A184808 n + floor(r*n), where r = sqrt(2/3); complement of A184809.

1, 3, 5, 7, 9, 10, 12, 14, 16, 18, 19, 21, 23, 25, 27, 29, 30, 32, 34, 36, 38, 39, 41, 43, 45, 47, 49, 50, 52, 54, 56, 58, 59, 61, 63, 65, 67, 69, 70, 72, 74, 76, 78, 79, 81, 83, 85, 87, 89, 90, 92, 94, 96, 98, 99, 101, 103, 105, 107, 108, 110, 112, 114, 116

Offset: 1

Author

Clark Kimberling, Jan 22 2011

Keywords

• nonn

Comments

This is the Beatty sequence for 1 + sqrt(2/3).

Also, a(n) is the position of 2*n^2 in the sequence obtained by arranging all the numbers in the sets {2*h^2, h >= 1} and {3*k^2, k >= 1} in increasing order. - Clark Kimberling, Oct 20 2014

Also, numbers n such that floor((n+1)*sqrt(6)) - floor(n*sqrt(6)) = 2. - Clark Kimberling, Jul 15 2015

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A184809, A182760 (comment about joint ranking),

A184810, A184811, A249096.

Programs

Magma

[n+Floor(n*Sqrt(2/3)): n in [1..70]]; // Vincenzo Librandi, Oct 23 2014

Mathematica

r=(2/3)^(1/2); s=(3/2)^(1/2);
a[n_]:=n+Floor [n*r];
b[n_]:=n+Floor [n*s];
Table[a[n],{n,1,120}]  (* A184808 *)
Table[b[n],{n,1,120}]  (* A184809 *)

PARI

main(size)={return(vector(size, n, n+floor(sqrt(2/3)*n)))} /* Anders Hellström, Jul 15 2015 */

Formula

a(n) = n + floor(r*n), where r = sqrt(2/3).