Avoiding abelian powers cyclically
aa r X i v : . [ c s . F L ] J u l Avoiding abelian powers cyclically
Jarkko Peltomäki ∗ ,1,2,3 and Markus A. Whiteland The Turku Collegium for Science and Medicine TCSM, University of Turku, Turku, Finland Turku Centre for Computer Science TUCS, Turku, Finland University of Turku, Department of Mathematics and Statistics, Turku, Finland Max Plank Institute for Software Systems, Saarbrücken, Germany
Abstract
We study a new notion of cyclic avoidance of abelian powers. A finite word w avoids abelian N -powers cyclically if for each abelian N -power of period m occurring in the infinite word w ω ,we have m ≥ | w | . Let A ( k ) be the least integer N such that for all n there exists a word oflength n over a k -letter alphabet that avoids abelian N -powers cyclically. Let A ∞ ( k ) be the leastinteger N such that there exist arbitrarily long words over a k -letter alphabet that avoid abelian N -powers cyclically.We prove that 5 ≤ A ( ) ≤
8, 3 ≤ A ( ) ≤
4, 2 ≤ A ( ) ≤
3, and A ( k ) = k ≥ A ∞ ( ) = A ∞ ( ) =
3, and A ∞ ( ) = Keywords : abelian equivalence, abelian power, abelian power avoidance, cyclic abelian power avoidance,circular word
Ever since the seminal work of A. Thue [36], repetitions or repetition avoidance in infinite wordshas been a central theme in the field of combinatorics on words. Thue showed that there ex-ists a ternary word which avoids squares, in symbols xx , that is, two identical blocks occurringadjacently in the word. Further, he showed that there exists a binary word avoiding cubes, i.e.,factors of the form xxx . These results are best possible concerning integral powers in terms ofthe size of the underlying alphabet. Thue’s results have inspired numerous papers on avoidingpowers culminating in the papers by Currie and Rampersad [12] and Rao [30] proving Dejean’sconjecture on repetition thresholds.An extremely prominent topic in combinatorics on words is the abelian equivalence of words.Two words u and v are abelian equivalent , in symbols u ∼ v , if each letter of the underlyingalphabet occurs equally many times in both words. This concept leads to that of an abelianpower: an abelian N -power of period m is a word u u · · · u N − such that u ∼ u ∼ . . . ∼ u N − and the words u i have common length m . Thus avoidance of abelian squares, abelian cubes, etc.can be considered. Erd˝os suggested in 1957 the problem of whether abelian squares are avoidableon four letters [14]. Thus the problem of searching for an alphabet of minimal cardinality overwhich an infinite word avoiding abelian squares exists was initiated. Evdokimov [15] gave thefirst upper bound of 25. Later, Pleasants [28] lowered the bound to five, and finally in 1992Keränen [21] answered Erd˝os’s question in the positive by constructing an appropriate infinite ∗ Corresponding author.E-mail addresses: [email protected] (J. Peltomäki), [email protected] (M. A. Whiteland). k -abelian equivalence and bino-mial equivalence. Of course, there is research on abelian equivalence beyond avoidance. We referthe reader to the recent survey [29]. Abelian equivalence has also been studied on graphs: thestudy of abelian 2-power-free graph colorings has been initiated under the term anagram-freecolorings in [19, 38]. A coloring of a graph is anagram-free if no sequence of colors correspond-ing to a path in the graph is an abelian 2-power. We remark that anagram-free colorings of cyclescorrespond to circular avoidance of abelian 2-powers (see below).A notion related to this paper is the notion of circular avoidance . A word w avoids N -powers circularly if no word in the conjugacy class of w contains an N -power as a factor. This is a morerestrictive type of avoidance and more difficult to study because the language of words avoiding N -powers circularly is not closed under taking factors. This results in interesting phenomena.For example, Currie shows in [9] that there exists a ternary word of length n avoiding squarescircularly for n ≥
18, but no such word of length 17 exists. For more on this notion, see, e.g.,[11] and references therein. According to our knowledge no research on the abelian analogue ofcircular avoidance exists.This paper introduces a stronger form of the circular avoidance called cyclic avoidance, andwe mainly study it in the abelian setting. A word w avoids abelian N -powers cyclically if anyabelian N -power occurring in w ω = ww · · · has period at least the length | w | of w . The differencebetween circular and cyclic avoidance is that, in cyclic avoidance, periods up to length | w | − ⌊| w | / N ⌋ are disallowed in w ω .Cyclic avoidance of abelian powers was introduced in the recent paper [27] by the authors of thispaper. There it served as a tool to construct infinite words with prescribed growth rate of abelianexponents. Due to the different focus, the abelian cyclic avoidance was only briefly studied in[27] and only to the extent that was necessary for the main result of that paper. The purpose ofthis paper is to extend this preliminary research by considering the question of what is the leastnumber of letters required to avoid abelian N -powers cyclically.Let A ( k ) be the least integer N such that for all n there exists a word of length n over a k -letteralphabet that avoids abelian N -powers cyclically. Similarly, let A ∞ ( k ) be the least integer N such that there exist arbitrarily long words over a k -letter alphabet that avoid abelian N -powerscyclically. The main results of this paper are as follows. Theorem 1.1.
We have ≤ A ( ) ≤ , ≤ A ( ) ≤ , ≤ A ( ) ≤ , and A ( k ) = for k ≥ . The lower bound for A ( ) might be a bit surprising at first sight. However, it can be checkedthat no binary word of length 8 avoids abelian 4-powers cyclically. The bounds for A ( ) , A ( ) ,and A ( ) are quite straightforward from the results of Dekking and Keränen mentioned previ-ously, but the upper bound for A ( ) requires an explicit construction.Extending the results of [27], we prove the following theorem. Theorem 1.2.
We have A ∞ ( ) = , A ∞ ( ) = , and A ∞ ( ) = . The last result of the theorem can be seen as progress in resolving a conjecture appearing in[38, p. 17], which reads: for all but finitely many n , there exists a four-letter word of length n avoiding abelian 2-powers circularly. 2e also extend previous results of Aberkane and Currie [1, 9] concerning the circular avoid-ance of powers to our cyclic setting. The results are as follows. Theorem 1.3.
If n / ∈ {
5, 7, 9, 10, 14, 17 } then there exists a word of length n over a -letter alphabet thatavoids -powers cyclically. Theorem 1.4.
For each n, there exists a word of length n over a -letter alphabet that avoids + -powerscyclically. The paper is structured as follows. In Section 2, we introduce notation and the main notions.We develop preliminary properties of cyclic abelian repetitions, and recall relevant results fromthe literature. In Section 3 we prove Theorem 1.2. The binary and ternary cases were provedalready in our previous work. The case of the four letter alphabet requires some technical de-velopments. In Section 4, we prove Theorem 1.1. The nonbinary results follow quite straight-forwardly from results in the literature when combined with our observations in Section 3. Theupper bound for the binary case requires an involved construction, splitting into even and oddlength words, and is the main technical part of the section. In Section 5, we extend known resultson circular avoidance of ordinary powers to our cyclic setting. We then conclude with futuredirections of research in Section 6.
We use standard terminology and notation of combinatorics on words; see [23, 24] for standardreferences. Let A be an alphabet , that is, a finite set of letters , or symbols . A word over the alphabet A is a sequence of letters of A obtained by concatenation. We denote the empty word by ε . Thelength of a word w is denoted by | w | , and the symbol | w | a stands for the number of occurrences ofthe letter a in w . If u and v are two words, then we denote their concatenation by uv . If w = uzv ,then z is a factor of w . If u = ε (resp. v = ε ), then z is a prefix (resp. suffix ) of w . A word z is a proper prefix (resp. proper suffix ) of w if z is a prefix (resp. suffix) of w and z = ε and z = w . If z isa factor of w , then we say that z occurs in w . If w = uv , then by u − w and wv − we respectivelymean the words v and u . If w = uu · · · u where u is repeated N times, we write w = u N and saythat w is an (ordinary) N-power of period | u | . A fractional power with exponent R , R >
1, is a wordof the form x N x ′ , where x ′ is a prefix of x and R = N + | x ′ | / | x | . The set of all words over A isdenoted by A ∗ . A language is a subset of A ∗ . A word w is primitive if w = u n only when n =
1. Ifthere exist words x and y such that u = xy and v = yx , then we say that u and v are conjugate . If w = a a · · · a n − , a i ∈ A , then the reversal of w is the word a n − · · · a a .An infinite word w is a mapping from N → A (we index words from 0). We refer to infinitewords in boldface symbols. We denote the infinite repetition of a finite word u by u ω .Let us define the Parikh mapping ψ : A ∗ → N | A | by setting ψ ( w ) = ( | w | a ) a ∈ A . We refer to thevector ψ ( w ) as the Parikh vector of w . Definition 2.1.
Let u , v ∈ A ∗ . We say that u and v are abelian equivalent if ψ ( u ) = ψ ( v ) .The following definition generalizes N -powers. Definition 2.2.
Let u , . . ., u N − , N ≥
2, be abelian equivalent and nonempty words of commonlength m . Then their concatenation u · · · u N − is an abelian N-power of period m and exponent N . If a word (finite or infinite) w does not contain as factors abelian N -powers, then we say that w avoids abelian N-powers or that w is abelian N-free .The next definition is central to this paper. Definition 2.3.
Let w be a word. Then w avoids abelian N -powers cyclically if for each abelian N -power of period m occurring in the infinite word w ω , we have m ≥ | w | .3 xample 2.4. Let w = w and w avoid abelian 5-powers. However, theword w has the abelian 5-power 100 · · · ·
001 of period 3 as a prefix. Therefore w does not avoid abelian 5-powers cyclically. Since w contains an abelian 6-power of period 4beginning from the second letter, the word w does not avoid abelian 6-powers cyclically either.By a straightforward inspection, it can be seen that it avoids abelian 7-powers cyclically. This factis immediate from the example following the next lemma.Notice that there might not be an integer N such that a word w avoids abelian N -powerscyclically. This happens when, e.g., w is conjugate to an abelian power. The following resultcharacterizes this situation. Lemma 2.5.
A word w avoids abelian | w | -powers cyclically if and only if for each k < | w | , w k is notconjugate to an abelian power with period less than | w | . Further, if w does not avoid abelian | w | -powerscyclically, then it does not avoid abelian N-powers cyclically for any N.Proof. If w is such that w k is conjugate to an abelian power of period m with m < | w | , then it isimmediate that w does not avoid abelian N -powers cyclically for any N . Suppose that w is suchthat for each k < | w | , the word w k is not conjugate to an abelian power with period less that | w | .Consider an abelian | w | -power u · · · u | w |− of period m occurring in w ω . By conjugating w ifnecessary, we may assume that u · · · u | w |− is a prefix of w ω . Let ℓ = m | w | / gcd ( m , | w | ) so that u · · · u ℓ / m − = w ℓ / | w | . The assumption then implies that m ≥ | w | or ℓ / | w | ≥ | w | . The latter alsoimplies that m ≥ | w | , so w avoids abelian | w | -powers cyclically.The previous example shows that the exponent | w | in the above characterization is tight. Weapply the above characterization to a subclass of words avoiding abelian | w | -powers cyclically. Example 2.6.
Let w be a word over A with gcd ( {| w | a : a ∈ A } ) =
1. This is satisfied, e.g., when w is not a power of a letter and | w | is a prime number. We claim that w avoids abelian | w | -powerscyclically. If w k is conjugate to an abelian N -power u · · · u N − with N > k , then k = gcd ( {| w k | a : a ∈ A } ) = gcd ( { N | u | a : a ∈ A } ) = N gcd ( {| u | a : a ∈ A } ) ≥ N > k which is impossible. Thus if w k is conjugate to an abelian power, the period of this abelian powermust be at least | w | . The claim follows from Lemma 2.5.The condition gcd ( {| w | a : a ∈ A } ) = Lemma 2.7.
Assume that x ω contains an abelian N-power of period m with | x | ≤ m < | x | . Then itcontains an abelian N-power with period | x | − m.Proof. There is nothing to prove when m = | x | , so we may assume that m > | x | . Withoutloss of generality, we may further assume that x ω begins with an abelian N -power u · · · u N − .We show, by induction on N , that if x ω begins with an abelian N -power with period m , then theword x N − ends with an abelian N -power s N − · · · s of period | x | − m .Consider first the base case N =
2. Since m satisfies | x | < m < | x | , we have | u | < | x | < | u u | . We may write x = u s and u = s p , where s is the length | x | − m suffix of x and p is aprefix of x . Notice that | p | < m , so we have u = ps for the suffix s of u of length | s | . We findthat 0 = ψ ( u ) − ψ ( u ) = ψ ( ps ) − ψ ( s p ) = ψ ( s ) − ψ ( s ) .Thus s is abelian equivalent to s , and x ends with the abelian 2-power s s .4et then N >
2. By proceeding as in the base case, we find that x ends with the abelian 2-power s s of period | x | − m . Consider the conjugate z = s u of x : the word z ω begins withthe abelian power u · · · u N − . By the induction hypothesis, z N − ends with the abelian power s N − · · · s of period | x | − m . To conclude the proof, we notice that x N − = u z N − s . The claimfollows. A ∞ ( k ) The aim of this section is to prove Theorem 1.2. Recall that A ∞ ( k ) is the least N such that thereexist arbitrarily long words over a k -letter alphabet that avoid abelian N -powers cyclically. Ourconstructions for the main result of the section involve building arbitrarily long words with mor-phisms. Next we recall the definition of a morphism and related abelian avoidance results.A morphism σ : A ∗ → B ∗ is a mapping such that σ ( uv ) = σ ( u ) σ ( v ) for all words u , v ∈ A ∗ .The morphism σ is prolongable on a letter a if σ ( a ) has prefix a and lim n → ∞ | σ n ( a ) | = ∞ . Thusiterating σ on the letter a produces an infinite word that is a fixed point of σ . We denote thisfixed point by σ ω ( a ) . The set { w : w is a factor of σ n ( a ) for some n ≥ a ∈ A } is called thelanguage of the morphism σ . Definition 3.1.
A morphism σ : A ∗ → B ∗ is abelian N-free if σ ( w ) is abelian N -free for all abelian N -free words w in A ∗ .Notice that | σ ( a ) | ≥ a when σ is a prolongable abelian N -free morphism.Indeed, if σ ( a ) = ε and σ ( b ) = ε , then the abelian N -free word bab N − has an abelian N -powerin its image.There are several results in the literature concerning abelian N -free morphisms. For example,Dekking gave sufficient conditions for a morphism to be abelian N -free in [13]. Later, Carpi ex-tended the results of Dekking by giving sharper sufficient conditions for a morphism to be abelian N -free in [6]. It is worth mentioning that Carpi’s general conditions, for abelian N -freeness, arenecessary and sufficient once the domain alphabet has cardinality at least 6 [6, Proposition 2].It remains open to this day, whether the conditions characterize abelian N -free morphisms forsmaller domain alphabets.Let us recall some morphisms that are abelian N -free for small values of N . The first is themorphism σ , found in [10], that satisfies Dekking’s conditions for the exponent 4. The morphism σ is different from the morphism of [13, Thm. 1]. We use this different morphism to reduce theamount of computations required to prove Theorem 4.3. See [10, Example 2] for the proof of thefollowing lemma. Lemma 3.2.
The morphism σ : 0 , satisfies Dekking’s conditions for the exponent . Itis thus abelian -free. By [13, Thm. 2], the morphism σ : 0
022 satisfies Dekking’s conditionsfor the exponent 3 and is thus abelian 3-free. The above two morphisms are prolongable onthe letter 0. It thus follows that the infinite words σ ω ( ) and σ ω ( ) avoid abelian 4-powers and3-powers respectively.Let π : 0
1, 1
2, 2
3, 3
0. Consider the morphism φ : {
0, 1, 2, 3 } ∗ → {
0, 1, 2, 3 } ∗ defined by setting φ ( ) = · φ ( ) = π ( φ ( )) , φ ( ) = π ( φ ( )) , φ ( ) = π ( φ ( )) . 5eränen proved in the breakthrough paper [21] that the fixed point φ ω ( ) of the morphism φ isabelian 2-free. See also his more recent paper [22] for additional morphisms with this property.Carpi simplified Keränen’s proof in [6] by showing that it satisfies Carpi’s conditions for theexponent 2, and it is thus abelian 2-free. It is noteworthy that the morphism does not satisfyDekking’s conditions for the exponent 2.Let us prove a general result related to abelian N -power cyclical avoidance and abelian N -freemorphisms. Proposition 3.3.
Let σ : A ∗ → B ∗ be an abelian N-free morphism, and assume that w in A ∗ is a wordthat avoids abelian N-powers cyclically. If N > , then σ ( w ) avoids abelian N-powers cyclically. IfN = and | w | ≥ , then σ ( w ) avoids abelian -powers cyclically.Proof. Suppose for a contradiction that σ ( w ) does not avoid abelian N -powers cyclically. As-sume thus that u · · · u N − is an abelian N -power occurring in σ ( w ) ω with | u | < | σ ( w ) | . ByLemma 2.7, we may assume that | u | ≤ ⌊| σ ( w ) | /2 ⌋ , so | u · · · u N − | ≤ N ⌊| σ ( w ) | /2 ⌋ ≤ N | σ ( w ) | /2 ≤ ⌈ N /2 ⌉| σ ( w ) | .We conclude that u · · · u N − is a factor of σ ( w ) ⌈ N /2 ⌉ + . Let a · · · a ℓ − , a i ∈ A , be a factorof w ⌈ N /2 ⌉ + of minimal length for which σ ( a · · · a ℓ − ) contains u · · · u N − . We may write σ ( a · · · a ℓ − ) = p u · · · u N − s ℓ − with σ ( a ) = p s and σ ( a ℓ − ) = p ℓ − s ℓ − . Since σ is abelian N -free, it follows that a · · · a ℓ − contains an abelian N -power v · · · v N − . As w avoids abelian N -powers cyclically, we have | v | ≥ | w | . Therefore the word v i has a conjugate of w as a factor,so | σ ( v i ) | ≥ | σ ( w ) | for all i . Thus N | σ ( w ) | ≤ | σ ( v · · · v N − ) | ≤ | σ ( a · · · a ℓ − ) | ≤ | σ ( w ) ⌈ N /2 ⌉ + | = ( ⌈ N /2 ⌉ + ) | σ ( w ) | .This inequality holds only when N ≤ N ≥
4, this contra-diction suffices for the claim. For the remainder of the proof, we operate under the assumption N ≤
3. Observe that the above computation shows that | σ ( v · · · v N − ) | = | σ ( a · · · a ℓ − ) | = N | σ ( w ) | . It follows that v i = w for all i and w N = a · · · a ℓ − .We claim that either | p u | ≥ | σ ( w ) | or | u N − s ℓ − | ≥ | σ ( w ) | . Indeed, this is clear if N = N = | p u | , | u s ℓ − | < | σ ( w ) | , then | u | > | σ ( w ) | contrary to our assumptions. Weassume that | p u | ≥ | σ ( w ) | ; the other case is symmetric.Next we claim that | p u u | ≤ | σ ( w ) | . If not, then | p | > | σ ( w ) | ≥ | σ ( a ) | because | u u | ≤ ⌊| σ ( w ) /2 |⌋ ≤ | σ ( w ) | by our assumption. Since p is a prefix of σ ( a ) , this is impossible. Wemay thus write σ ( w ) = pu s in such a way that p u = σ ( w ) p .Observe that ψ ( u ) = ψ ( σ ( w )) − ψ ( p ) + ψ ( p ) and ψ ( u ) = ψ ( σ ( w )) − ψ ( p ) − ψ ( s ) . Since ψ ( u ) = ψ ( u ) , we conclude that ψ ( p ) − ψ ( p ) = ψ ( p ) + ψ ( s ) . Since the Parikh vector ψ ( p ) + ψ ( s ) has nonnegative entries, we see that p is a prefix of p (both words are prefixes of σ ( w ) ). Weconclude that the words sp and p − p are abelian equivalent. Thus by writing sp = sp · p − p ,we see that sp is an abelian 2-power. Suppose now that N =
2. This implies that s = s ℓ − , so s ℓ − p is an abelian 2-power. Since s ℓ − p is a factor of σ ( a ℓ − a ) , it must be that a ℓ − = a as σ is abelian 2-free. Therefore w ω contains the abelian 2-power a ℓ − a of period 1. Since w avoidsabelian 2-powers cyclically, we infer that | w | =
1. This gives the latter claim.Suppose finally that N =
3. Then s σ ( w ) = u s ℓ − . We have ψ ( u ) = ψ ( σ ( w )) + ψ ( s ) − ψ ( s ℓ − ) . Since ψ ( u ) = ψ ( u ) = ψ ( u ) , we get ψ ( p ) − ψ ( p ) = ψ ( p ) + ψ ( s ) = ψ ( s ℓ − ) − ψ ( s ) .Since ψ ( p ) + ψ ( s ) has nonnegative entries, we conclude that s is a suffix of s ℓ − . We may nowwrite s ℓ − p = s ℓ − s − · sp · p − p and conclude that s ℓ − p is an abelian 3-power. Now s ℓ − p isa factor of σ ( a ℓ − a ) , so the image of the abelian 3-free word a ℓ − a contains an abelian 3-power.This contradicts the fact that σ is abelian 3-free. This proves the former claim.6otice that for the case N = | w | ≥ φ ( ) does not (here φ is Keränen’s morphism). This is evident from the fact that φ ( ) begins and ends with the letter0, so φ ( ) contains the abelian 2-power 00.The fact that A ∞ ( ) = A ∞ ( ) = A ∞ ( ) =
2, and we do this by iterating Keränen’s morphism φ on suitable words. Proof of Theorem 1.2.
Recall the morphisms σ and σ defined above. They are abelian 4-free andabelian 3-free, respectively. Now the word 0 avoids abelian N -powers cyclically for all N ≥ ( σ n ( )) n and ( σ n ( )) n avoid abelian N -powers cyclically for N = N =
3, respectively. As the morphisms are prolongable on 0,this establishes that A ∞ ( ) = A ∞ ( ) = ( φ n ( )) n avoid abelian 2-powers cyclically. Therefore A ∞ ( ) = φ is pro-longable on 0. A ( k ) Recall that A ( k ) is the least N such that for all n there exists a word of length n over a k -letteralphabet that avoids abelian N -powers cyclically. This section is devoted to proving Theorem 1.1.When k ≥
3, the idea is simply to add a new letter to a word avoiding abelian N -powers cyclically.For k =
2, this idea does not work, and we provide an explicit construction of the required words.
Lemma 4.1.
Let w be a word that avoids abelian N-powers and a letter that does not appear in w. Thenthe word w avoids abelian N-powers cyclically.Proof.
Set w = ( w ) ω , and assume for a contradiction that an abelian N -power u · · · u N − suchthat | u | < | w | occurs in w . By Lemma 2.7, we may assume that | u | ≤ | w | . Thus | u u | ≤| w | and u u at most once. Thus u , and so u · · · u N − mustbe a factor of w . This contradicts the assumption that w avoids abelian N -powers. Theorem 4.2.
We have ≤ A ( ) ≤ , ≤ A ( ) ≤ , and A ( k ) = for k ≥ .Proof. It is straightforward to verify that every ternary word of length 8 contains an abelian 2-power, so A ( ) ≥
3. Obviously A ( k ) ≥ k ≥ σ from Lemma 3.2. Taking w to be a factor of σ ω ( ) of length n −
1, we see by an application of Lemma 4.1 that the word w n over thealphabet {
0, 1, } avoids abelian 4-powers cyclically. In addition, the word 0 avoids abelian 4-powers cyclically, so A ( ) ≤ σ and φ , as defined in Section 3, are abelian 3-free and abelian 2-free, respec-tively. Similar to the previous paragraph, we see that A ( ) ≤ A ( ) ≤ Theorem 4.3.
We have ≤ A ( ) ≤ . We prove Theorem 4.3 by explicitly constructing the required words for each length. Ourconstruction is inspired by the proof of [4, Thm. 4]. Consider the morphism σ : 0 w of its fixed point σ ω ( ) of length n . Let h : 0
1, 1 w be7he reversal of h ( w ) . Set f = w ⋄ w , g = ww , and g = w • w ,where ⋄ ∈ {
0, 1 } and w • is obtained from w by changing its final letter to 0. We further define F = f ω , G = g ω , and G = g ω . Recall that the words w and w do not contain abelian 4-powersas factors. This follows from Lemma 3.2 and the discussion following it. Furthermore, w • avoidsabelian 5-powers.In Subsection 4.1, we prove that f avoids abelian 8-powers cyclically for all n . Subsection 4.2establishes that g avoids abelian 8-powers cyclically if n is odd and g avoids abelian 8-powerscyclically when n is even. These results establish that A ( ) ≤
8. Theorem 4.3 follows fromthe observation that there does not exist a binary word of length 8 avoiding abelian 4-powerscyclically. However, such a word exists in the circular sense (see the introduction): 00010011.The approach taken in Subsection 4.1 is identical to that of Subsection 4.2. Several of thestructural lemmas carry over with very minor modifications. In particular, we encourage thereader to notice that the presence of the symbol ⋄ does not often play any role. We shall makeuse of the following notion. Definition 4.4.
Let u be a binary word over the alphabet {
0, 1 } , and define ∆ ( u ) = | u | − | u | . If ∆ ( u ) > ∆ ( u ) < ∆ ( u ) = u is light (resp. heavy , neutral ).Let us first establish some properties of the fixed point σ ω ( ) of σ . In particular, we considerproperties of short factors of σ ω ( ) , which can be verified with the help of a computer.The word w below refers to the construction of the words f , g , and g . Lemma 4.5.
If u is a factor of w such that | u | ≥ , then u is light.Proof. It is straightforward to check that if u is a factor of the language of σ such that 29 ≤ | u | ≤ × =
58, then u is light. Any factor of length at least 58 can be written as a concatenation ofwords of length between 29 and 58, so it follows that all factors u with | u | ≥
29 are light.
Lemma 4.6.
If u is a factor of w such that | u | < , then ∆ ( u ) ≥ − .Proof. This is a finite check.
Lemma 4.7.
If u is a factor of w such that | u | ≥ , then ∆ ( u ) ≥ .Proof. Let u be a factor of w such that | u | ≥ × =
174 and factorize u = u · · · u in such a waythat | u i | ≥
29 for all i . Since | u i | ≥
29, we have ∆ ( u i ) > ∆ ( u ) = ∑ i = ∆ ( u i ) ≥
6. It can be verified with the help of a computer that if 64 ≤ | u | < ∆ ( u ) ≥ The aim of this subsection is to prove the following proposition.
Proposition 4.8.
The word f avoids abelian -powers cyclically. While the letter ⋄ can be freely chosen to be either 0 or 1, we use the symbol as a marker inthe proofs that follow. Proposition 4.8 can be verified to be true when | w | < × = | w | ≥ u · · · u N − be an abelian N -power occurring in F , and consider a word u i for some i . Weclassify the word u i as follows. 8 · · F : u u u α β α β α β w w Figure 1: A depiction of the structure of F . The words u and u are of type A and u is of type B.(A) u i = α i ⋄ β i for a suffix α i of w and a prefix β i of w ;(B) u i = α i β i for a nonempty suffix α i of w and a nonempty prefix β i of w .In the proofs, we implicitly use the above factorizations using the words α i and β i . Noticethat it is not necessary for u i to have type A or B. See Figure 1 for clarification.The following simple observation is very important in the subsequent proofs. Lemma 4.9.
Suppose that u and v are words of common length such that | u | ≥ . If u is a factor of wand v is a factor of w, then u and v are not abelian equivalent.Proof. If u is a factor of w and | u | ≥
29, then u is light by Lemma 4.5. If v is a factor of w , then v isa factor of w and must thus also be light. This means that v is heavy, so u and v cannot be abelianequivalent.Next we show that any abelian 8-power occurring in F must have a relatively large period. Lemma 4.10.
If an abelian -power of period m occurs in F , then m > | w | .Proof. Assume for a contradiction that F contains an abelian 8-power u · · · u such that | u | ≤ | w | . There exists u i such that u i is of type A or B because w and w avoid abelian 4-powers. Wesuppose that u i is of type A; the case that it is of type B is analogous. Suppose first that | u | < i ≤
3, then u i + u i + u i + u i + is a factor of w because | w | ≥ × = w avoids abelian 4-powers. Thus i ≥
4, but then u i − u i − u i − u i − is an abelian 4-power occurringin w . We conclude that | u | ≥ ≤ i ≤
6, so that u i − and u i + exist. Since | u | ≤ | w | , the word u i + endsbefore the end of w and the word u i − begins after the beginning of w . Therefore u i − is a factorof w and u i + is a factor of w . Lemma 4.9 shows that u i − and u i + cannot be abelian equivalent;a contradiction. Suppose then that i =
0. Then u is a factor of w since | u | ≤ | w | . Since w avoids abelian 4-powers, the word u u u u cannot be a factor of w . Thus u , u , or u is oftype B. Consequently, one of the words u , u , and u must be a factor w . This again contradictsLemma 4.9. The case i = ∆ ( u i ) may take fora u i of type A or B depending on the lengths of the corresponding words α i and β i . Lemma 4.11.
Suppose that the word F contains an abelian N-power u · · · u N − . Say u i is of type B andwrite u i = α i β i .(i) If | α i | ≥ | β i | , then ∆ ( u i ) ≥ − .(ii) If | α i | ≤ | β i | , then ∆ ( u i ) ≤ .Proof. Suppose that | α i | ≥ | β i | . Since β i is a prefix of w , the word β i is a suffix of w . We may thuswrite α i = z β i for some word z . Since | β i β i | = | β i β i | , we have ∆ ( u i ) = ∆ ( z ) . The word z isa factor of w , so if ∆ ( z ) ≤
0, then | z | <
29 by Lemma 4.5, and hence ∆ ( z ) ≥ − emma 4.12. Suppose that the word F contains an abelian N-power u · · · u N − . Say u i is of type A andwrite u i = α i ⋄ β i .(i) If | α i | ≥ | β i | , then ∆ ( u i ) − ∆ ( ⋄ ) ≤ .(ii) If | α i | ≤ | β i | , then ∆ ( u i ) − ∆ ( ⋄ ) ≥ − .Proof. This proof is similar to that of Lemma 4.11. Say | α i | ≥ | β i | . Then β i is a suffix of w , and wemay write α i = z β i . Thus ∆ ( α i β i ) = ∆ ( z ) . If ∆ ( z ) ≥
0, then ∆ ( z ) ≤ ∆ ( u i ) = ∆ ( α i ⋄ β i ) = ∆ ( z ) + ∆ ( ⋄ ) ≤ + ∆ ( ⋄ ) . Claim (ii) is analogous.We aim to combine Lemma 4.10 and the following observation. Together they imply that ifan abelian 8-power u · · · u occurs in F , then each of the factors u i has type A or type B. Lemma 4.13.
Let u u u be an abelian -power occurring in F . If(i) u occurs in w or w or(ii) u occurs in w or w,then | u | ≤ | w | .Proof. Assume on the contrary that | u | > | w | and u occurs in w . Now | u | ≥
29, so u islight, and thus u is also light. Since | u | > | w | , the word u is of type B. If | α | ≤ | β | , then ∆ ( u ) = ∆ ( u ) ≤ | w | ≥ | u | > | w | ≥ | α | > | β | . Since u α is a suffix of w , it followsthat | u α | ≤ | w | . Consequently, we have | β u | ≤ | w | which means that u is a factor of w . Thiscontradicts Lemma 4.9 since u is a factor of w .The remaining cases are proved by applying the analogous Lemma 4.12.We next prove the main technical lemma of this part. The proof of Proposition 4.8 is almostimmediate after this. Lemma 4.14.
The word F does not contain abelian -powers of period m such that m ≤ | w | .Proof. Assume for a contradiction that F contains an abelian 8-power u · · · u such that | u | ≤| w | . By Lemma 4.10, we may assume that | u | > | w | . By Lemma 4.13, the words u , . . ., u arenot factors of w or w . Therefore each u i is of type A or B. In fact, the words u , u , u , and u areof the same type, as are u , u , u , and u . Moreover, the word u is of type A if and only if u isof type B .Notice that v v v v , with v i = u i u i + , is an abelian 4-power of period 2 | u | occurring in F .Let M = | f | − | u | . Since | u | ≤ | w | < | f | /2, we have M >
0. By applying Lemma 2.7, we seethat F contains an abelian 4-power s s s s of period M . In fact, by inspecting the proof of theaforementioned lemma, the abelian 4-power s · · · s ends where v · · · v begins.Assume that u is of type A, the other case being symmetric. Let us write w = β − α for aword β − . Since u is of type B , we may write v = α ⋄ w β . Moreover, we have β − = β s ′ with | s ′ | = M . Since s · · · s ends where v · · · v begins, we see that s ′ = s . Repeating the argumentfor v i , i =
1, 2, 3, in place of v , we see that β i − = β i + s i . Hence β − = β s s s s . But now w contains the abelian 4-power s · · · s , which is absurd. Proof of Proposition 4.8.
Suppose for a contradiction that F contains an abelian 8-power of period m such that m < | f | = | w | +
1. By Lemma 2.7, we may suppose that m ≤ | w | . However,Lemma 4.14 indicates that no such abelian power exists. This is a contradiction.10 .2 Even Length Case In this section, we prove the following two propositions.
Proposition 4.15.
The word g avoids abelian -powers cyclically if | w | is odd. Proposition 4.16.
The word g avoids abelian -powers cyclically if | w | is even. As the reader might have observed, the letter ⋄ often did not play a particular role in theproofs of Subsection 4.1. This means that the previous lemmas transfer to the case of the words G and G mostly intact. Consequently, we omit repetitive details from the proofs of this sectionand indicate only what has changed.Similar to Subsection 4.1, let u · · · u N − be an abelian N -power occurring in G such that | u | ≤ | w | , and consider a word u i for some i . We classify the word u i as follows.(A) u i = α i β i for a nonempty suffix α i of w and a nonempty prefix β i of w .(B) u i = α i β i for a nonempty suffix α i of w and a nonempty prefix β i of w .For an abelian N -power u · · · u N − occurring in G such that | u | ≤ | w | , we define the type of u i as follows.(A) u i = α i β i for a nonempty suffix α i of w • and a nonempty prefix β i of w .(B) u i = α i β i for a nonempty suffix α i of w and a nonempty prefix β i of w • .Propositions 4.15 and 4.16 can be again verified when | w | < w haslength at least 145 for the remainder of this section. In the following lemmas, we shall makeno use of the parity of | g | or | g | . In fact, the parity shall only play a role in the proofs ofProposition 4.15 and Proposition 4.16 at the end of this section. Lemma 4.17.
If an abelian -power of period m occurs in G or G , then m > | w | .Proof. Assume for a contradiction that either of the words contains an abelian 8-power u · · · u with period m ≤ | w | . We first show that u i is of type A or type B for some i . Assume the contrarythat no u i is of type A or type B. Say the word u occurs in w and that w is followed by w ′ where w ′ ∈ { w , w • } . Since w avoids abelian 4-powers, one of the words u , u , or u , say u j , is a prefixof w ′ (since they do not have a type). Since j ≤
3, we see that u j + exists. There exists u k suchthat u k is a prefix of w and w ′ = u j u j + · · · u k − for otherwise the abelian 5-power u j u j + · · · u j + is a prefix of w ′ , but neither w nor w • can have such a factor. It follows that either w ′ is an abelian N -power for some N ≤ w ′ = u j . In the former case, we have | u | = | w ′ | / N ≥ ≥ u is light by Lemma 4.5. However, the word u j , a proper prefix of w ′ , is heavy by Lemma 4.5.Therefore it must be that w ′ = u j , but this contradicts the assumption that | u j | ≤ | w | . The casethat u occurs in w ′ is symmetric.To conclude the proof, we may now follow the proof of Lemma 4.10. Notice in particular thatif u i is of type A, then α i is nonempty, and thus the change of the final letter of w does not affect u i − .The following two lemmas are combinations of Lemmas 4.11 and 4.12 adjusted for the words G and G . Lemma 4.18.
Suppose that the word G contains an abelian N-power u · · · u N − . Suppose that u i is oftype A.(i) If | α i | ≥ | β i | , then ∆ ( u i ) ≤ .(ii) If | α i | ≤ | β i | , then ∆ ( u i ) ≥ − . uppose that u i is of type B.(i) If | α i | ≥ | β i | , then ∆ ( u i ) ≥ − .(ii) If | α i | ≤ | β i | , then ∆ ( u i ) ≤ .Proof. Follow the proof of Lemma 4.11.
Lemma 4.19.
Suppose that the word G contains an abelian N-power u · · · u N − . Suppose that u i is oftype A.(i) If | α i | ≥ | β i | , then ∆ ( u i ) ≤ .(ii) If | α i | ≤ | β i | , then ∆ ( u i ) ≥ − .Suppose that u i is of type B.(i) If | α i | ≥ | β i | , then ∆ ( u i ) ≥ − .(ii) If | α i | ≤ | β i | , then ∆ ( u i ) ≤ .Proof. We show how to handle the cases (i). Say u i is of type A and | α i | ≥ | β i | . We may write α i = z β i • for a word z (here we have | β i | > ∆ ( u i ) = ∆ ( z ) + z is a factor of w , we have ∆ ( z ) ≤
3, so ∆ ( u i ) ≤ u i is of type B and | α i | ≥ | β i | . Since | α i | > | u i | ≤ | w | , wesee that | β i | < | w | . It follows that α i = z β i for a word z . Thus ∆ ( u i ) = ∆ ( z ) . Since z is a factor of w , we see that ∆ ( z ) ≥ − Lemma 4.20.
Let u u u be an abelian -power occurring in G . If(i) u occurs in w or w or(ii) u occurs in w or w,then | u | ≤ | w | .Proof. Follow the proof of Lemma 4.13 and apply Lemma 4.18 appropriately.
Lemma 4.21.
Let u u u be an abelian -power occurring in G . If(i) u occurs in w or w • or(ii) u occurs in w or w • ,then | u | ≤ | w | .Proof. We show how to handle the case where u occurs in w • . Assume on the contrary that | u | > | w | and u occurs in w • . Suppose first that u is a suffix of w • . If | u | = | w | , then u = w and consequently w • and w are abelian equivalent. This means that | w | = | w | = | w • | + = | w | +
1. Thus ∆ ( w ) = −
1, and this contradicts Lemma 4.5. Therefore | u | < | w | implying that u is a factor of w . Thus ∆ ( u ) ≥ w • , Lemma 4.6 implies that ∆ ( u ) ≤ −
1, so it is not possible that ∆ ( u ) = ∆ ( u ) .We may thus assume that u is not a suffix of w • . Since | u | > | w | , it follows that the word u is of type A. If | α | ≤ | β | , then ∆ ( u ) = ∆ ( u ) ≥ − | α | > | β | . Since u α is a suffix of w • , it follows that | u α | ≤ | w | . Hence | β u | ≤ | w | and u is a factor of w . This contradicts Lemma 4.9. Lemma 4.22.
The word G does not contain abelian -powers of period m such that m < | w | . roof. The proof of Lemma 4.14 works mostly as it is for the word G . Indeed, Lemmas 4.17and 4.20 guarantee that each u i is of type A or B. Set M = | g | − | u | . Notice that we assume | u | < | w | = | g | /2, so M >
0. The remaining arguments are the same, only the ⋄ symbol isomitted. The conclusion is that w contains an abelian 4-power of period M ending at position | w | − | α | . This is impossible. Lemma 4.23.
The word G does not contain abelian -powers of period m such that m < | w | .Proof. We proceed as in the proof of Lemma 4.14. By Lemmas 4.17 and 4.21, we may suppose thateach u i is of type A or B. Set M = | g | − | u | . Again, | u | < | w | is assumed so M >
0. Followingthe arguments of Lemma 4.14 (omitting ⋄ ), we find that w • contains an abelian 4-power endingat position | w | − | α | . Observe that since the repetition is not a suffix of w • (as | α | > w . This is a contradiction. Proof of Proposition 4.15.
Say | w | is odd, and suppose for a contradiction that G contains anabelian 8-power of period m such that m < | g | = | w | . By Lemma 2.7, we may suppose that m ≤ | w | . Lemma 4.22 implies that m = | w | . By Lemma 4.20, it is not possible that u i = w or u i = w for some i . Therefore all u i are of type A or B. We handle the case that | α | ≥ | β | ; the case | α | ≤ | β | is symmetric. Write α = z β so that ∆ ( u ) = ∆ ( z ) . When u is of type A, the word w has prefix β z and suffix z β (here u = α β = z β β ). Since m = | w | , we have | β | = | β | .Since m = | β | + | z | = | β | + | z | , we conclude that z = z . The same conclusion is reachedif u is type B. Since ∆ ( u ) = ∆ ( u ) = ∆ ( z ) , we have ∆ ( z ) = ∆ ( z ) = ∆ ( z ) , so ∆ ( z ) = | z | is even. Since | w | = m = | β | + | z | , it follows that | w | is even. This is contrary toour hypothesis that | w | is odd. Proof of Proposition 4.16.
Suppose that | w | is even, and assume for a contradiction that G containsan abelian 8-power of period m with m < | w | . As in the proof of Proposition 4.15, we see thatit must be that m = | w | . Moreover, the words u i are of type A or B by Lemma 4.21. Supposethat u is of type A and | α | ≥ | β | . The remaining cases are similar. Write u = α • β = z β • β and u = α β = z β β for some words z and z of the same length. Therefore ∆ ( u ) = ∆ ( z ) + = ∆ ( u ) = ∆ ( z ) . Since | β | >
0, it is straightforward to see that z = z . Thus ∆ ( z ) + = ∆ ( z ) , that is, | z | − | z | + = | z | − | z | . It follows that | z | + = | z | , and so | z | = | z | + | z | = | z | +
1. Therefore | z | is odd, and consequently | w | = | β | + | z | is odd.This is a contradiction.Propositions 4.15 and 4.16 together with Proposition 4.8 imply Theorem 4.3. As mentioned in the introduction, previous research has considered the avoidance of ordinarypowers in circular words. A circular word is simply a conjugacy class of words, that is, a word w avoids N -powers circularly if none of the conjugates of w contains an N -power as a factor. Thisconstrains the periods to have length at most ⌊| w | / N ⌋ while our definition of cyclic avoidancedisallows periods up to length | w | −
1. The purpose of this section is to generalize the knownresults on circular avoidance of powers to our cyclic setting.
Definition 5.1.
Let w be a word. Then w avoids N -powers cyclically if for each N -power of period m occurring in the infinite word w ω , we have m ≥ | w | .The following analogue of Lemma 2.7 is straightforward to prove. Lemma 5.2.
Assume that x ω contains an N-power of period m with | x | ≤ m < | x | . Then it containsan N-power with period | x | − m. N >
2, this is not true. For example, the word 00 avoidsabelian N -powers circularly for N >
2, but it never avoids abelian N -powers cyclically.Currie proved in [9] that if n / ∈ {
5, 7, 9, 10, 14, 17 } then there exists a word of length n overa 3-letter alphabet that avoids 2-powers circularly. By the preceding paragraph, we have thefollowing result (notice that 2-powers cannot be avoided with just two letters). Theorem 5.3.
If n / ∈ {
5, 7, 9, 10, 14, 17 } then there exists a word of length n over a -letter alphabet thatavoids -powers cyclically. Notice that for n ∈ {
5, 7, 9, 10, 14, 17 } there exists a word of length n over a 4-letter alpha-bet avoiding 2-powers cyclically. Such words are, e.g., 01023, 0102013, 010201203, 0102010313,01020103010213, and 01020103010212313. Notice in addition that for each n there exists a wordof length n over a 3-letter alphabet that avoids 2 + -powers cyclically (see below for the defini-tion). To see this, it is sufficient to observe that the words 00102, 0010012, 001001102, 0010011202,00100112001002, and 00100112001001202 avoid 2 + -powers cyclically.What is left is to determine the least exponent N such that for all n there exists a binary word w of length n such that w avoids N -powers cyclically. In the context of ordinary powers, it isnatural to consider fractional exponents, and thus we give the following definition. We do notconsider fractional abelian exponents in this paper; for discussion on this concept, see [7, 32]. Definition 5.4.
Let w be a word and N be a rational number such that N >
1. Then w avoids N + -powers cyclically if for each N + -power of period m occurring in the infinite word w ω , wehave m ≥ | w | . A word u is an N + -power if u is an R -power for some R > N .Let t be the fixed point σ ω ( ) of the morphism σ : 0
01, 1
10. The word t is the famousThue-Morse word; see [3, Sect. 1.6]. Aberkane and Currie proved in [2] that the Thue-Morse word t contains a factor avoiding 5/2 + -powers circularly for all lengths. We generalize this result toour cyclic setting. This result implies Theorem 1.4. Theorem 5.5.
For each n, there exists a factor of length n of the Thue-Morse word avoiding + -powerscyclically. It can be shown that the exponent 5/2 is optimal for binary words by inspecting all binarywords of length 5.In order to prove Theorem 5.5, we employ the automatic theorem-proving software Walnut[25]. Properties of automatic sequences [3] that are expressible in a certain first-order logic are de-cidable, and Walnut implements the decision procedure. The Thue-Morse word t is a 2-automaticword, so Walnut is applicable. We wish to keep the discussion on the decision procedure and us-age of Walnut brief, so we merely describe the logical formulas necessary to encode our problemand refer the reader to [8] for a proof of Theorem 5.3 using Walnut. See also [34].Let w be a factor of the Thue-Morse word. If w ω contains an N -power of period m such that m < | w | and N >
3, then w ω contains a 3-power of period m . Therefore in order to show that w avoids 5/2 + -powers cyclically, we only need to consider N -powers with 5/2 < N ≤
3. Noticethat such a power u is necessarily a factor of w . We first write a predicate cRepK ( i , j , m , n , p ) , K =
1, . . . , 4, that evaluates to true if and only if the factor w of length n beginning at the position i of the Thue-Morse word t is such that w K has a factor u of length m beginning at position j , i ≤ j < i + n , such that u has period p and u is not a factor of w K − . The predicate needs to bewritten somewhat awkwardly as w K is not necessarily a factor of t . For example, we havecRep2 ( i , j , m , n , p ) = ( i ≤ j < i + n ) ∧ ( i + n ≤ j + m ≤ i + n ) ∧ ( ∀ k ( j ≤ k < i + n − p ) = ⇒ t [ k ] = t [ k + p ]) ∧ ( ∀ k ( i + n − p ≤ k < i + n ) = ⇒ t [ k ] = t [ k + p − n ]) ∧ ( ∀ k ( i + n ≤ k < j + m − p ) = ⇒ t [ k − n ] = t [ k + p − n ]) .14e can then write a predicate ncyc ( i , n ) that evaluates to true if and only if the factor w of t of length n starting at position i is such that w contains a factor that has period p with 5/2 < | u | / p ≤
3. Its definition is:ncyc ( i , n ) = ∃ j , m , p (( < p < n ) ∧ ( p < m ≤ p ) ∧ ( cRep ( i , j , m , n , p ) ∨ cRep ( i , j , m , n , p ) ∨ cRep ( i , j , m , n , p ) ∨ cRep ( i , j , m , n , p ))) .Finally the following predicate evaluates to true if and only if Theorem 5.5 is true: ∀ n (( n > ) = ⇒ ( ∃ i ¬ ncyc ( i , n ))) .Inputting the above predicates to Walnut produces an automaton accepting all inputs meaningthat Theorem 5.5 is true. Obviously the main question is what is the value of A ( k ) for k =
2, 3, 4. Theorem 1.2 seems tosupport the claims that A ( ) = A ( ) =
3, and A ( ) =
2, but the first claim is false as there is nobinary word of length 8 avoiding abelian 4-powers cyclically. This leads us to ask the followingquestions.
Question.
Is it the case that A ( ) = , A ( ) = , and A ( ) = ? Question.
If n = , does there exist a word of length n over a -letter alphabet avoiding abelian -powerscyclically? Our computer experiments have not found a counterexample to the above questions amonglengths less than 150. Notice that our question whether A ( ) = A ( ) = A ( ) . We remark that the particularconstruction given here cannot be used to improve the upper bound 8 in Theorem 4.3 as someof the words constructed contain abelian 7-powers with short period. If two words that avoidabelian 4-powers are concatenated, then a priori abelian 7-powers could appear. An improvedconstruction would need to take special care to concatenate the words in such a way that theirrespective abelian 3-powers of common period do not appear too close to each other. It seemsthat no precise information on the structure and location of abelian 3-powers in words that avoidabelian 4-powers is found in the literature. Even the sets of possible periods of abelian powersoccurring in infinite words have been studied very little. The only papers in this direction are thepapers [16, 26] concerning the abelian period sets of Sturmian words. This knowledge however isnot helpful in this context as Sturmian words contain abelian powers of arbitrarily high exponent[16, Proposition 4.10]. It seems that making such concatenation arguments work for the alphabetsizes 3 and 4 is even more difficult especially because there is less room for improvement.An alternative way to improve our results would be to find infinite words whose languagecontains the sought words. For example, Justin’s morphism 0 n avoiding abelian 5-powers cyclically for n =
1, . . . , 400. Wedo not know how to prove that such a factor exists for each length. Since the the fixed pointof Justin’s morphism is automatic, it might be possible to attack this problem via automatictheorem-proving as in Section 5. The problem in this plan is that this type of automatic theorem-proving requires the problem to be written in a certain restricted first order logic and generallyabelian properties of words cannot be expressed in this logic [33, Sect. 5.2].15e have dealt in this paper only with the question of existence. A significantly harder prob-lem would be to provide a lower bound, for example, for the number of binary words of length n that avoid abelian 4-powers cyclically. We have recorded this sequence as the sequence A334831in Sloane’s On-Line Encyclopedia of Integer Sequences [35]. The first values of the sequence are 2, 2,6, 8, 10, 6, 28, 0, 36, 120, 132, 168, 364, 112.
Acknowledgements
We thank the reviewer for remarks that improved the quality of the paper. We also thank him/herfor pointing out the conjecture in [38].
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