A369989 - OEIS

A369989

Infinite word formed by expanding terms of A372257 by the map h : 0 -> {1,0,1}, 1 -> {1,0,0,1}, 2 -> {1,0,0,0,1}, 3 -> {1,0,0,0,0,1}.

1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0

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OFFSET

COMMENTS

This is word "y" of Berstel et al.

They construct this sequence (and A372258 "z") as an example infinite word of 2 symbols which is uniformly recurrent, has subwords closed under reversal, but only has a finite set of palindrome subwords.

There are 24 palindrome subwords and they can be found using Berstel et al's proof that it's enough to look in all h(T) where T is a length 3 subword occurring in A372257.

Blocks of 18 terms a(18*i,...,18*i+17) can be calculated from blocks of 4 terms A372257(4*i,...,4*i+3) since the latter are 0,1 or 1,0 then 2,3 or 3,2 (by the construction there), and so always expand to 18 terms under h.

This is a morphic sequence, meaning a symbol to symbol mapping of the fixed point of a morphism (for instance a uniform morphism based on the bits of n and how they result in remainder n mod 18 and bits of the corresponding quotient).

LINKS

Kevin Ryde, Table of n, a(n) for n = 0..10000

Jean Berstel, Luc Boasson, Olivier Carton, and Isabelle Fagnot, Infinite words without palindrome, arXiv:0903.2382 [cs.DM], 2009, section 3 word "y".

Kevin Ryde, PARI/GP Code

FORMULA

With quotient q = floor(n/9) and remainder r = n mod 9:

a(n) = 1 - A014707(q) if r = 2;

a(n) = A014707(q) if r = 4;

a(n) = 1 if q even and r in {0, 3, 6, 7}, or q odd and r in {3, 8};

a(n) = 0 otherwise.

EXAMPLE

Expansion of the terms of A372257 by h begins: