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%I #31 Sep 09 2019 01:08:02
%S 1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1,0,1,1,
%T 1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,0,
%U 1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,1
%N Fixed point of the morphism 0 -> 10, 1 -> 1011.
%C From __Michel Dekking_, Sep 08 2019: (Start)
%C A short proof of Mathar's conjecture can be obtained by using the fact that sigma and tau are conjugate morphisms. For all words w one has
%C tau(w) = 1^{-1} sigma(w) 1.
%C The proof is again by induction. Suppose that
%C (A) : 0 sigma^n(0) = tau^n(0) 0
%C holds. Then, choosing w = tau^n(0):
%C tau^{n+1}(0) 0 = tau(tau^n(0)) 0 =
%C 1^{-1} sigma(tau^n(0)) 1 0 =
%C 1^{-1} sigma(0 sigma^n(0) 0^{-1}) 10 =
%C 1^{-1} 10 sigma^{n+1}(0) 0^{-1}1^{-1} 10 =
%C 0 sigma^{n+1}.
%C So (A) holds for n+1.
%C (End)
%H Clark Kimberling, <a href="/A285341/b285341.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Fi#FIXEDPOINTS">Index entries for sequences that are fixed points of mappings</a>
%F Conjecture: a(n) = A284893(n+1). - _R. J. Mathar_, May 08 2017
%F From _Michel Dekking_, Feb 05 2018: (Start)
%F Proof of this conjecture: let sigma be the morphism 0 -> 10, 1 -> 1011, and let tau be the morphism 0 -> 01, 1 -> 0111, which has A284893 as a fixed point. It clearly suffices to prove the relation, for all n=1,2,3,...:
%F (A) : 0 sigma^n(0) = tau^n(0) 0
%F To prove such a thing one needs a second relation, for all n=1,2,3,...:
%F (B) : 0 sigma^n(1) 0^{-1} = tau^n(0) tau^n(1) [tau^n(0)]^{-1}
%F Here 0^{-1} and [tau^n(0)]^{-1} are the free group inverses of 0 and tau^n(0).
%F For n=1, we do indeed have:
%F (a) 0 sigma(10) = 0101110 = 0101110 = tau(01)0
%F (b) 0 sigma(1) 0^{-1} tau(0) = 010111 = tau(0)tau(1).
%F Using the induction hypothesis with (A) and (B) in the second line, one obtains
%F 0 sigma^{n+1}(0) = 0 sigma^n(1) 0^{-1} 0 sigma^n(0)
%F = tau^n(0) tau^n(1) [tau^n(0)]^{-1} tau^n(0) 0
%F = tau^{n+1}(0) 0.
%F Similarly,
%F 0 sigma^{n+1}(1) 0^{-1}
%F = 0 sigma^n(1) 0^{-1} 0 sigma^n(0) 0^{-1} 0 sigma^n(11) 0^{-1}
%F = tau^n(0)tau^n(1)[tau^n(0)]^{-1} tau^n(0) 0 0^{-1}
%F tau^n(0) tau^n(11) [tau^n(0)]^{-1}
%F = tau^{n+1}(0) tau^{n+1}(1) [tau^n(1)]^{-1}[tau^n(0)]^{-1}
%F = tau^{n+1}(0) tau^{n+1}(1) [tau^{n+1}(0)]^{-1}. (End)
%e 0 -> 10-> 1011 -> 10111010111011.
%t s = Nest[Flatten[# /. {0 -> {1, 0}, 1 -> {1, 0, 1, 1}}] &, {0}, 10]; (* A285341 *)
%t u = Flatten[Position[s, 0]]; (* A285342 *)
%t Flatten[Position[s, 1]]; (* A285343 *)
%t u/2 (* A285344)
%Y Cf. A285342, A285343, A285344.
%K nonn,easy
%O 1
%A _Clark Kimberling_, Apr 25 2017
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