A190332 n + [n*s/r] + [n*t/r]; r=1, s=sqrt(3), t=1/s.

2, 6, 9, 12, 15, 19, 23, 25, 29, 32, 36, 38, 42, 46, 48, 52, 55, 59, 61, 65, 69, 72, 75, 78, 82, 86, 88, 92, 95, 98, 101, 105, 109, 111, 115, 118, 122, 124, 128, 132, 135, 138, 141, 145, 147, 151, 155, 158, 161, 164, 168, 172, 174, 178, 181, 184, 187, 191, 195, 197, 201, 204, 208, 210, 214, 218, 221, 224, 227, 231, 233, 237, 241, 244

Offset

1,1

Comments

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

f(n) = n + [n*s/r] + [n*t/r],

g(n) = n + [n*r/s] + [n*t/s],

h(n) = n + [n*r/t] + [n*s/t], where []=floor.

Taking r=1, s=sqrt(3), t=1/s gives

f=A190332, g=A190333, h=A190334.

Links

G. C. Greubel, Table of n, a(n) for n = 1..10000

Formula

A190332: f(n) = n + [n*sqrt(3)] + [n/sqrt(3)].

A190333: g(n) = n + [n/sqrt(3)] + [n/3].

A190334: h(n) = 4*n + [n*sqrt(3)].

Maple

r:=1: s:=sqrt(3): t:=1/s: seq(n+floor(n*s/r)+floor(n*t/r), n=1..80); # Muniru A Asiru, Apr 05 2018

Mathematica

r=1; s=3^(1/2); t=1/s;
f[n_] := n + Floor[n*s/r] + Floor[n*t/r];
g[n_] := n + Floor[n*r/s] + Floor[n*t/s];
h[n_] := n + Floor[n*r/t] + Floor[n*s/t];
Table[f[n], {n, 1, 120}]  (*A190332*)
Table[g[n], {n, 1, 120}]  (*A190333*)
Table[h[n], {n, 1, 120}]  (*A190334*)

Prog

(PARI) for(n=1, 100, print1(n + floor(n*sqrt(3)) + floor(n/sqrt(3)), ", ")) \\ G. C. Greubel, Apr 04 2018
(Magma) R:=RealField(); [n + Floor(n*Sqrt(3)) + Floor(n/Sqrt(3)): n in [1..100]]; // G. C. Greubel, Apr 04 2018

Crossrefs

Cf. A190333, A190334, A190329.

Sequence in context: A206813 A189371 A190059 * A187912 A186500 A190777

Adjacent sequences: A190329 A190330 A190331 * A190333 A190334 A190335

Keyword

nonn

Author

Clark Kimberling, May 08 2011

Status

approved