%I
%S 0,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,
%T 0,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,
%U 0,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0
%N a(n) = [(2n+4)r]  [(n+4)r]  [nr], where [ ] = floor and r = (1+sqrt(5))/2.
%H Clark Kimberling, <a href="/A327219/b327219.txt">Table of n, a(n) for n = 0..10000</a>
%t r = (1+Sqrt[5])/2; z = 200;
%t t = Table[Floor[(2 n + 4) r]  Floor[(n*r + 4 r)]  Floor[n*r], {n, 0, z}] (* fixed by _Jianing Song_, Sep 30 2019 *)
%Y The positions of 0's and 1's in {a(n) : n > 0} are given by A327220 and A327221.
%K nonn,easy
%O 0
%A _Clark Kimberling_, Sep 02 2019
