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Fixed point of the morphism 0 -> 10, 1 -> 1110.
5

%I #16 Jan 25 2018 03:02:39

%S 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,

%T 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,

%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 -> 1110.

%C From _Michel Dekking_, Jan 22 2018: (Start)

%C Proof of Mathar's conjecture of May 08 2017:

%C Let sigma be the morphism

%C sigma: 0 -> 10, 1 -> 1110,

%C and let tau be the morphism

%C tau: 0 -> 01, 1 -> 1101,

%C which has A284939 as a fixed point.

%C It clearly suffices to prove the relation

%C (A) : 0 sigma^n(0) = tau^n(0) 0 for all n=1,2,...

%C To prove such a thing one needs a second relation, for example,

%C (B) : sigma^n(10) = sigma^n(0) 0^{-1} tau^n(1) 0 for all n=1,2,...

%C Here 0^{-1} is the free group inverse of 0.

%C Note that (A) and (B) together imply

%C (C) : sigma^n(10) = 0^{-1} tau^n(01) 0 for all n=1,2,...

%C But, since sigma(0)=10, and tau(0)=01, relation (C) is equal to relation (A), with n replaced by n+1.

%C It is therefore enough to prove (B) by induction.

%C Well, (A), (B) and (C) are easily checked for n=1.

%C Furthermore, using the induction hypothesis with (B) and (C) in the first line, and again (C) in the third line, one obtains

%C sigma^{n+1}(10)

%C = sigma^n(11)sigma^n(01)sigma^n(01)

%C = sigma^n(1)sigma^n(1) sigma^n(0) 0^{-1} tau^n(1) 0 0^{-1} tau^n(01) 0

%C = sigma^n(1)sigma^n(10) 0^{-1} tau^n(101) 0

%C = sigma^n(1) sigma^n(0) 0^{-1} tau^n(1) 0 0^{-1} tau^n(101) 0

%C = sigma^n(10) 0^{-1} tau^n(1101) 0

%C = sigma^{n+1}(0) 0^{-1} tau^{n+1}(1) 0. (End)

%H Clark Kimberling, <a href="/A285373/b285373.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) = A284939(n+1). - _R. J. Mathar_, May 08 2017

%e 0 -> 10-> 1110 -> 11101110111010 ->

%t s = Nest[Flatten[# /. {0 -> {1, 0}, 1 -> {1, 1, 1, 0}}] &, {0}, 10] (* A285373 *)

%t Flatten[Position[s, 0]] (* A285374 *)

%t Flatten[Position[s, 1]] (* A285375 *)

%Y Cf. A284374, A285375.

%K nonn,easy

%O 1

%A _Clark Kimberling_, Apr 25 2017