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A189515 a(n) = n + [ns] + [nt]; s=Pi/2, t=2/Pi.

%I A189515 #12 Apr 09 2021 21:42:55
%S A189515 2,6,8,12,15,18,21,25,28,31,35,37,41,43,47,51,53,57,60,63,66,70,73,76,
%T A189515 79,82,86,88,92,96,98,102,105,108,111,114,118,121,124,127,131,133,137,
%U A189515 141,143,147,149,153,156,159,163,166,169,172,176,178,182,185,188,192,194,198,201,204,208,211,214,217,220,223,227,230,233,237,239,243,246,249,253,255,259,262,265
%N A189515 a(n) = n + [ns] + [nt]; s=Pi/2, t=2/Pi.
%C A189515 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 A189515   a(n) = n + [ns/r] + [nt/r],
%C A189515   b(n) = n + [nr/s] + [nt/s],
%C A189515   c(n) = n + [nr/t] + [ns/t], where []=floor.
%C A189515 Taking r=1, s=Pi/2, t=2/Pi gives a=A189515, b=A189516, c=A189517.
%F A189515 a(n) = n + A140758(n) + floor(2*n/Pi). - _R. J. Mathar_, Sep 30 2011
%t A189515 r=1; s=Pi/2; t=2/Pi;
%t A189515 a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
%t A189515 b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
%t A189515 c[n_] := n + Floor[n*r/t] + Floor[n*s/t];
%t A189515 Table[a[n], {n, 1, 120}]  (*A189515*)
%t A189515 Table[b[n], {n, 1, 120}]  (*A189516*)
%t A189515 Table[c[n], {n, 1, 120}]  (*A189517*)
%Y A189515 Cf. A189516, A189517.
%K A189515 nonn
%O A189515 1,1
%A A189515 _Clark Kimberling_, Apr 23 2011