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A173276 a(n) = a(n-2) + a(n-3) - floor(a(n-3)/2) - floor(a(n-4)/2).

%I A173276 #21 Jul 30 2023 16:26:45
%S A173276 1,1,1,2,2,3,3,3,4,4,5,5,5,6,6,7,7,7,8,8,9,9,9,10,10,11,11,11,12,12,
%T A173276 13,13,13,14,14,15,15,15,16,16,17,17,17,18,18,19,19,19,20,20,21,21,21,
%U A173276 22,22,23,23,23,24,24,25,25,25,26,26,27,27,27,28,28
%N A173276 a(n) = a(n-2) + a(n-3) - floor(a(n-3)/2) - floor(a(n-4)/2).
%C A173276 Instead of the Fibonacci sequence this has the base Padovan sequence.
%C A173276 The a(n+1)/a(n) ratio approaches one.
%H A173276 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,1,-1).
%F A173276 a(n) = a(n-2)+a(n-3)-floor(a(n-3)/2)-floor(a(n-4)/2).
%F A173276 Empirical g.f.: (x^3+1) / (x^6-x^5-x+1) = (x+1)*(x^2-x+1) / ((x-1)^2*(x^4+x^3+x^2+x+1)). - _Colin Barker_, Mar 23 2013
%F A173276 From _Wesley Ivan Hurt_, Mar 15 2015: (Start)
%F A173276 a(n) = a(n-1) + a(n-5) - a(n-6).
%F A173276 a(n) = floor( (2n+5)/5 ). (End)
%p A173276 A173276:=n->floor((2*n+5)/5): seq(A173276(n), n=0..50); # _Wesley Ivan Hurt_, Mar 15 2015
%t A173276 f[-2] = 0; f[-1] = 0; f[0] = 1; f[1] = 1;
%t A173276 f[n_] := f[n] = f[n - 2] + f[n - 3] - Floor[f[n - 3]/2] - Floor[f[n - 4]/2]
%t A173276 Table[f[n], {n, 0, 50}]
%t A173276 nxt[{a_,b_,c_,d_}]:={b,c,d,c+b-Floor[b/2]-Floor[a/2]}; NestList[nxt,{1,1,1,2},70][[;;,1]] (* _Harvey P. Dale_, Jul 30 2023 *)
%o A173276 (Magma) [Floor((2*n+5)/5) : n in [0..50]]; // _Wesley Ivan Hurt_, Mar 15 2015
%o A173276 (PARI) vector(100,n,(2*n+3)\5) \\ _Derek Orr_, Mar 21 2015
%Y A173276 Cf. A000931 (Padovan), A085269.
%K A173276 nonn,easy
%O A173276 0,4
%A A173276 _Roger L. Bagula_, Nov 22 2010