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a(n) = [(2n+3)r] - [(n+3)r] - [nr], where [ ] = floor and r = (1+sqrt(5))/2.
3

%I #11 Sep 30 2019 10:58:51

%S 0,1,0,1,0,1,1,0,1,0,0,1,0,0,1,0,1,0,1,1,0,1,1,0,1,0,0,1,0,1,0,0,1,0,

%T 0,1,0,1,0,1,1,0,1,1,0,1,0,0,1,0,1,0,1,1,0,1,1,0,1,0,1,1,0,1,0,0,1,0,

%U 0,1,0,1,0,1,1,0,1,1,0,1,0,0,1,0,1,0

%N a(n) = [(2n+3)r] - [(n+3)r] - [nr], where [ ] = floor and r = (1+sqrt(5))/2.

%H Clark Kimberling, <a href="/A327216/b327216.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 + 3) r] - Floor[(n*r + 3 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 A327217 and A327218.

%K nonn,easy

%O 0

%A _Clark Kimberling_, Sep 02 2019