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A327202 a(n) = [(2n+2)r] - [(n+2)r] - [nr], where [ ] = floor and r = (1+sqrt(5))/2. 3

%I #14 Sep 15 2019 20:10:29

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

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

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

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

%H Clark Kimberling, <a href="/A327202/b327202.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 = (1+sqrt(5))/2.

%t r = (1 + Sqrt[5])/2; 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 A327203 and A327204.

%K nonn,easy

%O 0

%A _Clark Kimberling_, Aug 26 2019

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