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A189677 n+[ns/r]+[nt/r]; r=1, s=arctan(1/2), t=arctan(2).

%I A189677 #4 Mar 30 2012 18:57:25
%S A189677 2,4,7,9,12,14,17,19,22,25,28,30,33,35,37,40,42,45,48,51,53,56,58,61,
%T A189677 63,66,68,71,74,76,79,81,84,86,89,91,94,97,100,102,105,107,109,112,
%U A189677 114,117,120,123,125,128,130,133,135,138,140,143,146,148,151,153,156,158,161,163,166,169,172,174,176,179,181,184,186,189,192,195,197,200,202,205,207,210,212,215
%N A189677 n+[ns/r]+[nt/r]; r=1, s=arctan(1/2), t=arctan(2).
%C A189677 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 A189677 a(n)=n+[ns/r]+[nt/r],
%C A189677 b(n)=n+[nr/s]+[nt/s],
%C A189677 c(n)=n+[nr/t]+[ns/t], where []=floor.
%C A189677 Taking r=1, s=arctan(1/2), t=arctan(2) gives
%C A189677 a=A189677, b=A189678, c=A189679.
%F A189677 a(n)=n+[n*arctan(1/2)]+[n*arctan(2)].
%t A189677 r=1; s=ArcTan[1/2]; t=ArcTan[2];
%t A189677 a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
%t A189677 b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
%t A189677 c[n_] := n + Floor[n*r/t] + Floor[n*s/t];
%t A189677 Table[a[n], {n, 1, 120}]  (*A189677*)
%t A189677 Table[b[n], {n, 1, 120}]  (*A189678*)
%t A189677 Table[c[n], {n, 1, 120}]  (*A189679*)
%Y A189677 Cf. A189678, A189679.
%K A189677 nonn
%O A189677 1,1
%A A189677 _Clark Kimberling_, Apr 25 2011