%I #10 Sep 08 2022 08:45:57

%S 2,5,8,11,15,18,20,24,27,31,33,36,39,43,46,49,51,56,58,62,64,67,71,74,

%T 77,80,84,87,89,93,95,100,102,105,108,112,115,118,120,124,127,131,133,

%U 136,140,143,146,149,153,156,158,162,164,169,171,174,177,181,184,187,189,193,196,200,202,205,209,212,215,218,220,225,227,231

%N n + [n*s/r] + [n*t/r] + [n*u/r]; r=sqrt(2), s=1/r, t=sqrt(3), u=1/t.

%C This is one of four sequences that partition the positive integers. In general, suppose that r, s, t, u are positive real numbers for which the sets {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1, {h/u: h>=1} are pairwise disjoint. Let a(n) be the rank of n/r when all the numbers in the four sets are jointly ranked. Define b(n), c(n), d(n) as the ranks of n/s, n/t, n/u, respectively. It is easy to prove that

%C a(n) = n + [n*s/r] + [n*t/r] + [n*u/r],

%C b(n) = n + [n*r/s] + [n*t/s] + [n*u/s],

%C c(n) = n + [n*r/t] + [n*s/t] + [n*u/t],

%C d(n) = n + [n*r/u] + [n*s/u] + [n*t/u], where []=floor.

%C Taking r=sqrt(2), s=1/r, t=sqrt(3), u=1/t gives

%C a=A190364, b=A190365, c=A190366, d=A190367.

%H G. C. Greubel, <a href="/A190364/b190364.txt">Table of n, a(n) for n = 1..10000</a>

%F A190364: a(n) = n + [n/2] + [n*sqrt(3/2)] + [n*sqrt(1/6)].

%F A190365: b(n) = 3*n + [n*sqrt(6)] + [n*sqrt(2/3)].

%F A190366: c(n) = n + [n*sqrt(2/3)] + [n*sqrt(1/6)] + [n/3].

%F A190367: d(n) = 4*n + [n*sqrt(6)] + [n*sqrt(3/2)].

%p r:=sqrt(2): s:=1/r: t:=sqrt(3): u:=1/t: seq(n + floor(n*s/r) + floor(n*t/r) + floor(n*u/r), n=1..10^3); # _Muniru A Asiru_, Feb 01 2018

%t Table[n + Floor[n/2] + Floor[n*Sqrt[3/2]] + Floor[n*Sqrt[1/6]], {n,1,30}] (* _G. C. Greubel_, Jan 31 2018 *)

%o (PARI) for(n=1,30, print1(n + floor(n/2) + floor(n*sqrt(3/2)) + floor(n*sqrt(1/6)), ", ")) \\ _G. C. Greubel_, Jan 31 2018

%o (Magma) [n + Floor(n/2) + Floor(n*Sqrt(3/2)) + Floor(n*Sqrt(1/6)): n in [1..30]]; // _G. C. Greubel_, Jan 31 2018

%Y Cf. A190365, A190366, A190367

%K nonn

%O 1,1

%A _Clark Kimberling_, May 09 2011