A190347 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sqrt(5/3), t=sqrt(3/5).

2, 5, 8, 12, 14, 17, 21, 24, 26, 29, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 64, 67, 69, 72, 76, 79, 81, 85, 88, 91, 95, 97, 100, 103, 107, 109, 112, 116, 119, 121, 124, 128, 131, 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

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(5/3), t=sqrt(3/5) gives f=A190347, g=A190348, h=A190349.

Links

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

Formula

A190347: f(n) = n + [n*sqrt(5/3)] + [n*sqrt(3/5)].

A190348: g(n) = n + [n*sqrt(3/5)] + [3*n/5].

A190349: h(n) = 2*n + [n*sqrt(5/3)] + [n/3].

Mathematica

r=1; s=(5/3)^(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}]  (* A190347 *)
Table[g[n], {n, 1, 120}]  (* A190348 *)
Table[h[n], {n, 1, 120}]  (* A190349 *)

Prog

(PARI) for(n=1, 100, print1(n + floor(n*sqrt(5/3)) + floor(n*sqrt(3/5)), ", ")) \\ G. C. Greubel, Apr 05 2018
(Magma) [n + Floor(n*Sqrt(5/3)) + Floor(n*Sqrt(3/5)): n in [1..100]]; // G. C. Greubel, Apr 05 2018

Crossrefs

Cf. A190348, A190349.

Sequence in context: A073837 A357505 A189531 * A193767 A209295 A184813

Adjacent sequences: A190344 A190345 A190346 * A190348 A190349 A190350

Keyword

nonn

Author

Clark Kimberling, May 09 2011

Status

approved