A285341 Fixed point of the morphism 0 -> 10, 1 -> 1011.

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

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