Extremal overlap-free and extremal $\beta$-free binary words

An overlap-free (or $\beta$-free) word $w$ over a fixed alphabet $\Sigma$ is extremal if every word obtained from $w$ by inserting a single letter from $\Sigma$ at any position contains an overlap (or a factor of exponent at least $\beta$, respectively). We find all lengths which admit an extremal overlap-free binary word. For every extended real number $\beta$ such that $2^+\leq\beta\leq 8/3$, we show that there are arbitrarily long extremal $\beta$-free binary words.


Introduction
Throughout, we use standard definitions and notations from combinatorics on words (see [11]). For every integer n ≥ 2, we let Σ n denote the alphabet {0, 1, . . . , n-1}. The word u is a factor of the word w if we can write w = xuy for some (possibly empty) words x, y. A square is a word of the form xx, where x is nonempty. An overlap is a word of the form axaxa, where a is a letter and x is a (possibly empty) word. A word is square-free if it contains no square as a factor, and overlap-free if it contains no overlap as a factor. Early in the twentieth century, Norwegian mathematician Axel Thue [21,22] demonstrated that one can construct arbitrarily long square-free words over a ternary alphabet, and arbitrarily long overlap-free words over a binary alphabet. For an English translation of Thue's work, see [2]. Thue's work is recognized as the beginning of the field of combinatorics on words [3].
Let w be a word over a fixed alphabet Σ. An extension of w is a word of the form w aw , where a ∈ Σ, and w w = w for some possibly empty words w , w ∈ Σ * . For example, over the English alphabet, the English word pans has extensions including the English words spans, plans, pawns, pants, and pansy. The word w is extremal square-free if w is square-free, and every extension of w contains a square. For example, the word abcabacbcabcbabcabacbcabc of length 25 is an extremal square-free word of minimum length over the alphabet {a, b, c}. The concept of extremal square-free word was recently introduced by Grytczuk et al. [10], who demonstrated that there are arbitrarily long extremal square-free words over a ternary alphabet. Two of the present authors [13] adapted their ideas to find all lengths admitting extremal square-free ternary words.
In this paper, we consider some variations of extremal square-free words, with a focus on the binary alphabet Σ 2 = {0, 1}. We begin by considering extremal overlap-free words, as suggested by Grytczuk et al. [10]. For a word w over a fixed alphabet Σ, we say that w is extremal overlap-free if w is overlap-free, and every extension of w contains an overlap. For example, the word 0010011011 of length 10 is an extremal overlap-free word of minimum length over Σ 2 .
While there is an extremal square-free ternary word of every sufficiently large length, the same cannot be said for extremal overlap-free binary words. Our first main result is the following characterization of the lengths of extremal overlap-free binary words. Theorem 1.1. Let n be a nonnegative number. Then there is an extremal overlap-free word of length n over the alphabet Σ 2 if and only if n is in the After proving Theorem 1.1, we consider a more general problem, which we now provide background for. Let w = w 1 w 2 · · · w n be a word, where the w i 's are letters. For an integer p ≥ 1, we say that w has period p if w i+p = w i for all 1 ≤ i ≤ n − p. Note that w may have many periods; the minimal period of w is called the period of w. The exponent of w is the length of w divided by the period of w. For a real number b, the word w is b-free if it contains no factor of exponent greater than or equal to b, and the word w is b + -free if it contains no factor of exponent greater than b. So 2-free words are exactly the square-free words, and 2 + -free words are exactly the overlap-free words.
For ease of writing, we unify the notions of b-free word and b + -free word by considering β-free words, where β belongs to the set of "extended real numbers". Let R ext denote the set of extended real numbers, consisting of all real numbers, together with all real numbers with a +, where x + covers x, and the inequalities y ≤ x and y < x + are equivalent. For β ∈ R ext , we say that w is β-free if no factor of w has exponent greater than or equal to β. Definition 1.2. Let w be a word over a fixed alphabet Σ, and let β ∈ R ext . We say that w is extremal β-free if w is β-free, and every extension of w contains a factor of exponent greater than or equal to β.
We consider the following problem. Problem 1.3. For which β ∈ R ext do there exist arbitrarily long extremal β-free words over Σ 2 ?
On the affirmative side, by Theorem 1.1, we know that there are arbitrarily long extremal 2 + -free words over Σ 2 . On the negative side, every binary word of length at least 4 contains a square, so it follows that for all β ≤ 2, there do not exist arbitrarily long extremal β-free words over Σ 2 . We make some further progress on Problem 1.3 on the affirmative side by establishing the following theorem. Theorem 1.4. Let β ∈ R ext satisfy 2 + ≤ β ≤ 8/3. Then there are arbitrarily long extremal β-free words over Σ 2 .
We also make the following conjecture. Conjecture 1.5. There is some number α ∈ R ext such that for all β ∈ R ext satisfying β ≥ α, there are no extremal β-free words over Σ 2 .
It is possible that Conjecture 1.5 is true with α = 8/3 + , but we have only very weak computational evidence supporting this. If one could show that Conjecture 1.5 is true with α = 8/3 + , then it would completely answer Problem 1.3.
The layout of the remainder of the paper is as follows. We prove Theorem 1.1 in Section 2 and Section 3. We consider the even lengths in Section 2, and the odd lengths in Section 3. We prove Theorem 1.4 in Section 4. We conclude with a discussion of some open problems and conjectures over larger alphabets.

Extremal overlap-free words of even length
In this section, we characterize the even lengths for which there are extremal overlap-free binary words. Throughout the remainder of the paper, we let µ : Σ * 2 → Σ * 2 denote the Thue-Morse morphism, defined by µ(0) = 01 and µ(1) = 10. The Thue-Morse word t is the unique fixed point of the morphism µ that begins with 0. In other words, we have t = µ ω (0). The Thue-Morse word is the prototypical example of a 2-automatic sequence. We begin with a lemma that is used frequently in the rest of the paper. Lemma 2.1. Let w ∈ Σ * 2 be an overlap-free word of length at least 10, and write w = w w with |w |, |w | ≥ 5. Then for every letter a ∈ Σ 2 , the extension w aw contains an overlap of period at most 3 (and hence a factor of exponent at least 7/3).
Proof. It suffices to check the lemma statement for all overlap-free words in Σ * 2 of length exactly 10, which is completed easily by computer.
Definition 2.2. A word w ∈ Σ * 2 is called earmarked if all of the following conditions are satisfied: (i) w is overlap-free; (ii) the length 4 prefix of w is in {0010, 1101}; and (iii) the length 4 suffix of w is 0100. Lemma 2.3. Let u be an earmarked word of length at least 8. Let w be the word obtained from v = µ(u) by complementing the first and last letters. Then w is both earmarked and extremal overlap-free.
Proof. Assume that u has prefix 0010; the case that u has prefix 1101 is handled similarly. So we may write u = 0010u 0100 for some word u ∈ Σ * 2 . It follows that w = 11011001µ(u )01100100 So w has length 4 prefix 1101 and length 4 suffix 0100.
We now show that w is overlap-free. First of all, since u and µ are overlapfree, we see that v is overlap-free. Now suppose that w contains the overlap x. Since v is overlap-free, we see that x must be either a prefix or a suffix of w. Assume that x is a prefix of w; the case that x is a suffix of w is handled similarly. Since the word 11011 may only appear as a prefix or a suffix of an overlap-free word, we conclude that the period of x is at most 4. But by inspection, there is no such overlap in w.
Finally, we show that w is extremal overlap-free. By Lemma 2.1, it suffices to check that every extension of w of the form w aw , where w = w w , a ∈ Σ 2 , and either |w | ≤ 4 or |w | ≤ 4, contains an overlap. We complete this check by inspection.
Lemma 2.4. Let n ≥ 10 be an integer satisfying n ≡ 0 (mod 4). Then there is an earmarked word of length n.
Proof. We use the automatic theorem-proving software Walnut [15] to show that the Thue-Morse word t contains a factor u of length n − 4 such that the word u0100 is earmarked. The interested reader can verify our results in Walnut; the complete code that we used can be found in Appendix A. We essentially adapt the predicates used by Clokie, Gabric, and Shallit [4, First, we create a predicate overlap(i, n, p, s) which evaluates to true if the word u0100 contains an overlap of period p with p ≥ 1 beginning at index i − s, where u = t[s..s + n − 5]. We use a straightforward modification of the method described by Clokie, Gabric, and Shallit [4, Proof of Theorem 1] to do so. Next, we create a predicate earmarked(n, s) which evaluates to true if the word u0100 defined above is earmarked: Finally, the predicate testEarmarked(n) := ∃s earmarked(n, s) evaluates to true if there is some length n − 4 factor v of the Thue-Morse word such that v0100 is earmarked. The automaton for testEarmarked(n) is shown in Figure 1. By inspection, this automaton accepts all integers n ≥ 10 such that n ≡ 0 (mod 4).
Lemma 2.5. Let n ≥ 10 be an integer that is not a power of two. Then there is an earmarked word of length n.
Proof. By Lemma 2.4, we may assume that n ≡ 0 (mod 4). Since n is not a power of two, we may write 2 k < n < 2 k+1 for some k ≥ 3. We proceed by induction on k. If k ≤ 4, then n ∈ {12, 20, 24, 28}. It is easily verified by computer that the following words (found by computer search) are earmarked: Length 12: 001001100100 Length 20: 00100110100101100100 Length 24: 110110010110100101100100 Length 28: 1101100110100101101001100100 So we may assume that k ≥ 5. Let m = n/2. Note that m is not a power of two, and that 10 < 2 k−1 < m < 2 k . If m ≡ 0 (mod 4), then there is an earmarked word of length m by Lemma 2.4. If m ≡ 0 (mod 4), then there is an earmarked word of length m by the induction hypothesis. So either way, there is an earmarked word of length m. By Lemma 2.3, there is an earmarked word of length 2m = n.
Corollary 2.6. Let n ≥ 20 be an even integer that is not a power of two. Then there is an extremal overlap-free word of length n.
Proof. By Lemma 2.5, there is an earmarked word u of length m = n/2. By Lemma 2.3, the word w of length n obtained from µ(u) by complementing the first and last letters is extremal overlap-free. Proof. Let s = 0µ(0011001)1 = (00101101) 2 . We claim that the word µ (s) is extremal overlap-free for every integer ≥ 1. Since s has length 16, the word µ (s) has length 2 +4 , and hence the theorem statement follows.
Fix ≥ 1, and let w = µ (s). If ≤ 2, then we verify that w is extremal overlap-free by computer, so we may assume that ≥ 3. First note that s is overlap-free, and hence w is overlap-free. It remains to show that every extension of w contains an overlap. Consider an extension w aw of w, where w = w w and a ∈ Σ 2 . By Lemma 2.1, we may assume that |w | ≤ 4 or |w | ≤ 4. We consider several cases.
Case I: |w | = 0. Note that w begins with the squares µ (00) and µ (s). If is even, then µ (00) ends with a 0, and µ (s) ends with a 1. If is odd, then µ (00) ends with a 1, and µ (s) ends with a 0. So either way, the extensions 0w and 1w both contain an overlap.
Case II: 1 ≤ |w | ≤ 4. Since ≥ 3, we see that w has prefix µ 3 (0) = 01101001. If |w | = 1, then the extension w 0w = 0w contains an overlap by Case I, and the extension w 1w contains the overlap 111. So we may assume that 2 ≤ |w | ≤ 4. By inspection, the extension w aw contains an overlap of period at most 3.
Case III: |w | = 0. Note that µ (s) and µ (101101) are square suffixes of µ (s) which begin in 0 and 1, respectively. So both of the extensions w0 and w1 contain an overlap.
Case IV: 1 ≤ |w | ≤ 4. Since ≥ 3, we see that w has suffix µ 3 (0) = 01101001 if is even, and suffix µ 3 (1) = 10010110 if is odd. Either way, the remainder of the proof is similar to that of Case II. Proof. If n ∈ {0, 2, 4, 6, 8, 14, 16, 18}, then an exhaustive backtracking search shows that no extremal overlap-free word of length n exists over Σ 2 . The words 0010011011 and 001001100100, of lengths 10 and 12, respectively, are extremal overlap-free. So suppose that n ≥ 20. If n is a power of two, then there is an extremal overlap-free word of length n by Corollary 2.7. If n is not a power of two, then there is an extremal overlap-free word of length n by Lemma 2.6.

Extremal overlap-free words of odd length
In this section, we characterize the odd lengths for which there are extremal overlap-free binary words. We need two classical results from the theory of overlap-free binary words. The first is the so-called factorization theorem of Restivo and Salemi [19] (see also [1, Proposition 1.7.5(a)]).  Proposition 3.4. Let u be an extremal overlap-free binary word of odd length. Then either |u| = 2 k + 1 or |u| = 3 · 2 k + 1 for some k.
Proof. By Theorem 3.1, we can, without loss of generality, consider two possible forms for u: either u = µ(y)a or u = bbµ(y)a for some a, b ∈ {0, 1}. If u is extremal overlap-free, then both ua and ua end in overlaps. Consequently, the word u ends in at least two distinct squares. Let vv be the longest square suffix of u. Suppose first that |vv| > 6. By Remark 3.3, we see that vv = aµ(z)a for some word z. If vv is a proper factor of µ(y)a, then vv is preceded by a in u; however, since v ends with a, the word avv is an overlap in u, which is a contradiction. We conclude that u = aaµ(z)a = avv, and hence that either |u| = 2 k + 1 or |u| = 3 · 2 k + 1 for some k, as required. Now consider the case |vv| ≤ 6. Since u ends in two distinct squares, these squares are both conjugates of words in A ∪ {0101}, and, since one must be a suffix of the other, we observe that the only possibilities for these two squares are aa and aaaaaa. However, aaaaaa is not a suffix of a word of either the form µ(y)a or the form bbµ(y)a. This contradiction completes the proof.
The proof of Lemma 3.4 tells us that any extremal overlap-free word of odd length can be obtained from an overlap-free square by adding a single letter at either the beginning or the end. This led us to the constructions of overlap-free words of odd length given in the next two lemmas. Lemma 3.5. For every integer k ≥ 5, there is an extremal overlap-free word of length 2 k + 1.
Proof. Fix k ≥ 5. Let u = (011) −1 µ k−1 (00)011. Note that u is a conjugate of µ k−1 (00). In particular, we have that u is a square of length 2 k , and by Theorem 3.2, we see that u is overlap-free. We claim that the word v = 0u is extremal overlap-free. We first show that v is overlap-free. Since u is overlap-free, it suffices to show that no prefix of v is an overlap. Since v has prefix 00100, which never appears again in v, it suffices to check that u does not begin with an overlap of period at most 4, which is easily done by inspection.
It remains to show that every extension of v contains an overlap. Consider an extension v av of v, where v = v v and a ∈ Σ 2 . By Lemma 2.1, we may assume that |v | ≤ 4 or |v | ≤ 4. First suppose that |v | ≤ 4. Note that v has prefix 0(011) −1 µ 4 (0) = 00100110010110. By inspection, the extension v av contains an overlap of period at most 4. Now suppose that |v | ≤ 4. Since u is a square with first letter 0, and u ends in the square 11, the extension va contains an overlap. Thus we may assume that 1 ≤ |v | ≤ 4. If k is even, then v has suffix µ 4 (0)011, and by inspection, the extension v av contains an overlap of period at most 6. If k is odd, then v has suffix µ 4 (1)011, and by inspection, the extension v av contains an overlap of period at most 6.
Lemma 3.6. For every integer k ≥ 3, there is an extremal overlap-free word of length 3 · 2 k + 1.
Proof. Fix k ≥ 3. Let u = (011) −1 µ k−1 (010010)011. Note that u is a conjugate of µ k−1 (010010). In particular, we have that u is a square of length 3 · 2 k , and by Theorem 3.2, we see that u is overlap-free. We claim that the word v = 0u is extremal overlap-free. The remainder of the proof is strictly analogous to the proof of Lemma 3.5.
We now prove the analogue of Proposition 2.8 for odd n. Proof. (⇐) Let n ∈ N . Since n is odd, we must have either n = 2 k + 1 for some k ≥ 5, or n = 3 · 2 k + 1 for some k ≥ 3. In the former case, there is an extremal overlap-free word of length n by Lemma 3.5, and in the latter case, there is an extremal overlap-free word of length n by Lemma 3.6.
(⇒) Suppose that there is an extremal overlap-free word of length n over the alphabet {0, 1}. By Proposition 3.4, we must have n = 2 k + 1 or n = 3·2 k +1 for some k. By exhaustive computer search, there is no extremal overlap-free word of length 2 k + 1 for k ≤ 4, and no extremal overlap-free word of length 3 · 2 k + 1 for k ≤ 2. Thus, we conclude that n ∈ N .

Extremal β-free binary words
This section is devoted to the proof of Theorem 1.4. Another definition facilitates our proof method.
Definition 4.1. Let w be a word over a fixed alphabet Σ, and let α, β ∈ R ext satisfy 1 < α ≤ β. We say that w is (α, β)-extremal if w is α-free, and every extension of w contains a factor of exponent greater than or equal to β.
If w is (α, β)-extremal, then for any γ ∈ R ext such that α ≤ γ ≤ β, the word w is extremal γ-free. Thus, the following result immediately implies Theorem 1.4.     We prove the first part of Proposition 4.2 now.
Proof of Proposition 4.2(a). Let u be a factor of the Thue-Morse word of the form 011v110, where v is a nonempty word. Note that there are arbitrarily long words of this form. We claim that the word x = 00µ 2 (11v11)00 is (2 + , 7/3)-extremal.
First we show that x is 2 + -free (or in other words, overlap-free). Since u is a factor of the Thue-Morse word, we have that u, and hence µ 2 (u), are overlap-free. Since the word µ 2 (u) contains the word 0µ 2 (11v11)0 as a factor, any overlap contained in x must be either a prefix or a suffix of x. Suppose without loss of generality that x contains an overlap z as a prefix. Since the factor 00100 does not appear in the Thue-Morse word, this factor appears only as a prefix and a suffix of x. So z must have period at most 4. But this is impossible by inspection.
It remains to show that every extension of x contains a factor of exponent at least 7/3. Consider an extension x ax of x, where x = x x and a ∈ Σ 2 . By Lemma 2.1, we may assume that |x | ≤ 4 or |x | ≤ 4. First suppose that |x | ≤ 4. Note that x has prefix 00µ 2 (11) = 0010011001. By inspection, the extension x ax contains a factor of exponent at least 7/3. The case that |x | ≤ 4 is handled by a symmetric argument.
One of the main tools that we use to prove Proposition 4.2 parts (b)-(e) is the following extension of a lemma due to Ochem [16,Lemma 2.1]. A morphism f : Σ * → ∆ * is called q-uniform if |f (a)| = q for all a ∈ Σ, and is called synchronizing if for any a, b, c ∈ Σ and u, v ∈ ∆ * , if f (ab) = uf (c)v, then either u = ε and a = c, or v = ε and b = c.
Let h : Σ * → ∆ * be a synchronizing q-uniform morphism. If h(w) is β-free for every α-free word w such that Proof. Suppose that there is an α-free word w such that the word W = h(w) contains a factor of exponent greater than or equal to β, and assume without loss of generality that w is a shortest word satisfying this property. We will show that |w| ≤ max 2b b−a , 2(q−1)(2b−1) , which gives the theorem statement. Let X be a factor of W of exponent greater than or equal to β. Let P be the period of X, and write X = U V , where |U | = P . Since X has period P , we can also write X = V U for some word U ∈ ∆ * . Let R = |V |. Then we have P +R P ≥ b, or equivalently P ≤ R b−1 . First suppose that R ≤ 2q − 2 = 2(q − 1). Then we have By the minimality of w, we must have |w| ≤ |X|−2 q + 2. Putting this together with the above bound on |X|, we find |w| ≤ 2(q−1)(2b−1) where the word V 1 is a proper suffix of a block of h, and the word V 2 is a proper prefix of a block of h. Let r = |v|. Since R ≥ 2q − 1, we must have r ≥ 1. Further, since |V 1 |, |V 2 | < q, we have R < qr + 2q. Similarly, write X = X 1 h(x)X 2 for some word x ∈ Σ * , where the word X 1 is a proper suffix of a block of h, and the word X 2 is a proper prefix of a block of h. Since X = U V = V U , and since h is synchronizing, it must be the case that X 1 = V 1 and X 2 = V 2 . It follows that we may write x = uv = vu for some words u, u ∈ Σ * , i.e., the word x has period |u|. Let p = |u|. Note that h(u) = V −1 1 U V 1 , so |h(u)| = |U | = P , and hence qp = P . Since w is α-free, we must have p+r p ≤ a, or equivalently r ≤ (a − 1)p. Now from which we conclude that p < (a−1)p+2 b−1 , or equivalently, that p < 2 b−a . Finally, by the minimality of w, we must have We conclude in either case that |w| ≤ max 2b b−a , 2(q−1)(2b−1) , as desired.
We are now ready to prove the remaining parts of Proposition 4.2. We use the following terminology in the proof. Let w be a word over a fixed alphabet Σ. A left extension of w is a word of the form aw, where a ∈ Σ. A right extension of w is a word of the form wa, where a ∈ Σ. An internal extension of w is a word of the form w aw , where |w |, |w | ≥ 1, we have a ∈ Σ, and w w = w.
Proof of Proposition 4.2(b). Let u ∈ Σ * 3 be a square-free word of length at least 3, and write u = avb, where a, b ∈ Σ 3 . Define f : Σ * 3 → Σ * 2 by Let r = 1100110010011 and s = 001001. We claim that the word w = rf (v)s is (7/3 + , 17/7)-extremal. First of all, we verify the following statements by computer for every letter c ∈ Σ 3 : • Every internal extension of the word f (c) contains a factor of exponent at least 17/7.
• Every left extension and every internal extension of the word rf (c) contains a factor of exponent at least 17/7.
• Every right extension and every internal extension of the word f (c)s contains a factor of exponent at least 17/7.
It now follows easily that every extension of the word w = rf (v)s contains a factor of exponent at least 17/7. The only extensions of w not checked above are those obtained by inserting a letter between two blocks of f . Since every block of f begins in 00 and ends in 11, every such extension contains a cube. It remains to show that w is 7/3 + -free. We first show that f (u) is 7/3 +free. Note that f is 36-uniform, and we verify by computer that f is synchronizing. Thus, by Lemma 4.3, it suffices to check that f (x) is 7/3 + -free for every square-free word x ∈ Σ * 3 such that |x| ≤ 14, which we verify by computer. Note that every block of f (u) has prefix s = 0010 and suffix r = 0110010011. So f (u) contains the word w = r f (v)s , and hence w is 7/3 + -free. Note that s = s 01 and r = 110r , so Suppose that w contains a factor z of exponent greater than 7/3. Then z begins at one of the first three letters of w, or ends at one of the last two letters of w. Suppose first that z begins at one of the first three letters of w. We claim that the factor t r = 00110010011, which occurs starting at the third letter of w, occurs only once in w. To establish this claim, we verify the following by computer: • For every c ∈ Σ 3 , the word t r occurs exactly once in the word rf (c), and does not occur in the word f (c)s.
• For every square-free word y ∈ Σ * 3 of length 2, the word t r does not occur in f (y).
So we see that the period of z is at most 13. However, this possibility is ruled out by computer check. So we may assume that z ends at one of the last two letters of w. By a computer check similar to the one used for t r , we verify that the factor t s = 001001100100, which occurs ending at the second last letter of w, occurs only once in w. So again, we see that the period of z is at most 13. This possibility is ruled out by computer check.

Conclusion
In this paper, we have focused on extremal β-free words over the binary alphabet Σ 2 . First, we characterized the lengths of extremal 2 + -free (i.e., overlap-free) words over Σ 2 . We then made some significant progress on Problem 1.3 by establishing that there are arbitrarily long extremal β-free words over Σ 2 for every β ∈ R ext such that 2 + ≤ β ≤ 8/3. Problem 1.3 remains open for β ≥ 8/3 + .
We close with a discussion of some related problems over larger alphabets. First of all, we have the following general problem which subsumes Problem 1.3.
Problem 5.1. Let n ≥ 2 be an integer. For which β ∈ R ext do there exist arbitrarily long extremal β-free words over Σ n ?
For every integer n ≥ 2, let B n denote the set of all β ∈ R ext such that there exist arbitrarily long extremal β-free words over Σ n . While it seems plausible that B n is an interval for every n, it is not immediately obvious to us that this is the case.
We note that Dejean's theorem gives us a partial answer to Problem 5.1. The repetition threshold for n letters, denoted RT(n), is defined by RT(n) = inf{b ∈ R : there are arbitrarily long b-free words over Σ n }.
In fact, for every n ≥ 2, it is known that there are only finitely many RT(n)free words over n letters, but infinitely many RT(n) + -free words over n letters. Thus, if there are arbitrarily long extremal β-free words over Σ n , then β > RT(n).
Conjecture 5.2. For every n ≥ 2, there are arbitrarily long extremal RT(n) + -free words over Σ n .
We define the extremal repetition threshold over n letters, denoted ERT(n), by ERT(n) = sup b ∈ R : there are arbitrarily long extremal b + -free words over Σ n .
If Conjecture 5.2 is true, then ERT(n) ≥ RT(n) for every n ≥ 2. We conjecture further that ERT(n) is finite for every n ≥ 2. In fact, we make the following stronger conjecture, which subsumes Conjecture 1.5. Conjecture 5.3. Let n ≥ 2 be an integer. Then there is some number α n ∈ R ext such that for all β ∈ R ext satisfying β ≥ α n , there are no extremal β-free words over Σ n .
We close with the following problem, which appears to be quite difficult.
Problem 5.4. For every n ≥ 2, find ERT(n) and the smallest number α n for which Conjecture 5.3 holds (if the conjecture is true). It is possible that we have α n = ERT(n) + for every n.