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A189395 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=1/sqrt(2), t=sqrt(3).

%I A189395 #12 Sep 22 2017 03:00:22
%S A189395 2,6,10,12,16,20,23,26,30,34,37,40,44,47,50,54,58,61,64,68,71,75,78,
%T A189395 81,85,89,92,95,99,102,105,109,113,116,119,123,127,129,133,137,140,
%U A189395 143,147,151,153,157,161,164,167,171,175,178,181,185,188,191,195,199,202,205,209,212,216,219,222,226,230,233,236,240,243,246,250,254,257,260,264,268,270,274,278,281,284
%N A189395 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=1/sqrt(2), t=sqrt(3).
%C A189395 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 A189395 a(n) = n + [n*s/r] + [n*t/r],
%C A189395 b(n) = n + [n*r/s] + [n*t/s],
%C A189395 c(n) = n + [n*r/t] + [n*s/t], where []=floor.
%C A189395 Taking r=1, s=1/sqrt(2), t=sqrt(3) gives
%C A189395 a=A189395, b=A189396, c=A189397.
%H A189395 G. C. Greubel, <a href="/A189395/b189395.txt">Table of n, a(n) for n = 1..1000</a>
%t A189395 r=1; s=2^(-1/2); t=3^(1/2);
%t A189395 a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
%t A189395 b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
%t A189395 c[n_] := n + Floor[n*r/t] + Floor[n*s/t]
%t A189395 Table[a[n], {n, 1, 120}]  (*A189395*)
%t A189395 Table[b[n], {n, 1, 120}]  (*A189396*)
%t A189395 Table[c[n], {n, 1, 120}]  (*A189397*)
%Y A189395 Cf. A189396, A189397, A189361, A189383, A189386.
%K A189395 nonn
%O A189395 1,1
%A A189395 _Clark Kimberling_, Apr 21 2011