Critical exponent of binary words with few distinct palindromes

We study inﬁnite binary words that contain few distinct palindromes. In particular, we classify such words according to their critical exponents. This extends results by Fici and Zamboni [TCS 2013]. Interestingly, the words with 18 and 20 palindromes happen to be morphic images of the ﬁxed point of the morphism 0 7→ 01 , 1 7→ 21 , 2 7→ 0 .


Introduction
We consider the trade-off between the number of distinct palindromes and the critical exponent in infinite binary words.For brevity, every mention of a number of palindromes will refer to a number of distinct palindromes, including the empty word.Fici and Zamboni [6] show that the least number of palindromes in an infinite binary word is 9 and this bound is reached by the word (001011) ω .At the other end of the spectrum, the famous Thue-Morse word T M , fixed point of the morphism 0 → 01, 1 → 10, has the least critical exponent and infinitely many palindromes.
Our results completely answer questions of the form "Is there an infinite β + -free binary word with at most p palindromes?"In each case, we also determine whether there are exponentially or polynomially many such words.The results are summarized in Table 1.A green (resp.red) cell means that there are exponentially (resp.polynomially) many words.We have labelled the cells that correspond to an item of Theorem 3 or 7. Fici and Zamboni [6] also show that an aperiodic binary word contains at least 11 palindromes and this bound is reached by the morphic image of the Fibonacci word by 0 → 0, 1 → 01101.This word contains in particular the factor (00110100011010011010001101000110100110100011010011010001101000110100110100011010001101) Theorem 1.(a) improves this exponent to 10   3   + .Fleischer and Shallit [7] have considered the number of binary words of length n with at most 11 palindromes (sequence A330127 in the OEIS) and proved that it is Θ (κ n ), where κ = 1.1127756842787 . . . is the root of X 7 = X + 1.

Preliminaries
An alphabet A is a finite set and its elements are called letters.A word u over A of length n is a finite string u = u 0 u 1 • • • u n−1 , where u j ∈ A for all j ∈ {0, 1, . . ., n − 1}.If A = {0, 1, . . ., d − 1}, the length of u is denoted |u| and |u| i denotes the number of occurrences of the letter i ∈ A in the word u.
The set of all finite words over A is denoted A * .The set A * equipped with concatenation as the operation forms a monoid with the empty word ε as the neutral element.We will also consider the set A ω of infinite words (that is, right-infinite words) and the set ω A ω of bi-infinite words.A word v is an epower of a word u if v is a prefix of the infinite periodic word uuu • • • = u ω and e = |v|/|u|.We write v = u e .We also call u e a repetition with period u and exponent e.For instance, the Czech word kapka (drop) can be written in this formalism as (kap) 5/3 .A word is α + -free (resp.α-free) if it contains no repetition with exponent β such that β > α (resp.β α).
The critical exponent E(u) of an infinite word u is defined as E(u) = sup{e ∈ Q : u e is a factor of u for a non-empty word u} .
The asymptotic critical exponent E * (u) of an infinite word u is defined as +∞ if E(u) = +∞, and E * (u) = lim sup n→∞ {e ∈ Q : u e is a factor of u for some u of length n} , otherwise.If each factor of u has infinitely many occurrences in u, then u is recurrent.Moreover, if for each factor the distances between its consecutive occurrences are bounded, then u is uniformly recurrent.The language L(u) is the set of factors occurring in u.The language L(u) is closed under reversal if for each factor w = w 0 w 1 • • • w n−1 , its reverse w R = w n−1 • • • w 1 w 0 is also a factor of u.A word w is a palindrome if w = w R .Let us denote 0 = 1 and 1 = 0, then for any binary word w its bit complement is w Consider a factor w of a recurrent infinite word u = u 0 u 1 u 2 • • • .Let j < ℓ be two consecutive occurrences of w in u.Then the word u j u j+1 • • • u ℓ−1 is a return word to w in u.
The (factor) complexity of an infinite word u is the mapping Given a word w ∈ L(u), we define the sets of left extensions, right extensions and bi-extensions of w in u over an alphabet A respectively as A morphism is a map ψ : A * → B * such that ψ(uv) = ψ(u)ψ(v) for all words u, v ∈ A * .The morphism ψ is non-erasing if ψ(i) = ε for each i ∈ A. Morphisms can be naturally extended to infinite words by setting ψ(u If there exists N ∈ N such that M N ψ has positive entries, then ψ is a primitive morphism.By definition, we have for each u ∈ A * the following relation for the Parikh vectors ψ(u) = M ψ u.
Let u be an infinite word over an alphabet A. Then the uniform frequency f i of the letter i ∈ A is equal to α if for any sequence (w n ) of factors of u with increasing lengths It is known that fixed points of primitive morphisms have uniform letter frequencies [9].Let u be an infinite word over an alphabet A and let ψ : A * → B * be a morphism.Consider a factor w of ψ(u).We say that (w 1 , w 2 ) is a synchronization point of w if w = w 1 w 2 and for all p, s ∈ L(ψ(u)) and v ∈ L(u) such that ψ(v) = pws there exists a factorization v = v 1 v 2 of v with ψ(v 1 ) = pw 1 and ψ(v 2 ) = w 2 s.We denote the synchronization point by Given a factorial language L and an integer ℓ, let L ℓ denote the words of length ℓ in L. The Rauzy graph of L of order ℓ is the directed graph whose vertices are the words of L ℓ−1 , the arcs are the words of L ℓ , and the arc corresponding to the word w goes from the vertex corresponding to the prefix of w of length ℓ − 1 to the vertex corresponding to the suffix of w of length ℓ − 1.
Finally, this paper mainly studies properties of the words µ(p) and ν(p) that are morphic images of the word p = ϕ ω (0) studied in [2], where 3 Fewest palindromes, least critical exponent, and factor complexity

General result
Theorem 1.There exists an infinite binary β + -free word containing only p palindromes for the following pairs (p, β).Moreover, this list of pairs is optimal.
(a) (11, 10 3 ) (b) (12, 23  7 ) (c) (13,3) (d) (15, 8  3 ) (e) (18, 28  11 ) (f ) (20, 5  2 ) (g) (25, 7  3 ) Proof.The optimality is obtained by backtracking.For example, the step between items (c) and (d) is obtained by showing that there exists no infinite cubefree word containing at most 14 palindromes.The proof of the positive results is split in two cases, depending on the factor complexity of the considered words, see Theorems 3 and 7. Let us already remark that, in any case, it is easy to check that the proposed word does not contain more than the claimed number of palindromes since it only requires to check the factors up to some finite length.

Exponential cases
We need some terminology and a lemma from [11].A morphism f : Σ * → ∆ * is q-uniform if |f (a)| = q for every a ∈ Σ, and is called synchronizing if for all a, b, c ∈ Σ and u, v .
If h(w) is β-free for every α-free word w with |w| ≤ t, then h(z) is β-free for every α-free word z ∈ Σ * .
The results in this subsection use the following steps.We find an appropriate uniform synchronizing morphism h by exhaustive search.We use Lemma 2 to show that h maps every binary 7 3 + -free word (resp.ternary squarefree word) to a suitable binary β + -free word.Since there are exponentially many binary 7 3 +free words [10] (resp.ternary squarefree words [12]), there are also exponentially many binary β + -free words.Theorem 3.There exist exponentially many infinite binary β + -free words containing at most p palindromes for the following pairs (p, β).

Polynomial cases
To show that v avoids F , we consider every f ∈ F and we show by contradiction that f is not a factor of v. Proof.Consider a bi-infinite binary cubefree word w avoiding F 20 .Since w is cubefree, w is in {011, 0, 01} ω .So w = ν(v) for some bi-infinite ternary word v. Since w is cubefree, its pre-image v is also cubefree.
To show that v avoids F , we consider every f ∈ F and we show by contradiction that f is not a factor of v.We construct the set S 20 18 defined as follows: a word v is in S 20 18 if and only if there exists a 13  5 -free binary word pvs containing at most 18 palindromes and such that |p| = |v| = |s| = 20.From S 20 18 , we construct the Rauzy graph R 20 18 such that the vertices are the factors of length 19 and the arcs are the factors of length 20.We notice that R 20 18 is disconnected.It contains four connected components that are symmetric with respect to reversal and bit complement.Let C 20 18 be the connected component which avoids the factor 1101.We check that C 20 18 is identical to the Rauzy graph of the factors of length 19 and 20 of µ(p).Now we consider a bi-infinite 13  5 -free binary word w with 18 palindromes.So w corresponds to a walk in one of the connected components of R 20 18 , say C 20 18 without loss of generality.By the previous remark, w has the same set of factors of length 20 as µ(p).Since max {|f |, f ∈ F 18 } = 19 20, w avoids every factor in F 18 .Moreover, w is cubefree since it is 13  5 -free.By Lemma 5, w has the same factor set as µ(p).
Then the proof is complete by symmetry by reversal and bit complement.
We construct the set S 78 20 defined as follows: a word v is in S 78 20 if and only if there exists a 28  11 -free binary word pvs containing at most 20 palindromes and such that |p| = |v| = |s| = 78.From S 78 20 , we construct the Rauzy graph R 78 20 such that the vertices are the factors of length 77 and the arcs are the factors of length 78.We notice that R 78 20 is not strongly connected.It contains four strongly connected components that are symmetric with respect to reversal and bit complement.Let C 78 20 be the strongly connected component which avoids the factor 1011.We check that C 78 20 is identical to the Rauzy graph of the factors of length 77 and 78 of ν(p).Now we consider a recurrent 28  11 -free binary word w with 20 palindromes.Since w is recurrent, w corresponds to a walk in one of the strongly connected components of R 78 20 , say C 78 20 without loss of generality.By the previous remark, w has the same set of factors of length 78 as ν(p).Since max {|f |, f ∈ F 20 } = 16 78, w avoids every factor in F 20 .Moreover, w is cubefree since it is 28  11 -free.By Lemma 6, w has the same factor set as ν(p).
Then the proof is complete by symmetry by reversal and bit complement.
Notice that item (b) requires recurrent words rather than bi-infinite words.That is because of, e.g., the bi-infinite word x = ν(p) R 010110ν(p).Obviously ν(p) and ν(p) R have the same set of 20 palindromes and it is easy to check that x contains no additional palindrome.We show that x is 5

The critical exponent of ν(p) and µ(p)
Before recalling the definition of the infinite words p, ν(p) and µ(p), let us underline that all of them are uniformly recurrent and ν(p) and µ(p) are morphic images of p. Hence in order to compute their (asymptotic) critical exponents, we will exploit the following two useful statements.See also [8].

Theorem 8 ([4]
).Let u be a uniformly recurrent aperiodic infinite word.Let (w n ) be a sequence of all bispecial factors ordered by their length.For every n ∈ N, let r n be a shortest return word to w n in u.Then Theorem 9. Let u be an infinite word over an alphabet A such that the uniform letter frequencies in u exist.Let ψ : A * → B * be an injective morphism and let L ∈ N be such that every factor v of ψ(u), |v| ≥ L, has a synchronization point.
Proof.The inequality E * (ψ(u)) ≥ E * (u) is proven in [5] for any non-erasing morphism under the assumption of existence of uniform letter frequencies in u.
Let us prove the opposite inequality.According to the definition of E * (ψ(u)), there exist sequences w n and v n such that 1. lim 2. w n is a factor of ψ(u) for each n ∈ N; 3. w n is a prefix of the periodic word v n ω for each n ∈ N; where we highlighted the first and the last synchronization point (not necessarily distinct) in w n and v n and where w ′ n and v ′ n are uniquely given factors of u and the lengths of x n , y n , z n are smaller than L.
By the third item, we have where u n is a proper prefix of v n and k ∈ N, k ≥ 1.
There are two possible cases for (u n ).(a) In the first case, since k ≥ 2 for large enough n, the factor w n starts with . By definition of synchronization points and injectivity of ψ, there exists a unique factor is a factor of u and it is a prefix of (t n ) ω and where the last equality holds thanks to boundedness of (|x n |), (|z n |) and (|u n |).
(b) In the second case, w jn = (x jn ψ(v ′ jn )z jn ) k x jn ψ(u ′ jn )y jn , where k ≥ 1.By definition of synchronization points and injectivity of ψ, there exists a unique factor t jn of u such that ψ(v ′ jn )z jn x jn = ψ(t jn ).Consequently, (t jn ) k u ′ jn is a factor of u and it is a prefix of (t jn ) ω and where the last equality holds thanks to boundedness of (x n ) and (y n ).
Combining two simple facts: |u| for each word u over A, where 1 is a vector with all coordinates equal to one; • for each sequence (s n ) of factors of u with lim n→∞ |s n | = ∞ we have, by uniform letter frequencies in u, lim n→∞ sn |sn| = f , where f is the vector of letter frequencies in u, Consequently, (a) in the first case, since lim n→∞ |t n | = ∞, we obtain using (2) where the last inequality follows from the fact that ( (b) in the second case, since lim |t jn | = ∞, we obtain using (2) where the last inequality follows from the fact that (t jn ) k u ′ jn ∈ L(u) and (t jn ) k u ′ jn is a power of t jn .

The infinite word p
In order to compute the critical exponent of morphic images of p, it is essential to describe bispecial factors and their return words in p and to determine the asymptotic critical exponent of p.
The infinite word p is the fixed point of the injective morphism ϕ, where Therefore, p has the following prefix It is readily seen that each non-empty factor of p has a synchronization point.
The following characteristics of p are known [2]: • The word p is not closed under reversal: 02 ∈ L(p), but 20 / ∈ L(p).
• The word p is uniformly recurrent and p has uniform letter frequencies because ϕ is primitive.

Bispecial factors in p
First, we will examine LS factors.Using the form of ϕ, we observe • 0 has only one left extension: 1, • 1 has two left extensions: 0 and 2, • 2 has two left extensions: 0 and 1.
Proof.It follows from the form of ϕ and the fact that p is the fixed point of the morphism ϕ, i.e., if u ∈ L(p), then ϕ(u) ∈ L(p).
Second, we will focus on RS factors.We observe • 0 has two right extensions: 1 and 2, • 1 has two right extensions: 0 and 2, • 2 has only one right extension: 1.
Therefore, every RS factor has right extensions either {1, 2}, or {0, 2}.Using similar arguments as for LS factors, we get the following statement.
It follows from the form of LS and RS factors that we have at most 4 possible kinds of non-empty BS factors in p. Proposition 13.Let v be a non-empty BS factor in p.
Proof.The implication (⇐) follows from Lemmata 11 and 12.We will prove the opposite implication for Item 1, the other cases may be proven analogously.
If v is a non-empty factor such that 0v, 2v, v0, v2 ∈ L(p), then v necessarily starts and ends with the letter 1.By the form of ϕ, we have the following synchronization points v = 1 • v• (v may be empty).Hence, by injectivity of ϕ, there exists a unique w in p such that v = 1ϕ(w).Thus, using again the form of ϕ and the knowledge of possible right extensions, the factor w is BS and 0w, 1w, w1, w2 ∈ L(p).
We can see that the only BS factor of length one is 1, it has left extensions 0, 2 and right extensions 0, 2. Applying Proposition 13 Item 2, we obtain that ϕ(1)0 is BS with left extensions 0, 1 and right extensions 1, 2. Proposition 13 Item 1 gives us that 1ϕ 2 (1)ϕ(0) is BS with left extensions 0, 2 and right extensions 0, 2. This process can be iterated providing us with infinitely many BS factors: The only BS factor of length two is 10, it has left extensions 0, 2 and right extensions 1, 2. Applying Proposition 13 Item 4, we obtain that ϕ(1)ϕ(0) is BS with left extensions 0, 1 and right extensions 0, 2. Proposition 13 Item 3 gives us that 1ϕ 2 (1)ϕ 2 (0)0 is BS with left extensions 0, 2 and right extensions 1, 2. This process can be iterated providing us again with infinitely many BS factors: Each BS factor v of length greater than two has at least two synchronization points and the corresponding BS factor w from Proposition 13 is non-empty.In other words, the BS factor v makes part of one of the sequences (3) and ( 4) of BS factors.
As a consequence of Proposition 13 and the above arguments, we get a complete description of BS factors in p. Corollary 14.Let w be a non-empty BS factor in p. Then it has one of the following forms: for n ≥ 1.If n = 0, then we set w (0)

The shortest return words to bispecial factors in p
Each factor of p has 3 return words.This claim follows from the next theorem.

Then each factor of u has exactly 3 return words if and only if C(n) = 2n + 1 and u has no weak BS factors.
Let us first comment on return words to the shortest BS factors -observe the prefix of p at the beginning of this section.
• The return words to ϕ(1)0 are 210 = ϕ(1)0, 21010, 2101.The shortest one is 210 and it is a prefix of all of them.
• The return words to 10 are 10, 102, 1012.The shortest one is 10 and it is a prefix of all of them.
Lemma 17.If w is a non-empty BS factor of p and v is a return word to w, then ϕ(v) is a return word to ϕ(w).
Proof.On one hand, since vw contains w as a prefix and as a suffix, ϕ(v)ϕ(w) contains ϕ(w) as a prefix and as a suffix, too.On the other hand, w starts in 1 or 2 and ends in 0 or 1, thus ϕ(w) starts in 0 or 2 and ends in 1, therefore it has the following synchronization points •ϕ(w)•.Consequently, ϕ(v)ϕ(w) cannot contain ϕ(w) somewhere in the middle because in such a case, by injectivity of ϕ, vw would contain w also somewhere in the middle.
The following observation is an immediate consequence of the definition of return words.
Observation 18.Let w be a factor of p and let v be its return word.If w has a unique right extension a, then v is a return word to wa, too.If w has a unique left extension b, then bvb −1 is a return word to bw.In particular, the Parikh vectors of the corresponding return words are the same.

The asymptotic critical exponent of p
Let us determine the asymptotic critical exponent of p using Theorem 8. We use the form of BS factors and their shortest return words determined above.We get E * (p) = 1 + max{A ′ , B ′ , C ′ , D ′ }, where By the Hamilton-Cayley theorem, we have 75488.By the Perron-Frobenius theorem, β is strictly larger than the modulus of the other roots of the characteristic polynomial.We thus obtain: Consequently,

The infinite word ν(p)
The morphism ν has the form: Therefore, and ν is injective.

The shortest return words to bispecial factors in ν(p)
Lemma 24.If w is a non-empty BS factor in p and v is its return word, then ν(v) is a return word to ν(w)0.
Proof.On one hand, consider any occurrence of vw and denote a the following letter, then ν(v)ν(w)0 is a prefix of ν(vwa).Since vw contains w as a prefix and as a suffix, then ν(v)ν(w)0 contains ν(w)0 as a prefix and as a suffix, too.
On the other hand, w starts in 1 or 2 and ends in 0 or 1, thus ν(w)0 starts in 0 and ends in 0110 or 00, therefore ν(w)0 has the following synchronization points •ν(w) • 0. Consequently, ν(v)ν(w)0 cannot contain ν(w)0 somewhere in the middle because in such a case, by injectivity of ν, vw would contain w also somewhere in the middle.
Applying Lemma 24 and Observation 18, we have the following description of the shortest return words to BS factors.
Corollary 25.The shortest return words to BS factors of length at least three in ν(p) have the following properties.
A has the same Parikh vector as ν(ϕ 2n−1 (012)) for n ≥ 1 and 100 is the shortest return word to v Proof.We will prove case (A).The other cases are similar.We know that the return words to w A = 1 in p are 12, 10, 102.Using Lemma 24, we obtain that ν(12) = 001, ν(10) = 0011 and ν(102) = 001101 are return words to ν(1)0.Since ν(1)0 has unique left and right extensions 1, using twice Observation 18, we obtain that 1ν(12)1 −1 = 100, 1001 and 100110 are return words to v 0 A = 1001.Therefore, the shortest return word to v (0) A = 1001 is 100 and it is a prefix of all of them.Now, let us consider n ≥ 1 and the bispecial factor Using Corollary 20, we know that the shortest return word to w A has the same Parikh vector as ϕ 2n−1 (012), moreover the shortest return word is a prefix of all other return words.
Using Lemma 24, and the fact that ν is non-erasing, we obtain that the shortest return word to ν(w (n) A )0 has the same Parikh vector as ν(ϕ 2n−1 (012)).Using Observation 18 twice, we obtain that the shortest return word to 1ν(w A )01 has the same Parikh vector as ν(ϕ 2n−1 (012)), since adding 1 at the beginning and erasing 1 at the end does not change the Parikh vector.

The critical exponent of ν(p)
Using Theorem 8 and the description of BS factors from Corollary 23 and of their shortest return words from Corollary 25, we obtain the following formula for the critical exponent of ν(p).
of length one or two and r its shortest return word .
Theorem 26.The critical exponent of ν(p) equals Proof.To evaluate the critical exponent of ν(p) using the above formula, we have to do several steps.
1. Determining the shortest return words of BS factors of length one and two in ν(p): • 0 is a BS factor with the shortest return word 0.
• 1 is a BS factor with the shortest return word 1.
• 01 is a BS factor with the shortest return words 010, 011.
• 10 is a BS factor with the shortest return word 10.
For the left side, we can write for n ≥ 1 For the right side, we can write for n ≥ 1 Since the inequality holds true for the given values, we obtain A ≤ 3 2 .Next, we will show that B ≤ 3  2 .For all n ≥ 0 we have to show that we need to prove the inequality in the form For the left side, we can write for n ≥ 0 For the right side, we can write for n ≥ 0 holds true for given values, we obtain B ≤ 3 2 .The explicit solution reads are the roots of the polynomial t 3 − 2t 2 + t − 1, and . = −0.106602784+ 0.24671731i ; First, we will show that C = we need to prove the inequality in the form Now, we need to be more careful with the approximations.For the left side, we can write for n ≥ 2 .
For the right side, we can write for n ≥ 2 Since the inequality holds true for given values, it remains to check the case for n = 1.
Therefore, we have proven that C = 3 2 .It remains to prove D ≤ 3  2 , however, the steps are the same as in the proof of the inequality A ≤ 3  2 .Thus, we dare to skip it.We have shown that max{A, B, C, D} = 3 2 , and F = 1.Consequently, E(ν(p)) = 1 + max{A, B, C, D, F } = 5  2 .

The infinite word µ(p)
The morphism µ has the form:  Proof.On one hand, since vw contains w as a prefix and as a suffix, then µ(v)µ(w) contains µ(w) as a prefix and as a suffix, too.On the other hand, w starts in 10 or 21 and ends in 0 or 1, therefore µ(w) has the following synchronization points •µ(w)•.Consequently, µ(v)µ(w) cannot contain µ(w) somewhere in the middle because in such a case, by injectivity of µ, vw would contain w also somewhere in the middle.
Applying Lemma 30 and Observation 18, we have the following description of the shortest return words to BS factors.Proof.We will prove case (A).The other cases are similar.Let us consider n ≥ 1 and the bispecial factor Using Corollary 20, we know that the shortest return word to w A has the same Parikh vector as ϕ 2n−1 (012), moreover the shortest return word is a prefix of all of the return words.
Using Lemma 30, and the fact that µ is non-erasing, we obtain that the shortest return word to µ(w (n) A ) has the same Parikh vector as µ(ϕ 2n−1 (012)).Using Observation 18 twice, we obtain that the shortest return word to µ(w (n) A )01 has the same Parikh vector as µ(ϕ 2n−1 (012)).are the roots of the polynomial t 3 − 2t 2 + t − 1, and Let us show that A ≤ 17 11 .We have to show for all n ≥ 1 that ).
For the left side, we can write for n ≥ 1 .Now, we need to be more careful with the approximations, we will therefore prove the inequality in the form 66 + 22 Re B 3 λ 2n 1 − 1 λ 2 1 − 1 For the left side, we can write for n ≥ 1 66 + 22 Re B 3 For the right side, we can write for n ≥ 1 1 and we distinguish ordinary BS factors with b(w) = 0, weak BS factors with b(w) < 0 and strong BS factors with b(w) > 0.

2 +
-free by checking the central factor of x of length 200.Then larger repetitions of exponent > 5 2are ruled out since the word 110011001001101 is a prefix of 110ν(p) but is neither a factor of ν(p) nor ν(p) R .By symmetry, this also holds for x R = ν(p) R 011010ν(p), x, and x R .

4 .
E * (ψ(u)) = lim n→∞ |wn| |vn| .If E * (ψ(u)) = 1, then, clearly, E * (ψ(u)) ≤ E * (u).Assume in the sequel that E * (ψ(u)) > 1, then we have for large enough n that |w n | > |v n | and moreover, by the first item, |v n | ≥ L. By assumption, both v n and w n have synchronization points and since v n is a prefix of w n for large enough n, we may write (a) Either (|u n |) is bounded, but as E * (ψ(u)) > 1, it follows that k ≥ 2 for large enough n.(b) Or there is a subsequence (u jn ) of (u n ) such that for all n ∈ N we have |u jn | ≥ L. Then by assumption, u jn has a synchronization point and we may write u jn = x jn • ψ(u ′ jn ) • y jn , where we highlighted the first and the last synchronization point in u jn and u ′ jn is a prefix of v ′ jn by injectivity of ψ.

4. 3 . 2
The shortest return words to bispecial factors in µ(p) Lemma 30.If w is a BS factor of p, |w| ≥ 2, and v is its return word, then µ(v) is a return word to µ(w).

Corollary 31 .
The shortest return words to BS factors of length greater than eight in µ(p) have the following properties.(A)The shortest return word r(n) A to v (n)A has the same Parikh vector as µ(ϕ 2n−1 (012)) for n ≥ 1. (B) The shortest return word r(n) B to v (n) B has the same Parikh vector as µ(ϕ 2n (012)).(C) The shortest return word r(n) C to v (n) C has the same Parikh vector as µ(ϕ 2n (01)).(D) The shortest return word r(n) D to v (n) D has the same Parikh vector as µ(ϕ 2n+1 (01)).

Table 1 :
Infinite β + -free binary words with at most p palindromes.