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A189383 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=1/sqrt(2), t=1/sqrt(3).

%I A189383 #13 Sep 22 2017 03:00:13
%S A189383 1,4,6,8,10,13,15,17,20,22,24,26,29,31,33,36,38,40,42,45,47,49,52,53,
%T A189383 56,59,61,63,65,68,69,72,75,77,79,81,84,85,88,91,92,95,97,100,101,104,
%U A189383 107,108,111,113,116,118,120,123,124,127,129,132,134,136,139,140,143,145,147,150,152,155,156,159,161,163,166,168,171,172,175,178,179,182,184,186,188,191
%N A189383 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=1/sqrt(2), t=1/sqrt(3).
%C A189383 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 A189383 a(n) = n + [n*s/r] + [n*t/r],
%C A189383 b(n) = n + [n*r/s] + [n*t/s],
%C A189383 c(n) = n + [n*r/t] + [n*s/t], where []=floor.
%C A189383 Taking r=1, s=1/sqrt(2), t=1/sqrt(3) gives
%C A189383 a=A189383, b=A189384, c=A189385.
%H A189383 G. C. Greubel, <a href="/A189383/b189383.txt">Table of n, a(n) for n = 1..1000</a>
%t A189383 r=1; s=2^(-1/2); t=3^(-1/2);
%t A189383 a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
%t A189383 b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
%t A189383 c[n_] := n + Floor[n*r/t] + Floor[n*s/t]
%t A189383 Table[a[n], {n, 1, 120}]  (*A189383*)
%t A189383 Table[b[n], {n, 1, 120}]  (*A189384*)
%t A189383 Table[c[n], {n, 1, 120}]  (*A189385*)
%Y A189383 Cf. A189384, A189385, A188386, A189361, A189386, A189395.
%K A189383 nonn
%O A189383 1,2
%A A189383 _Clark Kimberling_, Apr 21 2011