# Abelian Anti-Powers in Infinite Words

An abelian anti-power of order k (or simply an abelian k-anti-power) is a concatenation of k consecutive words of the same length having pairwise distinct Parikh vectors. This definition generalizes to the abelian setting the notion of a k-anti-power, as introduced in [G. Fici et al., Anti-powers in infinite words, J. Comb. Theory, Ser. A, 2018], that is a concatenation of k pairwise distinct words of the same length. We aim to study whether a word contains abelian k-anti-powers for arbitrarily large k. S. Holub proved that all paperfolding words contain abelian powers of every order [Abelian powers in paper-folding words. J. Comb. Theory, Ser. A, 2013]. We show that they also contain abelian anti-powers of every order.

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## 1 Introduction

Many of the classical definitions in combinatorics on words (e.g., period, power, factor complexity, etc.) have a counterpart in the abelian setting, though they may not enjoy the same properties.

Recall that the Parikh vector of a word over a finite ordered alphabet is the vector whose -th component is equal to the number of occurrences of the letter in , . For example, the Parikh vector of over is . This notion is at the basis of the abelian combinatorics on words, where two words are considered equivalent if and only if they have the same Parikh vector.

For example, the classical notion of factor complexity (the function that counts the number of distinct factors of length of a word, for every ) can be generalized by considering the so-called abelian factor complexity (or abelian complexity for short), that is the function that counts the number of distinct Parikh vectors of factors of length , for every .

Morse and Hedlund [8] proved that an infinite word is aperiodic if and only if its factor complexity is unbounded. This characterization does not have an analogous in the case of the abelian complexity, as there exist aperiodic words with bounded abelian complexity. For example, the well-known Thue-Morse word has abelian complexity bounded by , yet it is aperiodic.

Richomme et al. [9] proved that if a word has bounded abelian complexity, then it contains abelian powers of every order — an abelian power of order is a concatenation of words having the same Parikh vector. However, this is not a characterization of words with bounded abelian complexity. Indeed, Štěpán Holub [6] proved that all paperfolding words contain abelian powers of every order, and paperfolding words have unbounded abelian complexity (a property that by the way follows from the main result of this paper). The class of paperfolding words therefore constitutes an interesting example, as they are uniformly recurrent (every factor appears infinitely often and with bounded gaps) aperiodic words with linear factor complexity.

In a recent paper [5], the first and the third author, together with Antonio Restivo and Luca Zamboni, introduced the notion of an anti-power. An anti-power of order , or simply a –anti-power, is a concatenation of consecutive pairwise distinct words of the same length. E.g., is a –anti-power.

In [5], it is proved that the existence of powers of every order or anti-powers of every order is an unavoidable regularity for infinite words:

###### Theorem 1.

[5] Every infinite word contains powers of every order or anti-powers of every order.

Note that in the previous statement there is no hypothesis on the alphabet size.

Actually, in [5] a stronger result is proved (of which we omit the statement here for the seek of simplicity) from which it follows that every aperiodic uniformly recurrent word must contain anti-powers of every order.

In this paper, we extend the notion of an anti-power to the abelian setting.

###### Definition 2.

An abelian anti-power of order , or simply an abelian –anti-power, is a concatenation of consecutive words of the same length having pairwise distinct Parikh vectors.

For example, is an abelian –anti-power. Notice that an abelian –anti-power is a –anti-power but the converse does not necessarily holds (which is dual to the fact that a –power is an abelian –power but the converse does not necessarily holds).

We think that an analogous of Theorem 1 may still hold in the case of abelian anti-powers, but unfortunately the proof of Theorem 1 does not generalize to the abelian setting.

###### Problem 1.

Does every infinite word contain abelian powers of every order or abelian anti-powers of every order?

Clearly, if a word has bounded abelian complexity, then it cannot contain abelian anti-powers of every order. However, a word can avoid large abelian anti-powers even if its abelian complexity is unbounded. Indeed, in [5] it is shown an example of an aperiodic recurrent word avoiding -anti-powers (and therefore avoiding abelian -anti-powers), and from the construction it can be easily verified that the abelian complexity of this word is unbounded.

A similar situation can be illustrated with the well-known Sierpiǹski word. Recall that the Sierpiǹski word (also known as Cantor word) is the fixed point starting with of the substitution

 σ: a→aba b→bbb

so that the word begins as follows:

 ababbbababbbbbbbbbababbbabab27a⋯

Therefore, can be obtained as the limit, for , of the sequence of words defined by: , for . Notice that for every one has .

We show that the abelian complexity of is unbounded.

###### Theorem 3.

The Sierpiǹski word does not contain –anti-powers, hence it does not contain abelian –anti-powers.

An infinite word can contain both abelian powers of every order and abelian anti-powers of every order. This is the case, for example, of any word with full factor complexity. However, finding a class of uniformly recurrent words with linear factor complexity satisfying this property seems a more difficult task. Indeed, most of the well-known examples (Thue-Morse, Sturmian words, etc.) have bounded abelian complexity, hence they cannot contain abelian anti-powers of every order — whereas, by the aforementioned result of Richomme et al. [9], they contain abelian powers of every order. Building upon the framework that Štěpán Holub developed to prove that all paperfolding words contain abelian powers of every order [6], we prove in the next section that all paperfolding words contain also abelian anti-powers of every order.

## 2 Sierpiǹski Word

Blanchet-Sadri, Fox and Rampersad [2] characterized the asymptotic behavior of the abelian complexity of a morphic word. In the following proposition, we give the precise bounds of the abelian complexity of the Sierpiǹski word.

###### Proposition 4.

The abelian complexity of the Sierpiǹski word verifies .

###### Proof.

The Sierpiǹski word is prefix normal with respect to the letter (see [4] for the definition of prefix normal word), that is, for each length , no factor of of length contains more occurrences of the letter than the prefix of length . Since contains arbitrarily long blocks of s, the number of distinct Parikh vectors of factors of of a given length is given by plus the number of s in the prefix of length . It is easy to see that the values of for which the proportion of ’s is maximal in a prefix of length are of the form , while those for which the proportion of ’s is minimal are of the form , and in both cases the prefix of length contains s. With a standard algebraic manipulation, this gives

 nlog32/2log32≤a(n)≤nlog32.

Proof of Theorem 3. Suppose that contains an –anti-power , of length . Let us then consider the first occurrence of in . Let be the smallest integer such that occurs in but not in .

Let us first suppose that no is equal to for some . Then is a factor of , so hence . Then, by minimality of , there are only two possible cases: either starts before the block , or starts in the block and ends in .

In the first case, by minimality of , ends after the block , and since no equals , we get , which is in contradiction with .

If starts in the block and ends in , is a factor of and so hence . By minimality of , ends after the block . Again, since no equals , we get , which is in contradiction with .

Let us then suppose that , so that is a factor of . The same reasoning as before holds, since and . If , is a factor of with no and we can again apply the same reasoning.

Finally, suppose that with and . Hence, is a factor of , and . If starts before the block (and ends after by minimality of ), we get since otherwise would contain two blocks , and this contradicts . If does not start before the block , then by minimality of it starts in this block, so is a factor of which ends after the block , again by minimality of . This shows that , and at the same time , which produces a contradiction. ∎

## 3 Paperfolding Words

In what follows, we recall the combinatorial framework for dealing with paperfolding words introduced in [6], although we use the alphabet instead of .

A paperfolding word is the sequence of ridges and valleys obtained by unfolding a sheet of paper which has been folded infinitely many times. At each step, one can fold the paper in two different ways, thus generating uncountably many sequences. It is known that all the paperfolding words are uniformly recurrent and have the same factor complexity , and that for [1]. Madill and Rampersad [7] studied the abelian complexity of the regular paperfolding word and proved that it is a -regular sequence. The regular paperfolding word

 p=00100110001101100010011100110110⋯

is the paperfolding word obtained by folding at each step in the same way. It can be defined as a Toeplitz word (see [3] for a definition of Toeplitz words) as follows: Consider the infinite periodic word , defined over the alphabet . Then define and, for every , as the word obtained from by replacing the symbols with the letters of . So,

 p0 =0?1?0?1?0?1?0?1?0?1?0?1?0?1?⋯, p1 =001?011?001?011?001?011?001?⋯, p2 =0010011?0011011?0010011?0011⋯, p3 =001001100011011?001001110011⋯,

etc. Thus, , and hence p does not contain occurrences of the symbol .

More generally, one can define a paperfolding word f by considering the two infinite periodic words and . Then, let be an infinite word over , called the sequence of instructions. Define where, for every , if or if . The paperfolding word f associated with b is the limit of the sequence of words defined by and, for every , is obtained from by replacing the symbols with the letters of .

Recall that every positive integer can be uniquely written as , where is called the order of (a.k.a. the -adic valuation of ), and is called the odd part of . One can verify that the previous definition of f is equivalent to the following: for every define , where Then if and if . This is equivalent to

 fi=1iffi≡2k(2+bk)mod2k+2.
###### Remark 1.

The regular paperfolding word corresponds to the sequence of instructions .

###### Definition 5.

Let f be a paperfolding word. An occurrence of in f at position is said to be of order if the letter at position is in and in . We consider the ’s occurring in as of order .

Hence, in the paperfolding word f, the ’s of order appear at positions , , the ’s of order appear at positions , , and, in general, the ’s of order appear at positions , .

Let be a paperfolding word associated with the sequence . A factor of f of length starting at position , denoted by , contains a number of ’s that is given by the sum, for all , of the ’s of order in the interval . For each , since the ’s of order are at distance one from another, the number of occurrences of ’s of order in is given by

 ⌊n−ℓ2k+2⌋+εk,bk(ℓ,n),

where depends on the sequence b (in fact, determines the positions of the occurrences of the ’s of order in f). We set

 Δ(ℓ,n)=∑k≥0εk,bk(ℓ,n)

the number of “extra” ’s in .

For example, in the prefix of length of the regular paperfolding word, we know that there are at least ’s of rank , of rank and of rank . In the interval there are three ’s of rank (at positions , and ), two ’s of rank (at positions and ), and one of rank (at position ), so we have in no extra of rank , i.e., , one extra of rank , i.e., and one extra of rank , i.e., , so that .

We set

 Ek,bk(ℓ,d,m)=(εk,bk(ℓ,ℓ+d),…,εk,bk(ℓ+(m−1)d,ℓ+md))

and

 Δ(ℓ,d,m)=∑k≥0Ek,bk(ℓ,d,m)=(Δ(ℓ,ℓ+d),…,Δ(ℓ+(m−1)d,ℓ+md)).

The factor of f of length starting at position is an abelian -power if and only if the components of the vector are all equal, while it is an abelian -anti-power if and only if the components of the vector are pairwise distinct.

The next result (Lemma 4 of [6]) will be the fundamental ingredient for the construction of abelian anti-powers in paperfolding words.

###### Lemma 6 (Additivity Lemma).

Let and be integers with and both even. Let be such that , and for each the following implication holds: if then .

Then

 Δ(ℓ,d,m)+Δ(ℓ′,d′,m)=Δ(ℓ+2rℓ′,d+2rd′,m).

Using the Additivity Lemma, Holub [6] proved that all paperfolding words contain abelian powers of every order. We will use the Additivity Lemma to prove that all paperfolding words contain abelian anti-powers of every order. We start with the regular paperfolding word, then we extend the argument to all paperfolding words.

### 3.1 Regular paperfolding word

Let

 Φ:{0,1}2→{x,y,z}00↦x01↦y10↦y11↦z

be the morphism that identifies words of length over the alphabet that are abelian equivalent. We have the following lemma:

###### Lemma 7.

Let be an integer. Let be a factor of p of length . Then, no exists such that

 Φ(p)=Φ(u1v1)⋯Φ(u2n−1v2n−1)=Φ(uq+1vq+1)⋯Φ(u2n−1v2n−1)Φ(u1v1)⋯Φ(uqvq). (1)
###### Proof.

First, notice that if is the smallest solution of (1), then . Indeed, writing , we have

 w1⋯w2n−1 =w1⋯wq′wq′+1⋯w2n−1 =wq′+1⋯w2n−1w1⋯wq′,

and since two words commute if and only if they are powers of the same word, there exists a word and positive integers and such that

 w1⋯wq′=zs and wq′+1⋯w2n−1=zt.

This gives and . By the minimality of , we have that and so divides . Thus, for some integer .

By the Toeplitz construction of p, we immediately have that

 u1v1⋯u2n−1v2n−1=av1¯¯¯av2av3¯a⋯¯¯¯av2n−1

or

 u1v1⋯u2n−1v2n−1=u1au2¯¯¯au3au4¯¯¯a⋯u2n−1¯¯¯a

with and .

Suppose and . Since is even, we have that implies . But this cannot be the case, since two consecutive letters of order occur in p at distance . Since , we have , so the factor contains at least two consecutive letters of order . Suppose that the first of such letters is ; then is at distance , so , against the hypothesis that is a solution of (1).

Thus, we must have or . Since , contains two consecutive letters of order . Let us first suppose that is a of order and is a of order . Then, . The other cases would give with a of order 1 and a of order 1, and if is a of order 1 and a of order 1 or vice versa. Every case leads to . This implies and so . By minimality of , the only solution of (1) is . ∎

###### Theorem 8.

The regular paperfolding word contains abelian -anti-powers for every .

###### Proof.

The proof is mainly based on the Additivity Lemma. Let be fixed. To prove the result it is sufficient to find a vector having pairwise distinct components. Let be an integer such that . Consider the first factor of length containing a of order in the middle; our factor is then of the form

 w1w′

with . Since for every positive integers

 pj of order i⇒pj+2i+2=pj≠pj+2i+1 (2)

then, up to applying a translation, we can suppose . In fact, the equality is true for every letter of order smaller than by (2). Now, take the smallest order of a letter in or . It is the only letter of this order in our factor since two letters of order are distant of . If we consider the factor translated by , by (2) the letters of order smaller than are the same and the letter we considered becomes a . Since the length of is and the distance between two letters of order higher than is at least , we have that in less than steps we get with every letter of order greater than being a . Writing the starting position of an occurrence in p of the factor , we set if is even or otherwise. Consider the vectors

 Δ(ℓ′,2,2k),Δ(ℓ′+2,2,2k),Δ(ℓ′+4,2,2k),Δ(ℓ′+6,2,2k),…,Δ(ℓ′+2k+1−2,2,2k).

We claim that these vectors are pairwise distinct. By contradiction, if for some with , then we have that

 Φ(pℓ′+2p+1⋯pℓ′+2p+2k+1)=Φ(pℓ′+2q+1⋯pℓ′+2q+2k+1). (3)

Since the factor we are considering is , we have and so

 Φ(pℓ′+2q+1⋯pℓ′+2q+2k+1)=Φ(pℓ′+2q+1⋯pℓ′+2p+2k+1pℓ′+2p+1⋯pℓ′+2q)

but this and (3) contradicts Lemma 7.

Finally, as the vectors are different, we use the Additivity Lemma to obtain a vector whose components are pairwise distinct: applying times the Additivity Lemma on one can obtain . It then suffices to take a sequence of integers increasing enough to have

 Σ2k−1i=0αiΔ(s′+2i,2,2k),

a vector whose components are pairwise distinct. Indeed, labelling the -th component of this vector and the -th component of , we have

 aj=aj′⇔Σ2k−1i=0αixi,j=Σ2k−1i=0αixi,j′⇔Σ2k−1i=0αi(xi,j−xi,j′)=0.

By “increasing enough”, we precisely mean , so that by decreasing induction we have that for every , with , one has . In particular, this gives , which implies . Hence, all the components are pairwise distinct and the proof is complete. ∎

### 3.2 All paperfolding words

To generalize the result above to all paperfolding words, one has to take care of the condition in the Additivity Lemma.

Lemma 7 can be modified so that the translation is not by but by , for any . Let

 ϕ:{0,1}2u→Na1⋯a2u↦|{i∣ai=1}|

be the morphism that identifies words of length over that are abelian equivalent. Then we have the following lemma, analogous to Lemma 7:

###### Lemma 9.

Let be an integer and let f be a paperfolding word. Every factor of f of length satisfies the following property: If is such that

 ϕ(f)=ϕ(a1,1⋯a1,2u)⋯ϕ(a2n−1,1⋯a2n−1,2u)= ϕ(aq+1,1⋯aq+1,2u)⋯ϕ(a2n−1,1⋯a2n−1,2u)ϕ(a1,1⋯a1,2u)⋯ϕ(aq,1⋯aq,2u),

then .

###### Proof.

The proof of Lemma 7 mainly applies here; we only need to change the part where we use the Toeplitz construction to justify . Here, in each -tuple one can find one letter of order and one letter of higher order. Using (2), we then see that is totally determined by the letter of order and the letter of higher order in . Applying again (2) to the letter of order , we can apply exactly the same reasoning as in the proof of Lemma 7 (in a sense, our new is the previous one modulo the letters of order smaller than ). ∎

Now, we can prove the main theorem:

###### Theorem 10.

Every paperfolding word f contains abelian -anti-powers for every .

###### Proof.

Let be an integer such that . As before, we will prove that f contains abelian -anti-powers, hence it will contain abelian -anti-powers. Since the alphabet is finite, there must exist a factor of b that occurs infinitely often. As before, let us start with the first block of length containing a of order in the middle; our block is then

 w1w′

with . As before, in at most two steps, we can have , and the maximum order of a letter appearing in this factor is . Again, writing the starting position of an occurrence of this factor, we set if is even or otherwise. Consider the vectors

 Δ(ℓ′,2u,2k),Δ(ℓ′+2u,2u,2k),Δ(ℓ′+2u+1,2u,2k),…,Δ(ℓ′+2u+k+1−2u,2u,2k).

Here again, these vectors are pairwise distinct: if , we have that

 ϕ(pℓ′+2up+1⋯pℓ′+2u(p+2k))=ϕ(pℓ′+2uq+1⋯pℓ′+2u(q+2k))

and this contradicts Lemma 9 because, here again, and so

 pl′+2up+1⋯pl′+2uq=pl′+2u(p+2k)+1⋯pl′+2u(q+2k).

Moreover, , using (2) and the fact that no letter of order higher than appears in the factor . So, choosing such that and , we can apply the Additivity Lemma and, as for the regular paperfolding word, construct an abelian -anti-power that occurs as a factor in f. ∎

###### Remark 2.

From Theorem 10 it follows immediately that every paperfolding word has unbounded abelian complexity.

## References

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