Asymptotic repetitive threshold of balanced sequences
Abstract.
The critical exponent of an infinite sequence over a finite alphabet expresses the maximal repetition of a factor in . By the famous Dejean’s theorem, for every -ary sequence . We define the asymptotic critical exponent as the upper limit of the maximal repetition of factors of length . We show that for any there exists a -ary sequence having arbitrarily close to . Then we focus on the class of -ary balanced sequences. In this class, the values are bounded from below by a threshold strictly bigger than 1. We provide a method which enables us to find a -ary balanced sequence with the least asymptotic critical exponent for .
2020 Mathematics Subject Classification:
68R151. Introduction
The concatenation of copies of a non-empty word is usually abbreviated as . In 1972, Dejean extended this exponential notation to rational exponents. If is a non-empty word of length and is a positive rational number of the form , then denotes the prefix of length of the infinite periodic sequence . For instance, a Czech word (mayor) can be written in this formalism as . The rational exponent describes the repetition rate of in the string .
The critical exponent of an infinite sequence captures the maximal possible repetition rate of factors occurring in , formally,
The number can be rational or irrational. Krieger and Shallit [KrSh07] have shown that any positive real number larger than one is a critical exponent of some sequence over a finite alphabet. If the size of the alphabet is a fixed number , then the critical exponent of any sequence over this alphabet cannot be smaller than a threshold larger than one. This bound is denoted in [CuRa11] as and called the repetitive threshold, i.e.,
For instance, if is a binary sequence over , then or or appears in , and thus . The existence of an infinite binary sequence with was demonstrated by Thue in 1912. The binary sequence for which is nowadays known as the Thue-Morse sequence (or the Prouhet-Thue-Morse sequence). Therefore . Dejean [Dej72] proved that and this bound is attained. In the same paper she also conjectured:
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;
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for .
The conjecture had been proved step by step by many people [Pan84c, Mou92, MoCu07, Car07, CuRa11, Rao11].
Recently, Rampersad, Shallit and Vandome asked the same question for -ary balanced sequences. Let us recall that a sequence over a finite alphabet is balanced if, for any two of its factors and of the same length, the number of occurrences of each letter in and differs by at most 1. Binary balanced aperiodic sequences coincide with Sturmian sequences and the least critical exponent is equal to and it is reached by the Fibonacci sequence [CaLu2000]. Let us denote the repetitive threshold of balanced sequences over a -ary alphabet by , i.e.,
The following results have been proved so far:
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and [RSV19];
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for [Bar20, BaSh19, DDP21];
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for and all even numbers [DOPS2022].
It remains as an open problem to prove the conjecture also for all odd numbers .
In this paper we focus on the asymptotic critical exponent of . It is defined to be if and
otherwise. Obviously, and the equality holds true whenever is irrational. It is for instance the case of the Fibonacci sequence. Nevertheless, and can coincide even if is rational: it is the case of the Thue-Morse sequence.
While the value takes into account repetitions of all factors of , considers only repetitions of factors of length tending to infinity. There is a huge literature devoted to questions situated between these two extremes. For example, Shur and Tunev [ShTu] construct a -ary sequence whose all factors (except the trivial repetition of one letter in factors of the form ) have the exponent . Other results of this flavour can be found in [BaCr, De, Sh].
We provide a simple construction showing that for any and any , there exists a -ary sequence with . Therefore, if we denote
we have Then we restrict our study to balanced sequences and look for the threshold
This threshold is bounded from below by , see Corollary 15. It is known that and it is reached by the Fibonacci sequence. We introduce a new tool – graphs of admissible tails – for computation of the exact value of . Using it we obtain at once for , see Table 1. Let us point out that a similar result for had been done step by step by several research teams.
Comparing the results with the minimal critical exponent of balanced sequences, we can see that for , but for larger . Moreover, the precise values we have found for suggest that for some positive , whereas . Looking into Table 1 we can see that
, , , , .
The time and space complexity of our algorithm allowed us to determine only for . It seems that a new idea is needed to extend Table 1 for .
2. Preliminaries
An alphabet is a finite set of symbols called letters. A word over of length is a string , where for all . The length of is denoted by . The set of all finite words over together with the operation of concatenation forms a monoid, denoted . Its neutral element is the empty word and we denote . If for some , then is a prefix of , is a suffix of and is a factor of . To any word over with cardinality , we assign its Parikh vector defined as for all , where is the number of letters occurring in .
A sequence over is an infinite string , where for all . The notation stands for the set of all sequences over . We always denote sequences by bold letters. The shift operator maps any sequence to the sequence .
A sequence is eventually periodic if for some and . It is periodic if . In both cases, the number is a period of . We write for the minimal period of . If is not eventually periodic, then it is aperiodic. A factor of is a word such that for some , . The number is called an occurrence of the factor in . In particular, if , the factor is the empty word and any index is its occurrence. If , the factor is a prefix of . If each factor of has infinitely many occurrences in , the sequence is recurrent. Moreover, if for each factor the distances between its consecutive occurrences are bounded, is uniformly recurrent.
The language of a sequence is the set of all its factors. A factor of is right special if are in for at least two distinct letters . A left special factor is defined symmetrically. A factor is bispecial if it is both left and right special.
A sequence is balanced if for every letter and every pair of factors with , we have . Every recurrent balanced sequence over any alphabet is uniformly recurrent, see [DolceDP21]. An aperiodic balanced sequence over a binary alphabet is called Sturmian, [MoHe40]. There exist many equivalent definitions of Sturmian sequences, see for instance [BeSe02].
A morphism over is a mapping such that for all . Morphisms can be naturally extended to by setting . A fixed point of a morphism is a sequence such that .
Example 1.
The most famous Sturmian sequence is the Fibonacci sequence
defined as the fixed point of the morphism , . The critical exponent of is . As shown by Carpi and de Luca [CaLu2000], it is the least critical exponent for Sturmian sequences, i.e., .
Consider a factor of a recurrent sequence . Let be two consecutive occurrences of in . Then the word is a return word to in . The set of all return words to in is denoted by . If is uniformly recurrent, the set is finite for each factor . If is a prefix of , then can be written as a concatenation of return words to . The sequence over the alphabet of cardinality is called the derived sequence of to . If is not a prefix, then we construct the derived sequence in an analogous way starting from the first occurrence of in . The concept of derived sequences was introduced by Durand [Dur98].
3. Asymptotic critical exponent and its relation to return words
In [DDP21] a handy formula for computation of the critical exponent and asymptotic critical exponent of uniformly recurrent sequences is deduced. It uses the notion of return words to a factor of a sequence.
Theorem 2 ([DDP21]).
Let be a uniformly recurrent aperiodic sequence111Arseny M. Shur in private communication pointed out to us that the same formula remains valid for recurrent aperiodic sequences.. Let be a sequence of all bispecial factors ordered by their length. For every , let be a shortest return word to in . Then
Theorem 3.
Let be a finite alphabet of size at least 2. Then
Consequently, for every .
Proof.
It is enough to prove the statement for the alphabet . For every Fibonacci number222The Fibonacci sequence is defined recursively: , and for every . , with , we construct a binary sequence such that every bispecial factor of length at least and every return word to in satisfy . For the sequence , the second formula of Theorem 2 gives , which implies the theorem. Construction of follows.
Let , where . By [DOPS2022] there exists a balanced (and hence uniformly recurrent) -ary sequence having . Zeckendorf’s theorem [Ze72] says that every can be written in the form , where is a word over the alphabet , which does not contain two consecutive ’s. We denote the -tuple representing by and define a morphism and the binary sequence as follows:
The morphism is uniform since the image of any letter by has length . Moreover, is uniformly recurrent as is uniformly recurrent. is a coding since the factor occurs only as a prefix of each . Hence any factor longer than can be written uniquely in the form , where , is a proper suffix of and is a proper prefix of for some . Obviously, if is a left special factor of , i.e., and belong to the language of , then is a left special factor of . An analogous statement is true for right special factors of .
Let be a bispecial factor of with and be a return word to in . Then there exist a bispecial factor of length and such that and . Obviously, is a return word or concatenation of several return words to in and . As , Theorem 2 implies . Hence
as we wanted to show. ∎
4. Sturmian sequences
Sturmian sequences, i.e., aperiodic balanced sequences over a binary alphabet, are a principal tool in the study of balanced sequences over arbitrary alphabets. In the sequel, we will restrict our consideration to standard sequences. Let us recall that a Sturmian sequence is called a standard sequence if both sequences and are Sturmian. For each Sturmian sequence there exists a unique standard sequence having the same language and thus the same critical exponent and the asymptotic critical exponent. Sturmian sequences have well defined frequencies of letters. Let us recall that the frequency of a letter in a sequence is the limit if it exists.
We use the characterization of standard sequences by their directive sequences. To introduce them, we recall two morphisms
Proposition 4 ([JuPi02]).
For every standard sequence there is a uniquely given directive sequence of morphisms and a sequence of standard sequences such that
Both and occur in the sequence infinitely often.
If , then is the most frequent letter in . Otherwise, is the most frequent letter in . We adopt the convention that and thus the directive sequence of starts with . Let us write this sequence in the run-length encoded form , where all integers are positive. Then the number having the continued fraction expansion equals the ratio (see [BeSe02]) and is called the slope of .
The convergents to the continued fraction of , usually denoted , have a close relation to return words in a Sturmian sequence. Recall that the sequences and both satisfy the recurrence relation
(1) |
with initial conditions , and , . Two consecutive convergents satisfy for every .
Vuillon [Vui01] showed that an infinite recurrent sequence is Sturmian if and only if each of its factors has exactly two return words. Moreover, the derived sequence of a Sturmian sequence to any of its factors is also Sturmian.
All bispecial factors of any standard sequence are its prefixes. So, one of the return words to a bispecial factor of is a prefix of .
Proposition 5 ([DvMePe20]).
Suppose that is a standard sequence with slope and is a bispecial factor of . Let (resp., ) denote the return word to which is (resp., is not) a prefix of . Then
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there exists a unique pair with such that the Parikh vectors of , , and are respectively
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(2)
the slope of the derived sequence is
To express the length of bispecial factors and their return words in a Sturmian sequence with slope , we use for each the notation
, where is the convergent to ,
i.e., the sequence satisfies (1) with initial conditions .
Remark 6.
In the sequel, it will be necessary to recognize which vectors are Parikh vectors of some factors in a Sturmian sequence. The answer follows.
Lemma 7 ([DDP21]).
Let be a Sturmian sequence with slope . Denote . Then contains a factor such that and if and only if
(2) |
5. Balanced sequences
In 2000 Hubert [Hubert00] characterized balanced sequences over alphabets of cardinality bigger than 2 in terms of Sturmian sequences, colourings, and constant gap sequences.
Definition 8.
Let be a sequence over , and be arbitrary sequences. The colouring of by and is the sequence obtained from by replacing the subsequence of all ’s with and the subsequence of all ’s with .
Definition 9.
A sequence is a constant gap sequence if for each letter occurring in the distance between any consecutive occurrences of in is constant.
There is a rich literature on a notion equivalent to constant gap sequences, the so-called exact covering systems, see [Znam1969, PoSch2002, GoGrBrSh2019].
Example 10.
The periodic sequences and are constant gap sequences over binary and ternary alphabet, respectively. The sequence , where is the Fibonacci sequence defined in Example 1, looks as follows:
Theorem 11 ([Hubert00]).
A recurrent aperiodic sequence is balanced if and only if for some Sturmian sequence and constant gap sequences over two disjoint alphabets.
Let be two disjoint alphabets. The “discolouration map” is defined for any word or sequence over ; it replaces all letters from by and all letters from by . If , where , , then and for every .
Corollary 12 ([DDP21]).
Let and . For any , the word obtained by colouring with shifted constant gap sequences and is in . In particular, if a Sturmian sequence has the same language as , then, and .
Example 13.
Let and be as in Example 10. Let . The reader is invited to check that all words are factors of .
As a consequence of the previous corollary when studying the asymptotic critical exponent of balanced sequences, we can limit our consideration to colourings of standard Sturmian sequences.
Proposition 14 ([DDP21]).
Let and . One has:
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(1)
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depends on and , not on the structure of and .
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is finite is finite is a bounded sequence.
Hoffman et al. proved that any -ary constant gap sequence satisfies (see Theorem 2 in [Hof2011] based on Corollary 2 from [Si85]). Therefore, the first item of Proposition 14 leads to the following corollary.
Corollary 15.
for every positive integer .
The main task we solve in the paper is the following: for given and find a balanced sequence such that its asymptotic critical exponent has the least value among all balanced sequences which arise as colouring of Sturmian sequences by two constant gap sequences and with and .
Having a method for solving this task, we are able to determine for a fixed . We apply the method to all pairs of such that for some constant gap sequence over -ary alphabet and for some constant gap sequence over -ary alphabet, where .
For a fixed there are only finitely many pairs with the described property. It seems that to find the periods of all constant gap sequences over a -ary alphabet is a difficult problem. Nevertheless, it is known (see [Schnabel2015]) that constant gap sequences over letters with may be obtained by interlacing.
Definition 16.
Let be mutually disjoint alphabets and let be a constant gap sequence over the alphabet for every . The interlacing of is the sequence , where for every and .
In other words, the interlacing of is a sequence obtained by listing step by step the first letters of , then the second letters, the third letters etc.
The interlacing of satisfies
Example 17.
The interlacing of and equals . The reader may easily verify that it is again a constant gap sequence and its period equals .
Remark 18.
Using the fact that all constant gap sequences with at most 12 letters are obtained by interlacing, it is not difficult to verify that the periods of constant gap sequences over alphabet of size are as follows:
6. Formula for the asymptotic critical exponent of balanced sequences
In this section, denotes the continued fraction of the slope of a standard Sturmian sequence , and are periods of two constant gap sequences and . The computation of the asymptotic critical exponent of is based on the knowledge of bispecial factors of the Sturmian sequence and their return words. To provide an explicit formula let us first fix some notation.
Convention: Given positive .
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If , we write
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If for , we write
A formula for computation of is deduced in [DDP21]. To keep this paper self-contained, we provide a sketch of its proof. It is based on the following simple observation.
Observation 19.
Let be a factor of and , where . Then
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is bispecial in if and only if is bispecial in ;
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if is an occurrence of in , then is an occurrence of in .
Denote .
Let be a pair associated in Proposition 5 to a bispecial factor of . We assign to the sets:
Proposition 20.
Let be a Sturmian sequence with slope and and be two constant gap sequences. Put
(3) |
Then the asymptotic critical exponent of the balanced sequence equals .
Proof.
Let be a bispecial factor of satisfying assumptions of Observation 19 and be two consecutive occurrences of in and . By Item 1 of Observation 19 the factor is bispecial in . Let be the pair associated by Proposition 5 with and and be two return words to in . By Item 2 of Observation 19 the projection is concatenated from return words and return words for some Obviously, the vector is the Parikh vector of the derived sequence . By Proposition 5 the slope of the derived sequence is . The inverse of is . By Lemma 7 the pair satisfies the inequality . Hence belongs to .
Since and are occurrences of in , the number of letters , resp. in is a multiple of , resp. . Using the corresponding Parikh vectors we have . By Proposition 5, , hence belongs to and thus to .
Let us evaluate the ratio . We abbreviate . By Proposition 5
The previous equality is valid for each factor occurring between two occurrences of in . If is a shortest return word to , then
Hence expresses (up to the subtracted fraction ) the maximal value of the ratio among all possible pairs with a fixed . Since the length tends to infinity with growing , Theorem 2 concludes the proof. ∎
If the slope of a Sturmian sequence is quadratic irrational, then the asymptotic critical exponent can be computed explicitly. The algorithm is explained in details in [DDP21]. We implemented it and throughout this paper we use our computer program which for a given eventually periodic continued fraction and a pair , finds the exact value of .
7. Colouring with linked parameters
In this section we study how to restrict without loss of generality the periods of constant gap sequences when searching for balanced sequences with the minimal asymptotic critical exponent. Let us recall that in the whole paper we work without loss of generality with Sturmian sequences having the slope , i.e., the letter is the most frequent letter in .
Proposition 21.
Let and be two constant gap sequences and be a Sturmian sequence with being the most frequent letter in . Then there exists a Sturmian sequence with being the most frequent letter in such that
Proof.
Let be the slope of and be the convergent to . Define the slope of the Sturmian sequence by and for each . For the slope , we use the notation and as introduced in Section 6 and . Analogously, for the slope , we use the notation , and .
Using the recurrent relation (1), we get and for each . It follows then immediately for each and that
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;
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.
In the statement of Proposition 21, the asymptotic critical exponent cannot be replaced by the critical exponent. The following example demonstrates this fact.
Example 22.
Baranwal and Shallit show in [BaSh19] that the minimal critical exponent of -ary balanced sequences equals and it is reached by the sequence , where is a Sturmian sequence with slope and ’s are coloured by and ’s are coloured by .
Now we colour ’s by and ’s by . We will explain that for each sequence , where is a Sturmian sequence with slope , we have .
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If and , then is a factor of , in particular, . Hence and thus .
Lemma 23.
Let and . If is divisible by and is divisible by , then .
8. Equivalence on unimodular matrices
By Proposition 14, the asymptotic critical exponent depends only on the periods of constant gap sequences and , not on their structure. In the sequel we use for a fixed pair and the notation introduced by the following relations:
(4) |
Obviously, and are coprime. We always consider since for , the sequences and are the same and the minimal asymptotic critical exponent for Sturmian sequences is known.
In this section we will study the form of the set
which plays an essential role in the definition of (see Proposition 20), and consequently in the computation of the asymptotic critical exponent of balanced sequences. Note that the determinant of equals , i.e., the matrix is unimodular. A solution depends only on entries of the matrix counted in the first row and in the second row. Hence it is possible to group matrices into classes of the same behaviour with respect to the form of . The following lemma prepares such grouping.
Lemma 24.
Let be unimodular and . Then if and only if there exist such that
Proof.
We have for some . Since is unimodular, it is invertible in , hence
Moreover, implies
The reasoning is analogous. ∎
Remark 25.
To solve the equation , where is a unimodular matrix, we just need to know the remainders of division by , resp. , of the first row, resp. of the second row of . Therefore, we consider instead of a matrix the so-called
Obviously, if and only if . Hence, matrices with the same -name have the same solutions . By Lemma 24, the values , and the -name of capture all pieces of information we need to decide whether belongs to the set .
Nevertheless, matrices with distinct -names can have the same solutions as well. The following example illustrates it.
Example 26.
Let and . Consider the unimodular matrices and . Their -names are and . By the previous remark, a pair solves if and only if
which is equivalent to .
To group unimodular matrices into classes with the same pairs of solutions, we introduce an equivalence.
Definition 27.
Let and be unimodular matrices in . We say that is equivalent to , and write , if there exist coprime with and coprime with such that
(5) |
The equivalence class containing a matrix will be denoted .
Remark 28.
The relation is an equivalence. Evidently . For a fixed , the set of elements coprime with is closed under inverse modulo and multiplication modulo , which implies that is symmetric and transitive. Obviously, if and satisfy (5), then the -names and satisfy .
Lemma 29.
Let and let and be unimodular matrices in such that . Then
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for any unimodular matrix .
Proof.
Let and satisfy (5) and . Then
(6) |
Let us point out a simple fact: If is coprime with , then if and only if for every and analogously for the coprime values and . Hence
(7) |
(8) |
Item 2 is a direct consequence of Equation (6) in which represents the first column of the matrix and then its second column. ∎
9. A lower bound on the asymptotic critical exponent for fixed periods of constant gap sequences
We associate to a sequence of classes of equivalent matrices with representatives and a sequence of with . Let us stress that depends only of the first coefficients of the continued fraction of , whereas depends only on the remaining coefficients of . Representatives of two consecutive classes satisfy
(9) |
Definition 30.
Let be unimodular and . We say that is -forcing for the class if there exist , such that
- :
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- :
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and
- :
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if , then ;
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if , then ;
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if , then .
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The set of -forcing ’s for the class is denoted .
Note that the definition is correct because it does not depend on the choice of the representative from the class of equivalence. Indeed, only depends on and by Lemma 29 the solutions are the same for each matrix in .
Theorem 31.
Let , where is a Sturmian sequence with slope , and let be a fixed positive number.
Assume that there exist infinitely many such that is -forcing for the class . Then .
Proof.
First assume that the sequence of coefficients in the continued fraction expansion of is unbounded, then by Proposition 14 . In the sequel, assume that is bounded, say by .
Let be such that is -forcing for the class . We point out that and of Definition 30 together mean that and belongs to – the set used in the definition of in Proposition 20. Hence can be rewritten as
(10) |
where we abbreviate notation putting . Using , we get
(11) |
Hence for infinitely many . It implies .
To prove the strict inequality , we need to show more, namely, that there exists a positive number, say , such that for each , where is -forcing. Existence of such follows from the three facts:
1) . Indeed, the recurrence relation and the inequalities imply on one hand
and on the other hand
2) If , the function is strictly monotonous and thus the minimum of on the interval is strictly bigger than the minimum on . If , the strict inequality is required by .
3) There are only finitely many triplets satisfying , and . ∎
Example 32.
Using the previous proposition we show that and implies for every . In particular, we show that for any every sequence of contains infinitely many -forcing .
There are four classes of equivalent matrices with -names:
.
If has the name
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, then every is -forcing for , as and satisfy Properties , and with the triplet .
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, then every is -forcing for , the corresponding triplet is , .
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, then every is -forcing for , the corresponding triplet is , .
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, then every is -forcing for , the corresponding triplet is , .
Therefore, if is smaller than for some , then necessarily and is the name of for all . In particular, for all . However, if has the name , then the class containing the matrix has the name – a contradiction.
Example 33.
Now we apply Theorem 31 to balanced sequences obtained by colouring with two constant gap sequences both having the period 2, i.e., . Using notation (4), we have . According to Definition 27 all integer unimodular matrices belong to the same class of equivalence. By Lemma 24, the triplet , and satisfies Property for any unimodular matrix .
If we fix , then every with the same triplet satisfies and . In other words, is -forcing. Therefore, the only candidates for with are colourings of Sturmian sequences with slope , where is any finite preperiod. For every such , the formula (3) in Proposition 20 gives for sufficiently large , where fulfils the recurrence relation . Consequently, is the minimum value of the asymptotic critical exponent for and it is attained if and only if is a colouring of a Sturmian sequence with slope for a finite preperiod .
10. Admissible tails of continued fractions
In the previous section, we have associated with a continued fraction a sequence of classes of equivalent matrices . Since the number of classes is finite, there exists such that any class of equivalence either occurs in infinitely many times or does not occur at all in it. To find a balanced sequence with , we should at least guarantee that is not -forcing for the class for each . Formally, .
Remark 34.
In Definition 30 only Property depends on . Therefore, for each triplet satisfying and , we add to the interval of ’s satisfying , i.e., the interval
The set is a union of several open intervals. Boundaries of these intervals are rational. Rational ’s have finite continued fractions and do not occur as tails of the continued fraction expansion of slopes of Sturmian sequences.
Definition 35.
Let and be unimodular. We denote
Using the notation of , Theorem 31 can be rephrased as the following corollary.
Corollary 36.
Let be the slope of a Sturmian sequence and . If , then there exists such that for every the set contains .
Lemma 37.
Let and . Then the set is a subset of . In particular is bounded for each equivalence class .
Proof.
To compute easily , we collect several practical observations on triplets that may influence the form of . We will apply these rules in examples worked out in hand. They follow immediately from Definitions 30 and 35, from Remark 34 and Lemmas 24 and 37.
Remark 38.
Let be a unimodular matrix, denotes its -name. The following statements hold true.
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The number of triplets that may affect is bounded. In particular, by Definition 35 and Lemma 37 it holds . It together with Property forces to satisfy The number of such triplets is finite.
Each triplet adds to the set an interval as described in Remark 34. In fact, only few of the triplets add really new elements to , or equivalently, erase some elements of . Hence, when we present in examples the set , we list only the triplets which determine the set. It means that no other triplet reduces the set any more.
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(2)
is satisfied whenever and
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with does not fulfil for any and .
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(4)
If , then only may satisfy and for some and .
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(5)
Let and be fulfilled for some and . Then is fulfilled, too. Moreover, is satisfied for all .
Consequently,
for and for .
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(6)
If a triplet satisfies Properties and , then the triplet with , need not be taken into account when constructing .
Example 39.
Let , and . In this case , , , and there are three classes of equivalent matrices with -names:
To find we will apply Remark 38.
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Consider the triplet . is satisfied by Lemma 24. is satisfied for , by Item 2. As , says .Hence, any is -forcing and thus .
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Due to Lemma 24 we consider only triplets with , , where forces to be even. Item 4 gives the restriction . Hence by Items 1 and 6, we need to work only with and . The inequality (in case ) and (in case ) required by is fulfilled only if and .By Remark 34, the set of -forcing ’s is and thus .
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By Lemma 24 we need to check only the values , , where is even. Item 4 gives . Items 1 and 6 restrict and . The inequality in is fulfilled only for and .Analogously to the previous case, .
Note an interesting fact. If we replace the value by some , the triplets and extend the set of -forcing ’s and .
11. Graphs of admissible tails
In this section we create the main tool that enables us to find balanced sequences with a small value of . More precisely, for given and and , we will be able to identify candidates for suffixes (we call them “admissible tails”) of the continued fraction expansion of the slope of a Sturmian sequence whose colouring satisfies .
Definition 40.
Given and . An oriented graph is called the graph of -admissible tails, and denoted , if
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the set of vertices consists of classes of equivalence ;
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a pair labeled by belongs to the set of oriented edges if for some and .
The graph is finite: it has a finite number of vertices as the number of classes of equivalence is finite and a finite number of edges as by Lemma 37 the set is bounded for each class . Moreover, if , then for every unimodular matrix . Hence is a subgraph of .
Let us recall that an oriented infinite path in a graph is a sequence such that and for every .
The graph terminology allows us to rephrase Corollary 36 into the following theorem.
Theorem 41.
Let and , where is a Sturmian sequence with slope . Assume that .
Then there exists an infinite oriented path in and such that for every
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(1)
;
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(2)
is the label of the edge .
In the graph of admissible tails we are interested in infinite paths on which vertices occur infinitely many times. Therefore it is enough to consider strongly connected components, i.e., subgraphs with an oriented path from each vertex to each vertex. In particular, if a component has only one vertex, it has to have a loop.
Corollary 42.
Let and . If contains no oriented cycle, then for every colouring of a Sturmian sequence by constant gap sequences of periods and .
Example 43.
Let us construct the graph of -admissible tails for parameters and . By Example 39, the graph has three vertices with -names
and the corresponding sets
The graph and its only strongly connected component are depicted in Figure 1.
By Theorem 41 the suffix is the only candidate for the suffix of associated with a Sturmian sequence such that with has . And indeed, for . The reader is invited to check that in Proposition 20 (the maximum is attained for , ).
As we have noticed in Example 39, if we choose instead of the value , then and the corresponding graph contains no oriented cycle. In other words, we can see immediately that there is no with for .
Example 44.
The authors of [RSV19] show that the least critical exponent on 4 letters – in our notation – equals and it is reached for the colouring of the Fibonacci sequence by constant gap sequences with . Let us deduce that .
Thanks to Proposition 21, Examples 32 and 33, and Remark 18, it remains to inspect only the case and .
There are six classes of equivalent matrices with -names
Let us write down for each the triplets that influence the set (the other triplets satisfying Property and do not reduce any more):
In the graph with parameters and (see Figure 2) there is no strongly connected component. This completes the proof that .
12. Asymptotic repetitive threshold for binary to quinary balanced sequences
Using graphs of admissible tails we are able to list the least asymptotic critical exponent of -ary balanced sequences for .
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•
It is known that and it is reached for the Fibonacci sequence.
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: It suffices to consider and . There are three classes of equivalent matrices with -names:
Let us write down for each the triplets that influence the set :
There is a unique strongly connected component in the graph , which is the same as the one depicted in Figure 1. By Theorem 41 the suffix is the only candidate for the suffix of the continued fraction associated with a Sturmian sequence such that with has . And indeed, for . The reader is invited to check that (the maximum in (3) is attained for ) and for . Using Proposition 20 we have .
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: It was shown in Example 44.
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The method used to determine for does not work for . Let us explain why. To find we have to inspect the pairs . In the sequel, we will show that . If we construct the graph with the optimal value , we find out that in all cases besides there are no strongly connected components in .
Hence we focus on the case . For this pair, the strongly connected component of is depicted in Figure 3 (we will show later that the bold cycle corresponds to the unique -admissible tail).
Although is the label of an infinite path in , the asymptotic critical exponent for any colouring of a Sturmian sequence with slope having eventually periodic continued fraction with the period . In other words, the implication in Theorem 41 cannot be reversed. Our next goal is to reduce the graph in order to exclude paths corresponding to the asymptotic critical exponent .
13. Reduction of the graphs
We have seen that the existence of an infinite path in the graph of admissible tails does not guarantee a small value of . There may be even uncountably many paths in the graph corresponding to . However, we will show how to reduce the graph so that the number of unsuitable paths is diminished. This method enables us to find for .
We describe two reasons why the implication in Theorem 41 cannot be reversed.
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(1)
Let be labels of edges in an infinite path in and all vertices of the path occurs in it infinitely many times. If the path corresponds to an asymptotic critical exponent , then for each , the continued fraction belongs to , where . But the construction of the graph requires only for some .
Hence, if it is not possible to extend the edge starting in and labeled by to an infinite path whose labels make the continued fraction expansion of some , then this edge may be erased without influencing the validity of Theorem 41.
We call this process forward reduction.
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(2)
The terms used to describe Property in Definition 30 represent the minimal value of the function for , see (11). The actual value of we use in evaluation of in (10) is . It may happen that even if the minimum of on is smaller than .
To construct we use only the fact that . As , the value is given by the labels of edges of a path of length ending in . Assume that the structure of enables to deduce that all paths entering the vertex have . If for some satisfying Property the following inequality holds true
then satisfying Property is -forcing, i.e., all such ’s should be deleted from the set . The new set may cause that some edges starting in the vertex are deleted as well.
We call this process backward reduction.
We always apply the forward and backward reductions together with searching for strongly connected components as long as the graph changes.
Let us illustrate both kinds of reductions on the following example.
Example 45.
Let us construct for . Then and . There are 24 classes of equivalence with the following -names:
By Lemma 37, . We write down for each class the triplets that influence the set :
A unique strongly connected component of the graph is depicted in Figure 4. Using our computer program we can see that for the Sturmian sequence associated to , the balanced sequence with has .
If we search for such that , i.e., if we choose , then by Theorem 31 we have to exclude also the solution , , which reduces to , therefore no strongly connected component remains in the graph. To summarize, there is no with for .
Let us apply first the forward reduction on the graph from Figure 4. Consider the vertex and the outgoing edge labeled by 4. Each prolongation to an infinite path has the next edge label equal to 1. Therefore the corresponding , i.e., . Consequently, the edge labeled by 4 may be erased.
Next, we apply the backward reduction. Consider the vertex . The sequence of edge labels of each ingoing path ends in . For satisfying we have and . Therefore, by Remark 34 the triplet reduces to .
Finally, we apply once more the forward reduction. Consider again the vertex , now with , and its unique outgoing edge labeled by 3. Each prolongation to an infinite path has the next edge label equal to 1. Therefore the corresponding , i.e., . Thus, the edge labeled by 3 may be erased. Consequently, the graph is reduced so that it contains a unique cycle labeled by , which proves that it is the only -admissible suffix of .
14. Asymptotic repetitive threshold for senary to denary balanced sequences
In order to find , we first detect possible periods which give a -ary balanced sequence. For this task we use Remark 18. Then some pairs are eliminated due to Proposition 21 and Lemma 23.
The starting parameter for the construction of graphs of -admissible tails can be chosen to be the value of the minimal critical exponent since by definition for every sequence .
By Proposition 14 we have for every balanced sequence . Therefore when searching for we have to consider only the pairs such that .
As soon as we find with , we lower to and construct graphs of -admissible tails for the remaining pairs .
Using the above described procedure, we have found the minimal asymptotic critical exponent of -ary balanced sequences up to . They are listed in Table 1.
2 | 1 | 1 | ||
3 | 1 | 2 | ||
4 | 2 | 2 | ||
5 | 2 | 4 | ||
6 | 1 | 16 | ||
7 | 1 | 32 | ||
8 | 8 | 8 | ||
9 | 8 | 16 | ||
10 | 4 | 64 |
Let us comment properties of the graphs with the optimal value .
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It was sufficient to use the graphs of admissible tails without reduction, as explained in Section 12.
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The reduction was necessary. After reduction of the graph constructed for the pair from Table 1, there is always a unique strongly connected component in the form of a cycle, i.e., the period of the continued fraction was determined uniquely by the graph. For all other pairs of admissible periods the graph has no strongly connected component. Let us point out that for 6 letters the forward reduction suffices, while for more letters both the forward and the backward reduction is needed. For 6 letters, the unique -admissible tail corresponds to the bold cycle in Figure 3.
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For the graph has no strongly connected component. For , a unique strongly connected component of after reduction is depicted in Figure 5. Thus, there are two oriented cycles sharing two vertices. Hence a “human intervention” is needed to pick up the suitable continued fraction. Using our computer program we obtain: for the balanced sequence , where .
Figure 5. The strongly connected component of the graph of admissible tails for and . In Figure 5, corresponds to an infinite path going alternatively through the right hand cycle, then the left hand cycle, and again the right hand cycle, then the left hand cycle, and so on. The following two arguments show that if does not have a suffix corresponding to such alternation of the right and left hand cycle, then . First we observe, that every path in the graph from Figure 5 uses infinitely many times the vertices named and .
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Let us explain that any corresponding to an infinite path going infinitely many times twice consecutively through the left hand cycle gives rise to a colouring with .
Since , we have and , . Assume and there exist infinitely many such that belongs to the class with -name , where we arrive to using the left hand cycle and we leave using again the left hand cycle. In this case and . Then , therefore belongs to . It belongs to as the inequality is satisfied as well. Using (3) we obtain
By Proposition 20 .
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Let us show that also any corresponding to an infinite path going infinitely many times twice consecutively through the right hand cycle gives rise to a colouring with .
To sum up, only having the continued fraction with the same period as can give the minimal value of .
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15. Comments and open questions
Let us first make some comments on complexity of computation of the asymptotic repetitive threshold. We were not able to determine for with our computer because of time complexity of the computation. The bigger , the larger the number of pairs that have to be treated. The number of such pairs grows exponentially. Moreover, even the number of vertices in the graph grows exponentially with . For example, for every we have to consider and . The number of vertices in the corresponding graph = the number of classes of equivalence is equal to .
The closer the value to the more time consuming determination of . The reason is the fact that more triplets satisfy Properties and . Hence, it would be useful to have a good upper bound on so that we do not have to repeat the computation for more values .
Time complexity is not the only obstacle to revealing for . We are afraid that for larger our method will not give a unique period of the continued fraction corresponding to the minimal and that “human intervention” will be needed similarly as in the case .
Let us conclude with some other open problems connected to the topic.
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We conjecture that for some positive . But from the obtained values , we are not able to derive a conjecture on the precise value . We observe at least that if is an optimal pair of periods for even, then is optimal for . It works for .
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We believe that always belongs to a quadratic field, but we have no proof of this fact.
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is defined to be . We can see for that is minimum of the set and it is not its accumulation point. The question is whether may be an accumulation point. And if not, what is the second, third, etc. smallest element of the set.