A080607 Golomb's sequence using multiples of 3.

3, 3, 3, 6, 6, 6, 9, 9, 9, 12, 12, 12, 12, 12, 12, 15, 15, 15, 15, 15, 15, 18, 18, 18, 18, 18, 18, 21, 21, 21, 21, 21, 21, 21, 21, 21, 24, 24, 24, 24, 24, 24, 24, 24, 24, 27, 27, 27, 27, 27, 27, 27, 27, 27, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 33, 33, 33, 33, 33, 33

Offset: 1

Author

Benoit Cloitre, Feb 25 2003

Keywords

• nonn

Comments

More generally let b(k) be a sequence of integers in arithmetic progression: b(k) = A*k+B, then the Golomb's sequence a(n) using b(k) is asymptotic to tau^(2-tau)*(A*n)^(tau-1).

Examples

			Read 3,3,3,6,6,6,9,9,9,12,12,12,12,12,12,15 as (3,3,3),(6,6,6),(9,9,9),(12,12,12,12,12,12),... count occurrences between 2 parentheses, gives 3,3,3,6,... which is the sequence itself.
		

Links

• Ivan Neretin, Table of n, a(n) for n = 1..10062

Crossrefs

Cf. A001462, A080606, A080605.

Programs

Mathematica

a = {3, 3, 3}; Do[a = Join[a, Array[3i&, a[[i]]]], {i, 2, 11}]; a (* Ivan Neretin, Apr 03 2015 *)

Formula

a(n) is asymptotic to tau^(2-tau)*(3n)^(tau-1) and more precisely it seems that a(n) = round(tau^(2-tau)*(3n)^(tau-1)) +(-2, -1, +0, +1 or +1) where tau is the golden ratio.