A234322 Percentage of 1's among the first n terms of the Kolakoski sequence A000002.

100, 50, 33, 50, 60, 50, 57, 50, 44, 50, 45, 42, 46, 50, 47, 50, 53, 50, 47, 50, 48, 50, 52, 50, 52, 50, 48, 50, 52, 50, 52, 53, 52, 53, 51, 50, 51, 50, 49, 50, 51, 50, 51, 50, 49, 50, 49, 50, 51, 50, 51, 52, 51, 50, 51, 50, 49, 50, 51, 50, 51, 50, 49, 50, 49, 48, 49, 50, 49, 50, 51, 50

Offset: 1

Author

Jonathan Sondow, Jan 03 2014

Keywords

• nonn

Comments

Percentage of 1's (rounded) among the first n terms of A000002.

See A000002 for additional comments, references, links, formulas, examples, programs, and crossrefs.

It is conjectured that a(n) = 50 for all n sufficiently large.

Among the first 1500000 terms of the sequence, a(n) = 51 appears last at a(295) and a(n) = 49 appears last at a(471). - Paolo Bonzini, Jul 06 2016

Examples

			There are five 1's among the first twelve terms of A000002 = 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, ..., so a(12) = nearest integer to 100*5/12 = 41.666... rounded = 42.
		

Links

• Paolo Bonzini, Table of n, a(n) for n = 1..500
• W. Kolakoski and N. Ucoluk, Problem 5304: Self Generating Runs, Amer. Math. Monthly, 72 (1965), 674; 73 (1966), 681-682.
• J. Malkevitch, Periods, AMS Feature Column, Jan 2014.
• R. Oldenburger, Exponent trajectories in symbolic dynamics, Trans. Amer. Math. Soc., 46 (1939), 453-466.

Crossrefs

Cf. A000002, A156077.

Formula

a(n) = nearest integer to 100*A156077(n)/n.