A184808 - OEIS

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

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OFFSET

1,2

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

FORMULA

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

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 *)

PROG