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[nr+kr]-[nr]-[kr], where r=sqrt(3), k=6, [ ]=floor.
3

%I #4 Mar 30 2012 18:57:22

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

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

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

%N [nr+kr]-[nr]-[kr], where r=sqrt(3), k=6, [ ]=floor.

%C See A187950.

%F a(n)=[nr+6r]-[nr]-[6r], where r=sqrt(3).

%t r=3^(1/2); k=6;

%t seqA=Table[Floor[n*r+k*r]-Floor[n*r]-Floor[k*r], {n,1,220}] (* A188093 *)

%t Flatten[Position[seqA,0] ] (* A188094 *)

%t Flatten[Position[seqA,1] ] (* A188095 *)

%Y Cf. A187950, A188094, A188095.

%K nonn

%O 1

%A _Clark Kimberling_, Mar 20 2011