%I #4 Mar 30 2012 18:57:22
%S 0,1,1,0,0,1,1,0,1,1,1,0,1,1,0,0,1,1,0,1,1,1,0,1,1,1,0,1,1,0,0,1,1,0,
%T 1,1,1,0,1,1,1,0,1,1,0,0,1,1,0,1,1,1,0,1,1,0,0,1,1,0,1,1,1,0,1,1,1,0,
%U 1,1,0,0,1,1,0,1,1,1,0,1,1,1,0,1,1,0,0,1,1,0,1,1,1,0,1,1,1,0,1,1,0,0,1,1,0,1,1,1,0,1,1,0,0,1,1,0,1,1,1,0,1,1,1
%N [nr]-[kr]-[nr-kr], where r=sqrt(3), k=5, [ ]=floor.
%C See A188014.
%F a(n)=[nr]-[5r]-[nr-5r], where r=sqrt(3).
%t r=3^(1/2)); k=5;
%t t=Table[Floor[n*r]-Floor[(n-k)*r]-Floor[k*r],{n,1,220}] (*A188076*)
%t Flatten[Position[t,0]] (*A188077*)
%t Flatten[Position[t,1]] (*A188078*)
%Y Cf. A188014, A188077, A188078.
%K nonn
%O 1
%A _Clark Kimberling_, Mar 20 2011