OFFSET
1
COMMENTS
From Michel Dekking, Jan 22 2018: (Start)
Proof of Mathar's conjecture of May 08 2017:
Let sigma be the morphism
sigma: 0 -> 10, 1 -> 1110,
and let tau be the morphism
tau: 0 -> 01, 1 -> 1101,
which has A284939 as a fixed point.
It clearly suffices to prove the relation
(A) : 0 sigma^n(0) = tau^n(0) 0 for all n=1,2,...
To prove such a thing one needs a second relation, for example,
(B) : sigma^n(10) = sigma^n(0) 0^{-1} tau^n(1) 0 for all n=1,2,...
Here 0^{-1} is the free group inverse of 0.
Note that (A) and (B) together imply
(C) : sigma^n(10) = 0^{-1} tau^n(01) 0 for all n=1,2,...
But, since sigma(0)=10, and tau(0)=01, relation (C) is equal to relation (A), with n replaced by n+1.
It is therefore enough to prove (B) by induction.
Well, (A), (B) and (C) are easily checked for n=1.
Furthermore, using the induction hypothesis with (B) and (C) in the first line, and again (C) in the third line, one obtains
sigma^{n+1}(10)
= sigma^n(11)sigma^n(01)sigma^n(01)
= sigma^n(1)sigma^n(1) sigma^n(0) 0^{-1} tau^n(1) 0 0^{-1} tau^n(01) 0
= sigma^n(1)sigma^n(10) 0^{-1} tau^n(101) 0
= sigma^n(1) sigma^n(0) 0^{-1} tau^n(1) 0 0^{-1} tau^n(101) 0
= sigma^n(10) 0^{-1} tau^n(1101) 0
= sigma^{n+1}(0) 0^{-1} tau^{n+1}(1) 0. (End)
LINKS
FORMULA
Conjecture: a(n) = A284939(n+1). - R. J. Mathar, May 08 2017
EXAMPLE
0 -> 10-> 1110 -> 11101110111010 ->
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 25 2017
STATUS
approved