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%I #4 Mar 30 2012 18:57:22
%S 0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,1,
%T 0,0,0,1,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,
%U 0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0
%N [nr+kr]-[nr]-[kr], where r=sqrt(3), k=3, [ ]=floor.
%C See A187950
%F a(n)=[nr+3r]-[nr]-[3r], where r=sqrt(3).
%t r=3^(1/2); k=3;
%t seqA=Table[Floor[n*r+k*r]-Floor[n*r]-Floor[k*r], {n,1,220}] (* A188086 *)
%t Flatten[Position[seqA,0] ] (* A188087 *)
%t Flatten[Position[seqA,1] ] (* A188088 *)
%Y Cf. A187950, A188087, A188088.
%K nonn
%O 1
%A _Clark Kimberling_, Mar 20 2011
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