%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