# Repetition avoidance in products of factors

We consider a variation on a classical avoidance problem from combinatorics on words that has been introduced by Mousavi and Shallit at DLT 2013. Let pexp_i(w) be the supremum of the exponent over the products of i factors of the word w. The repetition threshold RT_i(k) is then the infimum of pexp_i(w) over all words w∈Σ^ω_k. Moussavi and Shallit obtained that RT_i(2)=2i and RT_2(3)=134. We show that RT_i(3)=3i2+14 if i is even and RT_i(3)=3i2+16 if i is odd and i>3.

## Authors

• 3 publications
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## 1 Introduction

A repetition in a word is a pair of words and such that is a factor of , is non-empty, and is a prefix of . If is a repetition, then its period is and its exponent is . A word is -free (resp. -free) if it contains no repetition with exponent such that (resp. ).

Given , Dejean [2] defined the repetition threshold for letters as the smallest such that there exists an infinite -free word over a -letter alphabet. Dejean initiated the study of in 1972 for and . Her work was followed by a series of papers which determine the exact value of for any .

•  [2];

•  [2];

•  [7];

• , for  [1, 4, 8].

Mousavi and Shallit [5] have considered two notions related to the repetition threshold.

The first notion considers repetitions in conjugates of factors of the infinite word. A word is circularly -free if it does not contain a factor such that is a repetition of exponent strictly greater than . Let . The smallest real number such that is circularly -free is denoted by . Let be the minimum of over every .

The second notion considers repetitions in concatenations of a fixed number of factors of the infinite word. Let be the smallest real number such that every product of factors of is -free. Let be the minimum of over every . Notice that generalizes the classical notion of repetition threshold which corresponds to the case , that is, for every .

Our first result shows that the case corresponds to the first notion of repetition avoidance in conjugates.

###### Theorem 1.

for every .

Mousavi and Shallit [5] have considered the binary alphabet and obtained that for every . Our second result considers the ternary alphabet and gives the value of for every . This extends the result of Dejean [2] that and the result of Mousavi and Shallit [5] that .

###### Theorem 2.
• if or is even.

• if is odd and .

## 2 Proofs

Proof of Theorem 1.
The language of words in avoiding circular repetitions of exponent at least (or strictly greater than ) is a factorial language. As it is well-known [3], if a factorial language is infinite, then it contains a uniformly recurrent word . By Proposition 14 in [5], . This implies that .

To obtain the two equalities of Theorem 2, we show the two lower bounds and then the two upper bounds.

Proof of for every even .
Mousavi and Shallit [5] have proved that , which settles the case . We have double checked their computation of the lower bound . Suppose that is a fixed even integer and that is an infinite ternary word. The lower bound for implies that there exists two factors and such that with . Thus, the prefix of is also a product of two factors of . So we can form the -terms product which is a repetition of the form with exponent . This is the desired lower bound.

Proof of for every odd .
Suppose that is a fixed odd integer, that is, . Suppose that is a recurrent ternary word such that the product of factors of is never a repetition of exponent at least . First, is square-free since otherwise there would exist an -terms product of exponent . Also, does not contain two factors and with the following properties:

• ,

• with .

Indeed, this would produce the -terms product which is a repetition of the form with exponent .

So if , , and are distinct letters, then does not contain both and and does not contain both and . A computer check shows that no infinite ternary square-free word satisfies this property. This proves the desired lower bound.

Proof of for every even .
Let be any even integer at least . To prove this upper bound, it is sufficient to construct a ternary word satisfying . The ternary morphic word used in [5] to obtain seems to satisfy the property. However, it is easier for us to consider another construction. Let us show that the image of every -free word over by the following -uniform morphism satisfies .

 0↦010201210212021012102010212012101% 2021012102121↦0102012102120121012021012102010212021012102122↦0102012101202120121021202101210212012101202123↦010201210120210121021201210120212012102010212

Recall that a word is -free if it does not contain a repetition with period at least and exponent strictly greater than . First, we check that such ternary images are -free using the method in [6]. By Lemma 2.1 in [6], it is sufficient to check this freeness property for the image of every -free word over of length smaller than . Since , the period of every repetition formed from pieces and with exponent at least must be at most . Then we check exhaustively by computer that the ternary images do not contain two factors and such that

• ,

• ,

• .

Thus, the period of every repetition formed from pieces and with exponent strictly greater than must be at most . So we only need to check that for -terms products that are repetitions of period at most .

Now the period is bounded, but can still be arbitrarily large, a priori. For every factor of length at most , we define as the length of a largest factor of that is a -terms product, divided by . We actually consider conjugacy classes, since if is a conjugate of , then . Let be such a factor. If, for some even , we have , then it means that by appending a -terms product to a -terms product that corresponds to a maximum factor of , that can only add a cube of period . This implies that for every , .

We have checked by computer that for every conjugacy class of words of length at most , there exists a (small) even such that . Thus we have in all cases.

Proof of for every odd .
Let us show that the image of every -free word over by the following -uniform morphism satisfies for every odd .

 0↦010201202101201021202102012101201% 0201202102012102120102012101202102012102120210120102012102120102012021020121012010212021020121021201020120210120102120210201210212021012010201210212010201210120210201210212021012010212021020121012010201202102012101201021202102012102120102012101202102012102120102012021012010212021020121012010201202102012102120210120102012101202102012102120102012021020121012010212021020121021201020121012021020121021202101201021202102012101201020120210201210120102120210201210212021012010201210212
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 2↦010201202101201021202102012101201% 0201202102012102120102012101202102012101201021202102012102120102012021012010212021020121021202101201020121021201020121012021020121021202101201021202102012101201020120210201210120102120210201210212010201210120210201210212010201202101201021202102012102120210120102012102120102012021020121012010212021020121021201020120210120102120210201210120102012021020121021202101201020121021201020121012021020121021202101201021202102012101201020120210201210120102120210201210212021012010201210212
 3↦010201202101201021202102012101201% 0201202102012102120102012101202102012101201021202102012102120102012021012010212021020121021202101201020121012021020121021201020120210201210120102120210201210212010201210120210201210212021012010201210212010201202101201021202102012101201020120210201210212010201210120210201210212010201202102012101201021202102012102120210120102012102120102012021012010212021020121021201020121012021020121021202101201021202102012101201020120210201210120102120210201210212021012010201210120210201210212

First, we check that such ternary images are -free using the method in [6]. By Lemma 2.1 in [6], it is sufficient to check this freeness property for the image of every -free word over of length smaller than . Thus, the period of every repetition formed from pieces and with exponent strictly greater than must be at most . Using the same argument as in the previous proof, we have checked by computer that for every conjugacy class of words of length at most , there exists a (small) odd such that . Thus we have in all cases.

## 3 Concluding remarks

The next step would be to consider the -letter alphabet. Obviously, for every and . Mousavi and Shallit [5] verified that , so that for every . We conjecture that this is best possible, i.e., that for every . However, a proof of an upper bound of the form cannot be similar to the proof of the upper bounds of Theorem 2. The multiplicative factor of , which drops from when to when , forbids that the constructed word is the morphic image of any (unspecified) Dejean word over a given alphabet.

## References

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• [5] H. Mousavi and J. Shallit. Repetition avoidance in circular factors. Developments in Language Theory 2013, 384–395.
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