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 A285341 Fixed point of the morphism 0 -> 10, 1 -> 1011. 5
 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, 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, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1 COMMENTS From _Michel Dekking, Sep 08 2019: (Start) 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 tau(w) = 1^{-1} sigma(w) 1. The proof is again by induction. Suppose that (A) : 0 sigma^n(0) = tau^n(0) 0 holds. Then, choosing w = tau^n(0): tau^{n+1}(0) 0 = tau(tau^n(0)) 0 = 1^{-1} sigma(tau^n(0)) 1 0 = 1^{-1} sigma(0 sigma^n(0) 0^{-1}) 10 = 1^{-1} 10 sigma^{n+1}(0) 0^{-1}1^{-1} 10 = 0 sigma^{n+1}. So (A) holds for n+1. (End) LINKS Clark Kimberling, Table of n, a(n) for n = 1..10000 Index entries for sequences that are fixed points of mappings FORMULA Conjecture: a(n) = A284893(n+1). - R. J. Mathar, May 08 2017 From Michel Dekking, Feb 05 2018: (Start) 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,...: (A) : 0 sigma^n(0) = tau^n(0) 0 To prove such a thing one needs a second relation, for all n=1,2,3,...: (B) : 0 sigma^n(1) 0^{-1} = tau^n(0) tau^n(1) [tau^n(0)]^{-1} Here 0^{-1} and [tau^n(0)]^{-1} are the free group inverses of 0 and tau^n(0). For n=1, we do indeed have: (a) 0 sigma(10) = 0101110 = 0101110 = tau(01)0 (b) 0 sigma(1) 0^{-1} tau(0) = 010111 = tau(0)tau(1). Using the induction hypothesis with (A) and (B) in the second line, one obtains 0 sigma^{n+1}(0) = 0 sigma^n(1) 0^{-1} 0 sigma^n(0) = tau^n(0) tau^n(1) [tau^n(0)]^{-1} tau^n(0) 0 = tau^{n+1}(0) 0. Similarly, 0 sigma^{n+1}(1) 0^{-1} = 0 sigma^n(1) 0^{-1} 0 sigma^n(0) 0^{-1} 0 sigma^n(11) 0^{-1} = tau^n(0)tau^n(1)[tau^n(0)]^{-1} tau^n(0) 0 0^{-1} tau^n(0) tau^n(11) [tau^n(0)]^{-1} = tau^{n+1}(0) tau^{n+1}(1) [tau^n(1)]^{-1}[tau^n(0)]^{-1} = tau^{n+1}(0) tau^{n+1}(1) [tau^{n+1}(0)]^{-1}. (End) EXAMPLE 0 -> 10-> 1011 -> 10111010111011. MATHEMATICA s = Nest[Flatten[# /. {0 -> {1, 0}, 1 -> {1, 0, 1, 1}}] &, {0}, 10]; (* A285341 *) u = Flatten[Position[s, 0]]; (* A285342 *) Flatten[Position[s, 1]]; (* A285343 *) u/2 (* A285344) CROSSREFS Cf. A285342, A285343, A285344. Sequence in context: A089045 A259599 A070749 * A059778 A355690 A359590 Adjacent sequences: A285338 A285339 A285340 * A285342 A285343 A285344 KEYWORD nonn,easy AUTHOR Clark Kimberling, Apr 25 2017 STATUS approved

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Last modified July 21 11:45 EDT 2024. Contains 374472 sequences. (Running on oeis4.)