%I #11 Sep 08 2022 08:45:56
%S 4,8,13,17,23,27,31,36,41,46,50,54,60,64,69,73,79,83,87,92,96,102,106,
%T 110,115,120,125,129,134,139,143,148,152,158,162,166,171,175,181,185,
%U 189,194,199,204,208,213,218,222,227,231,237,241,245,250,254,260,264,269,273,278,283,287,292,297,301,306,310,316,320,324,329,333,339,343,348,352,357,362,366,371,376,380,385
%N a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sqrt(2), t=sqrt(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(2), t=sqrt(5) gives f=A189370, g=A189371, h=A189372.
%H G. C. Greubel, <a href="/A189370/b189370.txt">Table of n, a(n) for n = 1..10000</a>
%t r = 1; s = Sqrt[2]; t = Sqrt[5];
%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}] (* A189370 *)
%t Table[g[n], {n, 1, 120}] (* A189371 *)
%t Table[h[n], {n, 1, 120}] (* A189372 *)
%o (PARI) for(n=1,100, print1(n + floor(n*sqrt(2)) + floor(n*sqrt(5)), ", ")) \\ _G. C. Greubel_, Apr 20 2018
%o (Magma) [n + Floor(n*Sqrt(2)) + Floor(n*Sqrt(5)): n in [1..100]]; // _G. C. Greubel_, Apr 20 2018
%Y Cf. A189371, A189372.
%K nonn
%O 1,1
%A _Clark Kimberling_, Apr 20 2011