1 Introduction
Throughout, we use standard definitions and notations from combinatorics on words (see [16]). The word is a factor of the word if we can write for some (possibly empty) words . If at least one of is nonempty, then we say that is a proper factor of . For a set of words , the word is a factor of if is a factor of some word in .
Let be a word, where the ’s are letters. A positive integer is a period of if for all . In this case, we say that is an exponent of , and the largest such number is called the exponent of . For a real number , a finite or infinite word is called free (free) if contains no factors of exponent greater than or equal to (strictly greater than , respectively).
Throughout, for every positive integer , let denote the letter alphabet . For every , the repetition threshold for letters, denoted , is defined by
Essentially, the repetition threshold describes the border between avoidable and unavoidable repetitions in words over an alphabet of letters. The repetition threshold was first defined by Dejean [13]. Her 1972 conjecture on the values of has now been confirmed through the work of many authors [13, 22, 19, 18, 5, 8, 7, 9, 23]:
The last cases of Dejean’s conjecture were confirmed in 2011 by Currie and the second author [9], and independently by Rao [23]
. However, probably the most important contribution was made by Carpi
[5], who confirmed the conjecture for all but finitely many values of .Here, we are interested in the notion of a repetition threshold for circular words on letters. Two (linear) words and are said to be conjugates if there are words and such that and . The conjugates of a word can be obtained by rotating the letters of cyclically. For a word , the circular word is the set of all conjugates of . Intuitively, one can think of a circular word as being obtained from a linear word by linking the ends, giving a cyclic sequence of letters. A circumnavigation of the circular word is a word of the form , where is a letter and is a conjugate of .
By the definition of a factor of a set of words, a word is a factor of a circular word if and only if it is a factor of some conjugate of . As for linear words, a circular word is free (free) if it has no factors of exponent greater than or equal to (strictly greater than , respectively).
While the factors of an free circular word must be free as linear words, they need not necessarily be free when taken as circular words. For example, the free circular word contains the factor 0120, and the circular word contains the square factor . This means that several equivalent definitions of the repetition threshold give rise to distinct notions of a repetition threshold for circular words.
In this paper, we will be most interested in the weak circular repetition threshold for letters, denoted , and defined by
The intermediate circular repetition threshold for letters, denoted , is defined by
and the strong circular repetition threshold for letters, denoted , is defined by
Evidently, we have
(1) 
for all . Table 1 contains all of the confirmed and conjectured values of , , , and . Note in particular that
which shows that the three notions of circular repetition threshold are not equivalent.
Through the work of several authors [1, 12, 24, 15, 10], all values of the strong circular repetition threshold are known:
The values of the weak and intermediate circular repetition thresholds are only known exactly for and :

(Thue [4]);
From (1) and the known values of and , we have
for every . These are currently the best known bounds on both and when . Currie and the present authors [10] recently made the following conjecture, which strengthens the second statement of an older conjecture of Shur [24, Conjecture 1].
Conjecture 1.
For every , we have .
Here, we prove a weak version (pun intended) of Conjecture 1 for all but finitely many values of .
Theorem 2.
For every , we have .
The layout of the remainder of the paper is as follows. In Section 2, we summarize the work of Carpi [5] in confirming all but finitely many cases of Dejean’s conjecture. In Section 3, we establish Theorem 2 with a construction that relies heavily on the work of Carpi. We conclude with a discussion of some related notions of repetition threshold for classes of graphs.
2 Carpi’s reduction to kernel repetitions
In this section, let be a fixed integer. Pansiot [22] was first to observe that if a word over the alphabet is free, then it can be encoded by a word over the binary alphabet . For consistency, we use the notation of Carpi [5] to describe this encoding. Let denote the symmetric group on , and define the morphism by
Now define the map by
where
for all . To be precise, Pansiot proved that if a word is free, then can be obtained from a word of the form , where , by renaming the letters.
Let , and let . Pansiot showed that if has a factor of exponent greater than , then either the word itself contains a short repetition, or the binary word contains a kernel repetition (see [22] for details). Carpi reformulated this statement so that both types of forbidden factors appear in the binary word . Let , and let . Then is called a stabilizing word (of order ) if fixes the points . Let denote the set of stabilizing words of order . The word is called a kernel repetition (of order ) if it has period and a factor of length such that and . Carpi’s reformulation of Pansiot’s result is the following.
Proposition 3 (Carpi [5, Proposition 3.2]).
Let . If a factor of has exponent larger than , then has a factor satisfying one of the following conditions:

and for some ; or

is a kernel repetition of order .
Now assume that , and define and . Carpi [5] defines an uniform morphism with the following extraordinary property.
Proposition 4 (Carpi [5, Proposition 7.3]).
Suppose that , and let . Then for every , the word contains no stabilizing word of length smaller than .
We note that Proposition 4 was proven by Carpi [5] in the case that in a computationfree manner. The improvement to stated here was achieved later by Currie and the second author [7], using lemmas of Carpi [5] along with a significant computer check.
Proposition 4 says that for every word , no factor of satisfies condition (i) of Proposition 3. Thus, we need only worry about factors satisfying condition (ii) of Proposition 3, i.e., kernel repetitions. To this end, define the morphism by for all . A word is called a kernel repetition if it has a period and a factor of length such that and . Carpi established the following result.
Proposition 5 (Carpi [5, Proposition 8.2]).
Let . If a factor of is a kernel repetition, then a proper factor of is a kernel repetition.
Remark.
In other words, if no proper factor of is a kernel repetition, then no factor of satisfies condition (ii) of Proposition 3. Finally, we note that the morphism is defined in such a way that the kernel of has a very simple structure.
Lemma 6 (Carpi [5, Lemma 9.1]).
If , then if and only if divides for every letter .
3 Constructing free circular words over letters
In this section, let be a fixed integer, and let and , as in the previous section. Finally, define . Since , we have and .
In order to prove Theorem 2, we will construct an free circular word of length over for every integer . We first show that we can restrict our attention to words over the smaller alphabet , just as Carpi did for linear words. We begin with an analogue of Proposition 3 for circular words.
Lemma 7.
Let . If a factor of the circular word has exponent larger than , then the circular word has a factor satisfying one of the following conditions:

and for some ; or

is a kernel repetition of order .
Proof.
Let , where for all . Let , where for all . First, we claim that for every conjugate of , the word is equal, up to a permutation of the letters, to the corresponding conjugate of . It suffices to show that is equal to , up to a permutation of the letters. By definition of , we have
for all . Let . Again, by definition of , we have
for all . Finally, using the definition of and the fact that , we have
Thus, we see that for all , and this completes the proof of the claim.
Suppose now that a factor of the circular word has exponent larger than . Then, up to permutation of the letters, the word is a factor of for some conjugate of . By Lemma 3, the word contains a factor satisfying either condition (i) or condition (ii). Since is a conjugate of , the word is a factor of , and this completes the proof. ∎
The next lemma is an analogue of Proposition 5 for circular words.
Lemma 8.
Let . If the circular word contains no kernel repetitions, then the circular word is free.
Proof.
Suppose, towards a contradiction, that contains no kernel repetitions, and that contains a factor of exponent greater than . Since , we have . Thus, by Lemma 7, some factor of satisfies either condition (i) or condition (ii) of Lemma 7. Now must be a factor of for some circumnavigation of (i.e., we have , where and is a conjugate of ). By Proposition 4, for every , the word contains no stabilizing words of length less than . So it must be the case that is a kernel repetition of order . Then by Proposition 5, some proper factor of must be a kernel repetition. Since every proper factor of the circumnavigation is a factor of the circular word , this is a contradiction. So we conclude that the circular word is free. ∎
By Lemma 8, in order to construct an free circular word of length over , it suffices to construct a word of length over that lies in and contains no kernel repetitions. Our construction of such a word uses many ideas of Carpi [5, Section 9], and we recommend that the reader reviews this section before proceeding.
Following Carpi, define , where
We note that can also be defined (c.f. [6, Example 4]) as the fixed point of the morphism defined by
1  
2 
We need two lemmas concerning the factors of .
Lemma 9.
For every , the word has a factor of length that begins and ends in the letter 2.
Proof.
The statement can be verified by a casebased proof depending on the value of , or by using the automatic theorem proving software Walnut [20]. We describe the latter approach.
After saving the automaton generating the fixed point of in the Word Automata Library folder as B.txt, we use the predicate
eval BeginsAndEndsInTwo "?msd_3 (k>=1 & (Ei (B[i]=@2 & B[i+k1]=@2)))": 
The automaton for this predicate is illustrated in Figure 1. The automaton clearly accepts all such that . ∎
Lemma 10 (Carpi [5, Lemma 9.2]).
Let be a factor of with period . For every , if , then divides .
For every , we define a set of words , all of length . We define by if and only if whenever (mod ), and whenever (mod ).
We prove two lemmas concerning the words in . The proof of the first lemma uses the main idea from Carpi’s proof of [5, Lemma 9.3].
Lemma 11.
Let , and let be a factor of the circular word . If , then divides .
Proof.
The statement is trivially true if , so assume . Set , where is the maximal power of dividing . Suppose, towards a contradiction, that . Since , by Lemma 6, we see that divides , meaning .
Write . Then we have for some , with indices taken modulo . Since is divisible by and , for all and , we see that divides if and only if divides . Since , we have , and hence implies (mod ). Thus, by definition, for any , we have if and only if divides . Thus, we have that the sum is exactly the number of integers in the set that are divisible by , which is exactly . Since , by Lemma 6, we conclude that divides , contradicting the maximality of . ∎
Lemma 12.
Let . For every letter , the number is a multiple of .
Proof.
Write , and let . Since , we have if and only if divides . Since , we have
Since , we see that divides . The fact that divides now follows by a straightforward inductive argument. ∎
We are now ready to construct a circular word of length over that belongs to and contains no kernel repetitions. The proof of the next proposition uses some arguments that were first used by Carpi [5, Proposition 9.4].
Proposition 13.
Suppose that . For every integer , there is a word of length such that the circular word contains no kernel repetitions.
Proof.
Fix . Since , we have and . We use these facts frequently. Let be a factor of of length that begins and ends in the letter 2; such a factor is guaranteed to exist by Lemma 9. Let be the morphism defined by
1  
2 
Write .
Let , and define by
Finally, define by
Evidently, we have . We will show that , and that the circular word contains no kernel repetitions, thus proving the proposition statement.
First, we show that . By Lemma 5, it suffices to show that divides for all . Since , by Lemma 12, we have that divides for all . Next, note that
which is clearly divisible by . Since
we see that divides as well. Finally, we have
Since divides both and for all , we conclude that divides .
It remains to show that the circular word contains no kernel repetitions. Suppose towards a contradiction that some factor of is a kernel repetition of period . Then by definition, we have
(2) 
Since , we must also have .
First of all, consider the case . Applying this inequality to the right side of (2) and then rearranging, we obtain
By Lemma 11, we must have . Together with the fact that , this gives
But this last inequality implies , a contradiction.
So we may assume that . Then one has for some integer . Using (2) and the inequality , we obtain
(3) 
Deleting the letters of even index in , we obtain a factor of the circular word . By Lemma 11, we know that divides , so is certainly divisible by , and has period . Since , we have
(4) 
Write , where . From (4), we see that .
By construction, we may write and , where and . So we have
Thus, in the circular word , the factors 23 and 32 both appear exactly once. Further, every factor of length of contains a 2, and every factor of length of contains a 3. Therefore, either every appearance of in is contained entirely in , or every appearance of in is contained entirely in . Without loss of generality, assume that every appearance of in is contained entirely in . Now we consider two cases.
Case 1: The factor is contained entirely in .
Then is a factor of with period . By Lemma 11, we know that divides . Recall from (4) that . Thus, by Lemma 10, we have that divides . Therefore, we have . Together with (3), this gives
But we have already seen that this inequality implies , a contradiction.
Case 2: The factor contains the entire word as a factor.
Then we can write , where , and contains the entire word . Let be a conjugate of . See Figure 2 for an illustration of the circular word .
Now let be the conjugate of corresponding to , i.e., we have that is the length prefix of , , , and . We claim that is a kernel repetition with period . First of all, note that since is a factor of , while contains all of , we have . Next, note that both and . Now let . By Lemma 5, both and are multiples of . It follows that
is also a multiple of , and hence . By the definition of kernel repetition, we have
Rearranging the above inequality, and then using the fact that , we have
which implies that is also a kernel repetition. But this is impossible by Case 1. ∎
Finally, Theorem 2 follows directly from the next result.
Proposition 14.
Suppose that . For every integer , there is a word of length such that the circular word is free.
4 Conclusion
We have shown that for all . The conjecture that remains open for all , and the stronger conjecture that remains open for all .
To conclude, we will place the notion of weak circular repetition threshold in a broader context, and discuss a more general problem. Let be a graph, and let be an colouring of . A word is called a factor of if for some path in that contains no repeated vertices. An colouring of the graph is called free if it contains no factors of exponent greater than or equal to . By this definition, the free colourings of are exactly the nonrepetitive colourings of , first defined by Alon et al. [3]. Nonrepetive colourings have been widely studied in the last two decades. In particular, Dujmović et al. [14] recently confirmed what was probably the most important conjecture on nonrepetitive colourings, namely that every planar graph can be nonrepetitively coloured with a bounded number of colours. Their paper also contains an extensive list of references to other work on nonrepetitive colourings and related notions.
To date, most work on free colourings has concerned the problem of fixing a number (most commonly ) and determining the minimum number of colours necessary for an free colouring of a given graph. Ochem and Vaslet [21] introduced a notion of repetition threshold for classes of graphs, which considers the problem the other way around – for a fixed number of colours, find the smallest value of such that there is an free colouring of a given graph. Formally, the repetition threshold for letters on , denoted , is defined by:
For a collection of graphs , the repetition threshold for is defined by . Note that the strong circular repetition threshold is equivalent to the repetition threshold , where is the collection of cycles.
Ochem and Vaslet determined all values of , where is the collection of all trees. Lužar, Ochem, and Pinlou [17] determined all values of and , where is the collection of all caterpillars, and is the collection of all caterpillars of maximum degree . They also gave upper and lower bounds on , where is the collection of all trees of maximum degree .
Ochem and Vaslet [21] also defined a notion of repetition threshold for “sufficiently large subdivisions” of all graphs. For a graph , let denote the collection of subdivisions of (i.e., those graphs obtained from by a sequence of edge subdivisions). For a collection of graphs , we define the weak repetition threshold for , denoted , by
The repetition threshold for subdivided graphs, as defined by Ochem and Vaslet, is then equivalent to the weak repetition threshold for the collection of all graphs. Ochem and Vaslet proved that
For all , the lower bound follows from the somewhat trivial fact that any colouring of a graph with a vertex of degree must contain a factor of exponent . This suggests restricting to classes of graphs with bounded maximum degree, as was done for the repetition thresholds of caterpillars and trees [17]. Let denote the collection of graphs with maximum degree . For all , it is easy to see that we have
since every graph of maximum degree is a disjoint union of paths and cycles. So by Theorem 2, we have for all . In addition to determining the unknown values of , it would be interesting to determine values of for .
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