n - 2] + 2 a[n - 3]; a[0] = a[1] = 1; a[2] = 2; Table[a@n, {n, 0, 33}] (* Or *)
%t CoefficientList[ Series[(1 - 2 x + 2 x^2)/(1 - 3 x + 3 x^2 - 2 x^3), {x, 0, 33}], x] (* _Robert G. Wilson v_, Sep 08 2007 *)
%t LinearRecurrence[{3,-3,2},{1,1,2},40] (* _Harvey P. Dale_, Sep 17 2013 *)
%Y See A130750, A130752, A130755, A129339.
%Y Essentially a duplicate of A024493.
%K nonn,easy
%O 0,3
%A _Paul Curtz_, Jul 14 2007, Jul 18 2007
%E Edited by _N. J. A. Sloane_, Jul 28 2007
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