How many letters suffice to construct words with no repetition?Bits and orbitsExtension of Tao-Green TheoremA property of unimodal sequencesPartitioning finite ordered setsRepresentability of sets of infinite sequences sharing common prefixes and factors (i.e. infixes)“Nyldon words”: understanding a class of words factorizing the free monoid increasinglyProbability of no $k$ 1's in arithmetic progression in binary sequence of length $n$Reference for one-sided subshiftsSequences with 3 LettersMinimum number of operations necessary to arrive at any configuration

How many letters suffice to construct words with no repetition?


Bits and orbitsExtension of Tao-Green TheoremA property of unimodal sequencesPartitioning finite ordered setsRepresentability of sets of infinite sequences sharing common prefixes and factors (i.e. infixes)“Nyldon words”: understanding a class of words factorizing the free monoid increasinglyProbability of no $k$ 1's in arithmetic progression in binary sequence of length $n$Reference for one-sided subshiftsSequences with 3 LettersMinimum number of operations necessary to arrive at any configuration













8












$begingroup$


Given a finite set $A=a_1,ldots , a_k$, consider the sequences of any length that can be constructed using the elements of $A$ and which contain no repetition, a repetition being a pair of consecutive subsequences (of any length) that are equal. Is it true that $k = 4$ is the minimum number of elements in $A$ that allows the construction of sequences of any length containing no repetition? Can anyone indicate a reference for this result, if true?










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    8












    $begingroup$


    Given a finite set $A=a_1,ldots , a_k$, consider the sequences of any length that can be constructed using the elements of $A$ and which contain no repetition, a repetition being a pair of consecutive subsequences (of any length) that are equal. Is it true that $k = 4$ is the minimum number of elements in $A$ that allows the construction of sequences of any length containing no repetition? Can anyone indicate a reference for this result, if true?










    share|cite|improve this question









    New contributor




    PiCo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      8












      8








      8


      2



      $begingroup$


      Given a finite set $A=a_1,ldots , a_k$, consider the sequences of any length that can be constructed using the elements of $A$ and which contain no repetition, a repetition being a pair of consecutive subsequences (of any length) that are equal. Is it true that $k = 4$ is the minimum number of elements in $A$ that allows the construction of sequences of any length containing no repetition? Can anyone indicate a reference for this result, if true?










      share|cite|improve this question









      New contributor




      PiCo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Given a finite set $A=a_1,ldots , a_k$, consider the sequences of any length that can be constructed using the elements of $A$ and which contain no repetition, a repetition being a pair of consecutive subsequences (of any length) that are equal. Is it true that $k = 4$ is the minimum number of elements in $A$ that allows the construction of sequences of any length containing no repetition? Can anyone indicate a reference for this result, if true?







      co.combinatorics symbolic-dynamics






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      edited 6 hours ago









      YCor

      29k486140




      29k486140






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      asked 8 hours ago









      PiCoPiCo

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          1 Answer
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          $begingroup$

          Wikipedia has some examples of square-free sequences of infinite length (and therefore square-free words of arbitrary length) over alphabets with 3 letters.
          https://en.wikipedia.org/wiki/Square-free_word




          One example of an infinite square-free word over an alphabet of size 3 is the word over the alphabet 0,±1 obtained by taking the first difference of the Thue–Morse sequence.[6][7] That is, from the Thue–Morse sequence



          0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, ...



          one forms a new sequence in which each term is the difference of two consecutive terms of the Thue–Morse sequence. The resulting square-free word is



          1, 0, −1, 1, −1, 0, 1, 0, −1, 0, 1, −1, 1, 0, −1, ... (sequence A029883 in the OEIS).







          share|cite|improve this answer











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            $begingroup$

            Wikipedia has some examples of square-free sequences of infinite length (and therefore square-free words of arbitrary length) over alphabets with 3 letters.
            https://en.wikipedia.org/wiki/Square-free_word




            One example of an infinite square-free word over an alphabet of size 3 is the word over the alphabet 0,±1 obtained by taking the first difference of the Thue–Morse sequence.[6][7] That is, from the Thue–Morse sequence



            0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, ...



            one forms a new sequence in which each term is the difference of two consecutive terms of the Thue–Morse sequence. The resulting square-free word is



            1, 0, −1, 1, −1, 0, 1, 0, −1, 0, 1, −1, 1, 0, −1, ... (sequence A029883 in the OEIS).







            share|cite|improve this answer











            $endgroup$

















              15












              $begingroup$

              Wikipedia has some examples of square-free sequences of infinite length (and therefore square-free words of arbitrary length) over alphabets with 3 letters.
              https://en.wikipedia.org/wiki/Square-free_word




              One example of an infinite square-free word over an alphabet of size 3 is the word over the alphabet 0,±1 obtained by taking the first difference of the Thue–Morse sequence.[6][7] That is, from the Thue–Morse sequence



              0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, ...



              one forms a new sequence in which each term is the difference of two consecutive terms of the Thue–Morse sequence. The resulting square-free word is



              1, 0, −1, 1, −1, 0, 1, 0, −1, 0, 1, −1, 1, 0, −1, ... (sequence A029883 in the OEIS).







              share|cite|improve this answer











              $endgroup$















                15












                15








                15





                $begingroup$

                Wikipedia has some examples of square-free sequences of infinite length (and therefore square-free words of arbitrary length) over alphabets with 3 letters.
                https://en.wikipedia.org/wiki/Square-free_word




                One example of an infinite square-free word over an alphabet of size 3 is the word over the alphabet 0,±1 obtained by taking the first difference of the Thue–Morse sequence.[6][7] That is, from the Thue–Morse sequence



                0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, ...



                one forms a new sequence in which each term is the difference of two consecutive terms of the Thue–Morse sequence. The resulting square-free word is



                1, 0, −1, 1, −1, 0, 1, 0, −1, 0, 1, −1, 1, 0, −1, ... (sequence A029883 in the OEIS).







                share|cite|improve this answer











                $endgroup$



                Wikipedia has some examples of square-free sequences of infinite length (and therefore square-free words of arbitrary length) over alphabets with 3 letters.
                https://en.wikipedia.org/wiki/Square-free_word




                One example of an infinite square-free word over an alphabet of size 3 is the word over the alphabet 0,±1 obtained by taking the first difference of the Thue–Morse sequence.[6][7] That is, from the Thue–Morse sequence



                0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, ...



                one forms a new sequence in which each term is the difference of two consecutive terms of the Thue–Morse sequence. The resulting square-free word is



                1, 0, −1, 1, −1, 0, 1, 0, −1, 0, 1, −1, 1, 0, −1, ... (sequence A029883 in the OEIS).








                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 hours ago

























                answered 8 hours ago









                user44191user44191

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