%I #12 Sep 08 2022 08:45:57
%S 2,5,8,12,14,17,21,24,26,29,33,36,39,42,45,48,51,54,57,60,64,67,69,72,
%T 76,79,81,85,88,91,95,97,100,103,107,109,112,116,119,121,124,128,131,
%U 134,137,140,143,146,149,152,155,159,162,164,168,171,174,176,180,183,186,190,192,195,198,202,204,207,211,214,216,219,223,226
%N a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sqrt(5/3), t=sqrt(3/5).
%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 f(n) = n + [n*s/r] + [n*t/r],
%C g(n) = n + [n*r/s] + [n*t/s],
%C h(n) = n + [n*r/t] + [n*s/t], where []=floor.
%C Taking r=1, s=sqrt(5/3), t=sqrt(3/5) gives f=A190347, g=A190348, h=A190349.
%H G. C. Greubel, <a href="/A190347/b190347.txt">Table of n, a(n) for n = 1..10000</a>
%F A190347: f(n) = n + [n*sqrt(5/3)] + [n*sqrt(3/5)].
%F A190348: g(n) = n + [n*sqrt(3/5)] + [3*n/5].
%F A190349: h(n) = 2*n + [n*sqrt(5/3)] + [n/3].
%t r=1; s=(5/3)^(1/2); t=1/s;
%t f[n_] := n + Floor[n*s/r] + Floor[n*t/r];
%t g[n_] := n + Floor[n*r/s] + Floor[n*t/s];
%t h[n_] := n + Floor[n*r/t] + Floor[n*s/t];
%t Table[f[n], {n, 1, 120}] (* A190347 *)
%t Table[g[n], {n, 1, 120}] (* A190348 *)
%t Table[h[n], {n, 1, 120}] (* A190349 *)
%o (PARI) for(n=1,100, print1(n + floor(n*sqrt(5/3)) + floor(n*sqrt(3/5)), ", ")) \\ _G. C. Greubel_, Apr 05 2018
%o (Magma) [n + Floor(n*Sqrt(5/3)) + Floor(n*Sqrt(3/5)): n in [1..100]]; // _G. C. Greubel_, Apr 05 2018
%Y Cf. A190348, A190349.
%K nonn
%O 1,1
%A _Clark Kimberling_, May 09 2011