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A071408 a(n+1) - 2*a(n) + a(n-1) = (2/3)*(1 + w^(n+1) + w^(2*n+2)) with a(1)=0, a(2)=1, and where w is the imaginary cubic root of unity.

%I A071408 #25 Mar 02 2017 11:39:31
%S A071408 0,1,4,7,10,15,20,25,32,39,46,55,64,73,84,95,106,119,132,145,160,175,
%T A071408 190,207,224,241,260,279,298,319,340,361,384,407,430,455,480,505,532,
%U A071408 559,586,615,644,673,704,735,766,799,832,865,900,935,970,1007,1044
%N A071408 a(n+1) - 2*a(n) + a(n-1) = (2/3)*(1 + w^(n+1) + w^(2*n+2)) with a(1)=0, a(2)=1, and where w is the imaginary cubic root of unity.
%C A071408 w = exp(2*Pi*i/3)= (-1 - sqrt(-3))/2. Beginning with a(2) the first differences are 3,3,3,5,5,5,7,7,7,9,9,9,11, etc.
%H A071408 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,1,-2,1).
%F A071408 a(n) = A032765(n)-1.
%F A071408 a(n) = floor((n-1)*(n+1)*(n+3)/(3*n+3)). - _Gary Detlefs_, Jul 13 2010
%F A071408 a(n) = (n-1)^2 - A030511(n-1). - _Wesley Ivan Hurt_, Jun 19 2013
%F A071408 G.f.: x^2*(1+x)*(x^2-x-1) / ( (1+x+x^2)*(x-1)^3 ). - _R. J. Mathar_, Jun 23 2013
%F A071408 a(n) = n + floor(n*(n-1)/3) - 1. - _Bruno Berselli_, Mar 02 2017
%t A071408 a[1] = 0; a[2] = 1; w = Exp[2Pi*I/3]; a[n_] := a[n] = Simplify[(2/3)(1 + w^n + w^(2n)) + 2a[n - 1] - a[n - 2]]; Table[ a[n], {n, 1, 60}]
%t A071408 Table[If[n<3,n-1,Floor[((n+1)^2-4)/3]],{n,1,100}] (*  _Vladimir Joseph Stephan Orlovsky_, Jan 30 2012 *)
%t A071408 LinearRecurrence[{2,-1,1,-2,1},{0,1,4,7,10},60] (* _Harvey P. Dale_, Jun 10 2016 *)
%o A071408 (PARI) a(n)=n*(n+2)\3 - 1 \\ _Charles R Greathouse IV_, Mar 02 2017
%Y A071408 Cf. A071618.
%K A071408 nonn,easy
%O A071408 1,3
%A A071408 _Robert G. Wilson v_, Jun 24 2002