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%I
%S 2,4,6,9,12,14,17,19,22,25,27,29,33,35,37,40,43,45,48,50,53,56,58,60,
%T 64,66,68,71,74,76,78,81,84,86,89,91,94,97,99,101,105,107,109,112,115,
%U 117,120,122,125,128,130,132,136,138,140,143,146,148,150,153,156,158,161,163,166,169,171,173,177,179,181,184,187,189,192,194,197,200,202,204,208,210
%N n+[ns/r]+[nt/r]; r=1, s=arctan(1/3), t=arctan(3).
%C This is one of three sequences that partition the positive integers. In general, suppose that r, s, t are positive real numbers for which the sets {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1} are pairwise disjoint. Let a(n) be the rank of n/r when all the numbers in the three sets are jointly ranked. Define b(n) and c(n) as the ranks of n/s and n/t. It is easy to prove that
%C a(n)=n+[ns/r]+[nt/r],
%C b(n)=n+[nr/s]+[nt/s],
%C c(n)=n+[nr/t]+[ns/t], where []=floor.
%C Taking r=1, s=arctan(1/3), t=arctan(3) gives
%C a=A189753, b=A189754, c=A189755.
%t r=1; s=ArcTan[1/3]; t=ArcTan[3];
%t a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
%t b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
%t c[n_] := n + Floor[n*r/t] + Floor[n*s/t];
%t Table[a[n], {n, 1, 120}] (*A189753*)
%t Table[b[n], {n, 1, 120}] (*A189754*)
%t Table[c[n], {n, 1, 120}] (*A189755*)
%Y Cf. A189754, A189755.
%K nonn
%O 1,1
%A _Clark Kimberling_, Apr 26 2011
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