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a(n) = [(2*n+1)*r] - [(n+1)*r] - [n*r], where [ ] = floor and r = (1+sqrt(5))/2.
4

%I #17 Sep 15 2019 20:10:14

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

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

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

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

%H Clark Kimberling, <a href="/A327174/b327174.txt">Table of n, a(n) for n = 0..34999</a>

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

%t r = (1 + Sqrt[5])/2; z = 200;

%t t = Table[Floor[(2 n + 1)*r] - Floor[n*r + r] - Floor[n*r], {n, 0, z}]

%Y The positions of 0's and 1's in {a(n) : n > 0} are given by A327175 and A327176.

%K nonn,easy

%O 0

%A _Clark Kimberling_, Aug 25 2019