%I #24 Jun 26 2022 22:46:52
%S 1,0,0,1,1,0,1,0,1,0,0,1,1,0,0,1,1,0,1,0,0,1,1,0,1,0,0,1,1,0,1,0,1,0,
%T 0,1,1,0,0,1,1,0,1,0,1,0,0,1,1,0,0,1,1,0,1,0,0,1,1,0,1,0,1,0,0,1,1,0,
%U 0,1,1,0,1,0,0,1,1,0,1,0,1,0,0,1,1,0
%N Fixed point of the morphism 0 -> 10, 1 -> 1001.
%C This is a 3-automatic sequence. See Allouche et al. link. - _Michel Dekking_, Oct 05 2020
%H Clark Kimberling, <a href="/A285305/b285305.txt">Table of n, a(n) for n = 1..10000</a>
%H J.-P. Allouche, F. M. Dekking, and M. Queffélec, <a href="https://arxiv.org/abs/2010.00920">Hidden automatic sequences</a>, arXiv:2010.00920 [math.NT], 2020.
%H <a href="/index/Fi#FIXEDPOINTS">Index entries for sequences that are fixed points of mappings</a>
%F Conjecture: a(n) = A284775(n+1). - _R. J. Mathar_, May 08 2017
%F From _Michel Dekking_, Feb 20 2021: (Start)
%F Proof of this conjecture. Let mu: 0 -> 10, 1 -> 1001 be the defining morphism of (a(n)), and let tau: 0 -> 01, 1 -> 0011 be the defining morphism of A284775.
%F Since mu^n(0) starts with 1 for n>1, mu^n(0) tends to (a(n)) as n tends to infinity. So the conjecture will follow directly from the following claim.
%F CLAIM: 0 mu^n(0) = tau^n(0) 0 for n>0.
%F Proof: By induction over two levels, exploiting the obvious equality tau(1) = 0 tau(0) 1 to go from the third to the fourth line below.
%F For n=1: 0 mu(0)= 010 = tau(0) 0.
%F For n=2: 0 mu^2(0)= 0100110 = tau^2(0) 0.
%F Suppose true for n-1 and n. Then
%F tau^{n+1}(0) =
%F tau^n(tau(0)) =
%F tau^n(0) tau^n(1) =
%F tau^n(0) tau^{n-1)(0) tau^n(0) tau^{n-1}(1) =
%F 0 mu^n(0)0^{-1} 0 mu^{n-1}(0)0^{-1}0mu^n(0)0^{-1}tau^{n-1)(1)=
%F 0 mu^{n-1}(mu(0))mu^{n-1}(0)mu^{n-1}(mu(0))0^{-1}0tau^{n-)(1)=
%F 0 mu^{n-1}(10) mu^{n-1}(0) mu^{n-1}(10) 0^{-1} tau^{n-1)(1) =
%F 0 mu^{n-1}(1001) mu^{n-1}(0) 0^{-1} tau^{n-1)(1) =
%F 0 mu^n(1) 0^{-1} tau^{n-1)(0) 0 0^{-1} tau^{n-1)(1) =
%F 0 mu^n(1) 0^{-1} tau^{n-1}(01) =
%F 0 mu^n(1) 0^{-1} tau^n(0) =
%F 0 mu^n(1) mu^n(0) 0^{-1} =
%F 0 mu^n(mu(0)) 0^{-1} =
%F 0 mu^{n+1}(0) 0^{-1}.
%F So we proved 0 mu^{n+1}(0) = tau^{n+1}(0) 0.
%F (End)
%e 0 -> 10-> 1001 -> 100110101001 ->
%t s = Nest[Flatten[# /. {0 -> {1, 0}, 1 -> {1, 0, 0, 1}}] &, {0}, 10]; (* A285305 *)
%t u = Flatten[Position[s, 0]]; (* A285306 *)
%t v = Flatten[Position[s, 1]]; (* A285307 *)
%Y Cf. A284306, A285307, A284775.
%Y Partial sums of A284793.
%K nonn,easy
%O 1
%A _Clark Kimberling_, Apr 25 2017