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

%I #11 Sep 15 2019 20:10:48

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

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

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

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

%H Clark Kimberling, <a href="/A327208/b327208.txt">Table of n, a(n) for n = 0..10000</a>

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

%t r = Sqrt[3]; z = 200;

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

%Y The positions of 0's and 1's in {a(n) : n > 0} are given by A327209 and A327210.

%K nonn,easy

%O 0

%A _Clark Kimberling_, Aug 26 2019