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%I #17 Feb 15 2021 01:58:27
%S 1,2,4,5,7,8,10,11,17,18,20,21,23,24,26,27,29,30,36,37,39,40,42,43,45,
%T 46,48,49,55,56,58,59,61,62,64,65,67,68,74,75,81,82,88,89,95,96,102,
%U 103,105,106,108,109,111,112,114,115,121,122,124,125,127,128
%N Positions of 0 in A285177; complement of A285402.
%C Conjecture: a(n)/n -> (61-sqrt(3))/26 = 2.279...
%C From _Michel Dekking_, Feb 10 2021: (Start)
%C This conjecture is false. In fact,
%C a(n)/n --> (5+sqrt(17))/4 = 2.28077...
%C Let mu be the defining morphism for A285177, i.e,
%C mu(0) = 11, mu(1) = 001.
%C The sequence A285177 is the fixed point x = 0010010010011111... starting with 0 of mu^2:
%C mu^2(0) = 001001, mu^2(1) = 1111001.
%C The 0's in x are at positions a(1)=1, a(2)=2, a(3)=4, etc.
%C Now suppose that N_0(K) = n is the number of 0's in a prefix x[1,K] of length K of x. Then obviously a(n) = K +/- 6.
%C Also N_0(K) + N_1(K) = K, where N_1(K) is the number of 1's in x[1,K].
%C So
%C K/N_0(K) = a(n)/n +/- 6/n.
%C Letting n tend to infinity, we find that
%C a(n)/n --> 1/f0,
%C where f0 is the frequency of 0's in x.
%C It is well known that these exist and are equal to the normalized eigenvector of the Perron-Frobenius eigenvalue of the incidence matrix of the morphism mu.
%C A simple computation yields that f0 = 4/(5+sqrt(17)).
%C It follows that a(n)/n --> (5+sqrt(17))/4.
%C (End)
%H Clark Kimberling, <a href="/A285401/b285401.txt">Table of n, a(n) for n = 1..10000</a>
%e As a word, A285177 = 001001..., in which 0 is in positions 1,2,4,5,7,...
%t s = Nest[Flatten[# /. {0 -> {1, 1}, 1 -> {0, 0, 1}}] &, {0}, 10] (* A285117 *)
%t Flatten[Position[s, 0]] (* A285401 *)
%t Flatten[Position[s, 1]] (* A285402 *)
%Y Cf. A285177, A285402, A285403.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Apr 26 2017
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