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%I #20 Nov 22 2017 01:21:56
%S 100,50,33,50,60,50,57,50,44,50,45,42,46,50,47,50,53,50,47,50,48,50,
%T 52,50,52,50,48,50,52,50,52,53,52,53,51,50,51,50,49,50,51,50,51,50,49,
%U 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
%N Percentage of 1's among the first n terms of the Kolakoski sequence A000002.
%C Percentage of 1's (rounded) among the first n terms of A000002.
%C See A000002 for additional comments, references, links, formulas, examples, programs, and crossrefs.
%C It is conjectured that a(n) = 50 for all n sufficiently large.
%C 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
%H Paolo Bonzini, <a href="/A234322/b234322.txt">Table of n, a(n) for n = 1..500</a>
%H W. Kolakoski and N. Ucoluk, <a href="http://www.jstor.org/stable/2314839">Problem 5304: Self Generating Runs</a>, Amer. Math. Monthly, 72 (1965), 674; 73 (1966), 681-682.
%H J. Malkevitch, <a href="http://www.ams.org/samplings/feature-column/fc-2014-01">Periods</a>, AMS Feature Column, Jan 2014.
%H R. Oldenburger, <a href="http://dx.doi.org/10.1090/S0002-9947-1939-0000352-9">Exponent trajectories in symbolic dynamics</a>, Trans. Amer. Math. Soc., 46 (1939), 453-466.
%F a(n) = nearest integer to 100*A156077(n)/n.
%e 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.
%Y Cf. A000002, A156077.
%K nonn
%O 1,1
%A _Jonathan Sondow_, Jan 03 2014
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