%I #4 Mar 30 2012 18:57:25

%S 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 63,66,68,71,74,76,79,81,84,86,89,91,94,97,100,102,105,107,109,112,

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

%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(1/2), t=arctan(2) gives

%C a=A189677, b=A189678, c=A189679.

%F a(n)=n+[n*arctan(1/2)]+[n*arctan(2)].

%t r=1; s=ArcTan[1/2]; t=ArcTan[2];

%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}] (*A189677*)

%t Table[b[n], {n, 1, 120}] (*A189678*)

%t Table[c[n], {n, 1, 120}] (*A189679*)

%Y Cf. A189678, A189679.

%K nonn

%O 1,1

%A _Clark Kimberling_, Apr 25 2011