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%I #15 Jun 04 2017 08:04:38
%S 2,4,6,9,11,14,16,18,21,23,26,28,30,33,35,38,40,43,45,47,50,52,55,57,
%T 59,62,64,67,69,72,74,76,79,81,84,86,88,91,93,96,98,100,103,105,108,
%U 110,113,115,117,120,122,125,127,129,132,134,137,139,142,144,146
%N Positions of 1 in A285073; complement of A285074.
%C Conjecture: -1 + sqrt(2) < n*r - a(n) < sqrt(2) for n>=1, where r = 1 + sqrt(2).
%C From _Michel Dekking_, May 29 2017: (Start)
%C Proof of the conjecture: the sequence d:=A285073 can be written as d=01c, where c is the homogeneous Sturmian sequence with slope sqrt(2)-1 (see comments of A285073).
%C In general, a homogeneous Sturmian sequence (floor((n+1)r)-floor(nr)) gives the positions of 1 in the Beatty sequence b=(floor((n+1)s)), where s=1/r.
%C In our case s=1+sqrt(2). It follows that for n=0,1,… one has a(n+2) = floor((n+1)(1+sqrt(2)) + 2, which directly implies the conjecture. (End)
%H Clark Kimberling, <a href="/A285075/b285075.txt">Table of n, a(n) for n = 1..10000</a>
%e As a word, A285073 = 01010...,..., in which 1 is in positions 2,4,6,9,11,...
%t s = Nest[Flatten[# /. {0 -> {1, 0}, 1 -> {0, 1, 0}}] &, {0}, 14]; (* A285073 *)
%t Flatten[Position[s, 0]]; (* A285074 *)
%t Flatten[Position[s, 1]]; (* A285075 *)
%Y Cf. A285073, A285074, A285076.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Apr 19 2017
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