For what $n$ is $W_n$ finite?












6












$begingroup$


Suppose, $W_n$ is the set of all words formed by letters '$a$' and '$b$', that do not contain $n$ same consecutive nonempty subwords (that means that for any nonempty word $u$, the word $u^n$ is not a subword of words from $W_n$) For example "$bababab$" is not in $W_3$, as it contains three consecutive "$ba$" subwords, but it obviously is in $W_4$.
For what $n$ is $W_n$ finite?



It is easy to see, that $W_n subset W_{n+1}$ and thus the sequence of cardinals ${|W_n|}_{n=1}^{infty}$ is monotonously non-decreasing.
Thus either $W_n$ is finite for all $n$, or it is infinite for all $n$, or there exists $n_0$, such that $W_n$ is finite for all $n < n_0$ and infinite for all $n geq n_0$.



One can also see, that $W_2 = {a, b, ab, ba, aba, bab}$ is finite. One can prove that just by looking at all 16 words of length 4 and seeing that none of them lies in $W_2$.



However, $W_{665}$ is already infinite. Suppose $G$ is an infinite $2$-generated group of exponent $665$ (such group exists according to Adyan-Novikov theorem). Then any element of it can be expressed as a multiple product of those two generators (which can be written as a word formed by letters '$a$' and '$b$' (that denote the first and the second generator respectively). Due to the group having exponent $665$, any such word can be "reduced" to a word from $W_{665}$. One can see that two elements that can be written as the same word are equal. And in $G$ there is infinitely many pairwise not equal elements. Thus $W_{665}$ is infinite by pigeonhole principle.



So we can say that there exists such $n_0$, that $W_n$ is finite for all $n < n_0$ and infinite for all $n geq n_0$. And that aforementioned $n_0$ satisfies the inequality $2 < n_0 leq 665$. However I failed to determine anything else about that number.



Any help will be appreciated.






combinatorics group-theory dynamical-systems formal-languages combinatorics-on-words







share|cite|improve this question











$endgroup$












  • $begingroup$
    The main subject of the question is symbolic dynamics.
    $endgroup$
    – YCor
    Dec 13 '18 at 22:59
















6












$begingroup$


Suppose, $W_n$ is the set of all words formed by letters '$a$' and '$b$', that do not contain $n$ same consecutive nonempty subwords (that means that for any nonempty word $u$, the word $u^n$ is not a subword of words from $W_n$) For example "$bababab$" is not in $W_3$, as it contains three consecutive "$ba$" subwords, but it obviously is in $W_4$.
For what $n$ is $W_n$ finite?



It is easy to see, that $W_n subset W_{n+1}$ and thus the sequence of cardinals ${|W_n|}_{n=1}^{infty}$ is monotonously non-decreasing.
Thus either $W_n$ is finite for all $n$, or it is infinite for all $n$, or there exists $n_0$, such that $W_n$ is finite for all $n < n_0$ and infinite for all $n geq n_0$.



One can also see, that $W_2 = {a, b, ab, ba, aba, bab}$ is finite. One can prove that just by looking at all 16 words of length 4 and seeing that none of them lies in $W_2$.



However, $W_{665}$ is already infinite. Suppose $G$ is an infinite $2$-generated group of exponent $665$ (such group exists according to Adyan-Novikov theorem). Then any element of it can be expressed as a multiple product of those two generators (which can be written as a word formed by letters '$a$' and '$b$' (that denote the first and the second generator respectively). Due to the group having exponent $665$, any such word can be "reduced" to a word from $W_{665}$. One can see that two elements that can be written as the same word are equal. And in $G$ there is infinitely many pairwise not equal elements. Thus $W_{665}$ is infinite by pigeonhole principle.



So we can say that there exists such $n_0$, that $W_n$ is finite for all $n < n_0$ and infinite for all $n geq n_0$. And that aforementioned $n_0$ satisfies the inequality $2 < n_0 leq 665$. However I failed to determine anything else about that number.



Any help will be appreciated.






combinatorics group-theory dynamical-systems formal-languages combinatorics-on-words







share|cite|improve this question











$endgroup$












  • $begingroup$
    The main subject of the question is symbolic dynamics.
    $endgroup$
    – YCor
    Dec 13 '18 at 22:59














6












6








6


0



$begingroup$


Suppose, $W_n$ is the set of all words formed by letters '$a$' and '$b$', that do not contain $n$ same consecutive nonempty subwords (that means that for any nonempty word $u$, the word $u^n$ is not a subword of words from $W_n$) For example "$bababab$" is not in $W_3$, as it contains three consecutive "$ba$" subwords, but it obviously is in $W_4$.
For what $n$ is $W_n$ finite?



It is easy to see, that $W_n subset W_{n+1}$ and thus the sequence of cardinals ${|W_n|}_{n=1}^{infty}$ is monotonously non-decreasing.
Thus either $W_n$ is finite for all $n$, or it is infinite for all $n$, or there exists $n_0$, such that $W_n$ is finite for all $n < n_0$ and infinite for all $n geq n_0$.



One can also see, that $W_2 = {a, b, ab, ba, aba, bab}$ is finite. One can prove that just by looking at all 16 words of length 4 and seeing that none of them lies in $W_2$.



However, $W_{665}$ is already infinite. Suppose $G$ is an infinite $2$-generated group of exponent $665$ (such group exists according to Adyan-Novikov theorem). Then any element of it can be expressed as a multiple product of those two generators (which can be written as a word formed by letters '$a$' and '$b$' (that denote the first and the second generator respectively). Due to the group having exponent $665$, any such word can be "reduced" to a word from $W_{665}$. One can see that two elements that can be written as the same word are equal. And in $G$ there is infinitely many pairwise not equal elements. Thus $W_{665}$ is infinite by pigeonhole principle.



So we can say that there exists such $n_0$, that $W_n$ is finite for all $n < n_0$ and infinite for all $n geq n_0$. And that aforementioned $n_0$ satisfies the inequality $2 < n_0 leq 665$. However I failed to determine anything else about that number.



Any help will be appreciated.






combinatorics group-theory dynamical-systems formal-languages combinatorics-on-words







share|cite|improve this question











$endgroup$




Suppose, $W_n$ is the set of all words formed by letters '$a$' and '$b$', that do not contain $n$ same consecutive nonempty subwords (that means that for any nonempty word $u$, the word $u^n$ is not a subword of words from $W_n$) For example "$bababab$" is not in $W_3$, as it contains three consecutive "$ba$" subwords, but it obviously is in $W_4$.
For what $n$ is $W_n$ finite?



It is easy to see, that $W_n subset W_{n+1}$ and thus the sequence of cardinals ${|W_n|}_{n=1}^{infty}$ is monotonously non-decreasing.
Thus either $W_n$ is finite for all $n$, or it is infinite for all $n$, or there exists $n_0$, such that $W_n$ is finite for all $n < n_0$ and infinite for all $n geq n_0$.



One can also see, that $W_2 = {a, b, ab, ba, aba, bab}$ is finite. One can prove that just by looking at all 16 words of length 4 and seeing that none of them lies in $W_2$.



However, $W_{665}$ is already infinite. Suppose $G$ is an infinite $2$-generated group of exponent $665$ (such group exists according to Adyan-Novikov theorem). Then any element of it can be expressed as a multiple product of those two generators (which can be written as a word formed by letters '$a$' and '$b$' (that denote the first and the second generator respectively). Due to the group having exponent $665$, any such word can be "reduced" to a word from $W_{665}$. One can see that two elements that can be written as the same word are equal. And in $G$ there is infinitely many pairwise not equal elements. Thus $W_{665}$ is infinite by pigeonhole principle.



So we can say that there exists such $n_0$, that $W_n$ is finite for all $n < n_0$ and infinite for all $n geq n_0$. And that aforementioned $n_0$ satisfies the inequality $2 < n_0 leq 665$. However I failed to determine anything else about that number.



Any help will be appreciated.






combinatorics group-theory dynamical-systems formal-languages combinatorics-on-words




combinatorics group-theory dynamical-systems formal-languages combinatorics-on-words






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 22:58









YCor

7,788929




7,788929










asked Dec 13 '18 at 8:39









Yanior WegYanior Weg

2,34711144




2,34711144












  • $begingroup$
    The main subject of the question is symbolic dynamics.
    $endgroup$
    – YCor
    Dec 13 '18 at 22:59


















  • $begingroup$
    The main subject of the question is symbolic dynamics.
    $endgroup$
    – YCor
    Dec 13 '18 at 22:59
















$begingroup$
The main subject of the question is symbolic dynamics.
$endgroup$
– YCor
Dec 13 '18 at 22:59




$begingroup$
The main subject of the question is symbolic dynamics.
$endgroup$
– YCor
Dec 13 '18 at 22:59










1 Answer
1






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5












$begingroup$

$W_n$ is infinite for all $n geq 3$.



There is an infinite binary word called the Thue–Morse sequence which is cube-free (among other properties). It can be constructed recursively as follows:




  • $w_1 = a$


  • $w_{n+1} = w_n overline{w_n}$ where for word $w in { a,b }^*$ by $overline{w}$ we denote the "Boolean complement" of $w$ (e.g., $overline{a} = b$, $overline{ab} = ba$, $overline{aabaa} = bbabb$).


The Thue–Morse sequence is the natural limit of these finite words.






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    $begingroup$

    $W_n$ is infinite for all $n geq 3$.



    There is an infinite binary word called the Thue–Morse sequence which is cube-free (among other properties). It can be constructed recursively as follows:




    • $w_1 = a$


    • $w_{n+1} = w_n overline{w_n}$ where for word $w in { a,b }^*$ by $overline{w}$ we denote the "Boolean complement" of $w$ (e.g., $overline{a} = b$, $overline{ab} = ba$, $overline{aabaa} = bbabb$).


    The Thue–Morse sequence is the natural limit of these finite words.






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      $W_n$ is infinite for all $n geq 3$.



      There is an infinite binary word called the Thue–Morse sequence which is cube-free (among other properties). It can be constructed recursively as follows:




      • $w_1 = a$


      • $w_{n+1} = w_n overline{w_n}$ where for word $w in { a,b }^*$ by $overline{w}$ we denote the "Boolean complement" of $w$ (e.g., $overline{a} = b$, $overline{ab} = ba$, $overline{aabaa} = bbabb$).


      The Thue–Morse sequence is the natural limit of these finite words.






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        $W_n$ is infinite for all $n geq 3$.



        There is an infinite binary word called the Thue–Morse sequence which is cube-free (among other properties). It can be constructed recursively as follows:




        • $w_1 = a$


        • $w_{n+1} = w_n overline{w_n}$ where for word $w in { a,b }^*$ by $overline{w}$ we denote the "Boolean complement" of $w$ (e.g., $overline{a} = b$, $overline{ab} = ba$, $overline{aabaa} = bbabb$).


        The Thue–Morse sequence is the natural limit of these finite words.






        share|cite|improve this answer









        $endgroup$



        $W_n$ is infinite for all $n geq 3$.



        There is an infinite binary word called the Thue–Morse sequence which is cube-free (among other properties). It can be constructed recursively as follows:




        • $w_1 = a$


        • $w_{n+1} = w_n overline{w_n}$ where for word $w in { a,b }^*$ by $overline{w}$ we denote the "Boolean complement" of $w$ (e.g., $overline{a} = b$, $overline{ab} = ba$, $overline{aabaa} = bbabb$).


        The Thue–Morse sequence is the natural limit of these finite words.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 9:09









        Meta-мета-μετα-meta-мета-μεταMeta-мета-μετα-meta-мета-μετα

        4,705927




        4,705927






























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