%I #15 Sep 08 2022 08:45:57
%S 2,6,8,12,15,18,22,25,28,31,34,37,41,44,47,51,53,57,61,63,67,69,73,76,
%T 79,83,86,89,92,95,99,102,105,108,112,114,118,122,124,128,130,134,137,
%U 140,144,147,150,153,156,160,163,166,169,173,175,179,183,185,189,191,195,199,201,205,208,211,214,218,221,224,227,230,234,237
%N a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sqrt(5/2), t=sqrt(2/5).
%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(5/2), t=sqrt(2/5) gives f=A190344, g=A190345, h=A190346.
%H G. C. Greubel, <a href="/A190344/b190344.txt">Table of n, a(n) for n = 1..10000</a>
%F A190344: f(n) = n + [n*sqrt(5/2)] + [n*sqrt(2/5)].
%F A190345: g(n) = n + [n*sqrt(2/5)] + [2*n/5].
%F A190346: h(n) = 3*n + [n*sqrt(5/2)] + [n/2].
%t r=1; s=(5/2)^(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}] (* A190344 *)
%t Table[g[n], {n, 1, 120}] (* A190345 *)
%t Table[h[n], {n, 1, 120}] (* A190346 *)
%o (PARI) for(n=1,100, print1(n + floor(n*sqrt(5/2)) + floor(n*sqrt(2/5)), ", ")) \\ _G. C. Greubel_, Apr 05 2018
%o (Magma) [n + Floor(n*Sqrt(5/2)) + Floor(n*Sqrt(2/5)): n in [1..100]]; // _G. C. Greubel_, Apr 05 2018
%Y Cf. A190345, A190346.
%K nonn
%O 1,1
%A _Clark Kimberling_, May 09 2011