Abstract 1 Introduction 2 Preliminaries 3 Overlapping Net Occurrence Cover 5 Occurrences of Thue-Morse Words of Smaller Order 6 Net Occurrences in Fibonacci Words 7 Net Occurrences in Thue-Morse Words 8 Conclusion and Future Work References Appendix D A Factorization of Thue-Morse Word

Net Occurrences in Fibonacci and Thue-Morse Words

Peaker Guo ORCID School of Computing and Information Systems, The University of Melbourne, Parkville, Australia Kaisei Kishi Department of Information Science and Technology, Kyushu University, Fukuoka, Japan
Abstract

A net occurrence of a repeated string in a text is an occurrence with unique left and right extensions, and the net frequency of the string is the number of its net occurrences in the text. Originally introduced for applications in Natural Language Processing, net frequency has recently gained attention for its algorithmic aspects. Guo et al. [CPM 2024] and Ohlebusch et al. [SPIRE 2024] focus on its computation in the offline setting, while Guo et al. [SPIRE 2024], Inenaga [arXiv 2024], and Mieno and Inenaga [CPM 2025] tackle the online counterpart. Mieno and Inenaga also characterize net occurrences in terms of the minimal unique substrings of the text. Additionally, Guo et al. [CPM 2024] initiate the study of net occurrences in Fibonacci words to establish a lower bound on the asymptotic running time of algorithms. Although there has been notable progress in algorithmic developments and some initial combinatorial insights, the combinatorial aspects of net occurrences have yet to be thoroughly examined. In this work, we make two key contributions. First, we confirm the conjecture that each Fibonacci word contains exactly three net occurrences. Second, we show that each Thue-Morse word contains exactly nine net occurrences. To achieve these results, we introduce the notion of overlapping net occurrence cover, which narrows down the candidate net occurrences in any text. Furthermore, we provide a precise characterization of occurrences of Fibonacci and Thue-Morse words of smaller order, offering structural insights that may have independent interest and potential applications in algorithm analysis and combinatorial properties of these words.

Keywords and phrases:
Fibonacci words, Thue-Morse words, net occurrence, net frequency, factorization
Copyright and License:
[Uncaptioned image] © Peaker Guo and Kaisei Kishi; licensed under Creative Commons License CC-BY 4.0
2012 ACM Subject Classification:
Mathematics of computing → Combinatorics on words
Acknowledgements:
The authors thank Hideo Bannai and the organizers of the StringMasters workshop at CPM 2024 for initiating the collaboration between the authors during the workshop. The authors also thank Shunsuke Inenaga and William Umboh for their helpful advice.
Funding:
Peaker Guo: Supported by an Australian Government Research Training Program Scholarship.
Editors:
Paola Bonizzoni and Veli Mäkinen

1 Introduction

The work by Axel Thue at the beginning of the 20th century marked the beginning of the field of combinatorics on words [6]. Central to the field are two key objects that have attracted extensive research: Fibonacci words and Thue-Morse words [30]. These objects are remarkable for their rich combinatorial properties and applications in seemingly unrelated fields beyond combinatorics on words. Fibonacci words, for instance, have been used to establish lower bounds and analyze behaviors of string algorithms [24], while Thue-Morse words appear in diverse areas such as group theory, physics, and even chess [2]. They have also been used to prove properties related to repetitiveness measures [3, 27, 11, 32, 5].

Another key aspect of combinatorics on words involves identifying significant strings in a text. Different definitions of significance lead to different problem formulations. These significant strings could be repetitions [10], tandem repeats [20], or runs [4]. There is also a rich literature on the study of these significant strings in Fibonacci and Thue-Morse words [7, 12, 22, 13]. For many applications, frequency serves as a basis for significance measure. However, frequency alone can be misleading, as it may be inflated by occurrences of longer repeated strings. Consider the text the␣theoretical␣theme as an example. The string the is the most frequent string of length three, but this is due to the fact that two of its occurrences are contained by the longer repeated string ␣the.

To address this issue, Lin and Yu [28, 29] introduced the notion of net frequency (NF), motivated by Natural Language Processing tasks. As reconceptualized by Guo et al. [18], a net occurrence of a repeated string in a text is an occurrence with unique left and right extensions, and the NF of the string is the number of its net occurrences in the text. In the earlier example, only the first occurrence of the is a net occurrence, reflecting the only occurrence that is not contained by a longer repeated string.

There has been a recent surge of interest in the computation of NF. Guo et al. [18] and Ohlebusch et al. [33] focus on the offline setting, while Guo et al. [19], Inenaga [23], and Mieno and Inenaga [31] extend the computation to the online setting. Mieno and Inenaga also characterize net occurrences in terms of the minimal unique substrings of the text. Additionally, Guo et al. [18] study net occurrences in Fibonacci words to establish a lower bound on the asymptotic running time of algorithms. Despite these advances, the combinatorial aspect of net occurrences has yet to be thoroughly investigated. It has been shown that there are at least three net occurrences in each Fibonacci word [18]. However, proving that these are the only three is more challenging and was only conjectured. Meanwhile, the net occurrences in each Thue-Morse word had not been investigated before – both of which we address in this work.

Our results.

In this work, our main contribution is twofold. First, we confirm the conjecture by Guo et al. [18] that there are exactly three net occurrences in each Fibonacci word (Theorem 34). Second, we show that there are exactly nine net occurrences in each Thue-Morse word (Theorem 41). To achieve these results, we first introduce the concept of an overlapping net occurrence cover, which drastically reduces the number of occurrences that need to be examined when proving certain net occurrences are the only ones (Lemma 12). Additionally, we provide a precise characterization of occurrences of smaller-order Fibonacci and Thue-Morse words (Theorem 17 and Theorem 19). These findings could also be of independent interest, providing tools and insights for analyzing algorithms and exploring the combinatorial properties of these words. For example, they lead to methods to count the smaller-order occurrences (Corollary 18 and Corollary 21).

Other related work.

Occurrences of Fibonacci and Thue-Morse words of smaller order have been previously studied. For Fibonacci words, these occurrences have been shown to be related to the Fibonacci representation of positive integers [22, 35]. For Thue-Morse words, these occurrences have been investigated using the binary representation of numbers and properties of the compact directed acyclic word graph (CDAWG) of each Thue-Morse word [34]. We emphasize that our work addresses occurrences of Fibonacci and Thue-Morse words of smaller order from a different angle than prior work: we provide a recurrence relation that precisely characterizes the occurrences, bypassing the need for other representations.

2 Preliminaries

Strings.

Throughout, we consider the binary alphabet Σ:={a,b}. A string is an element of Σ∗. The length of a string S is denoted as |S|. Let ϵ denote the empty string of length 0. We use S⁢[i] to denote the ith character of a string S. Let [n] denote the set {1,2,…,n}. Let S⁢T be the concatenation of two strings, S and T. A substring of a string T of length n, starting at position i∈[n] and ending at position j∈[n], is written as T⁢[i⁢…⁢j]. A substring T⁢[1⁢…⁢j] is called a prefix of T, while T⁢[i⁢…⁢n] is called a suffix of T. A substring S of T is a proper substring if S≠T. An occurrence in the text T of length n is a pair of starting and ending positions (i,j)∈[n]×[n]. We say (i,j) is an occurrence of string S if S=T⁢[i⁢…⁢j], and i is an occurrence of S if S=T⁢[i⁢…⁢i+|S|−1]. An occurrence (i′,j′) is a sub-occurrence of (i,j) if i≤i′≤j′≤j. An occurrence (i,j) is a super-occurrence of (i′,j′) if (i′,j′) is a sub-occurrence of (i,j). Moreover, (i′,j′) is a proper sub-occurrence (or super-occurrence) of (i,j) if (i′,j′) is a sub-occurrence (or super-occurrence) of (i,j) and i≠i′ or j′≠j. Two occurrences (i,j) and (i′,j′) overlap if there exists a position k such that i≤k≤j and i′≤k≤j′. For a non-empty string S, a sequence of non-empty strings ℱ=(xk)k=1m=(x1,x2,…,xm) is referred to as a factorization of S if S=x1⁢x2⁢⋯⁢xm. Each string xk is called a factor of ℱ. The size of ℱ, denoted by |ℱ|, is the number of factors in the factorization.

Net frequency and net occurrences.

In a text T, the net frequency (NF) of a unique string in T is defined to be zero. The NF of a repeated string is the number of net occurrences in T.

Definition 1 (Net occurrence [18]).

In a text T, an occurrence (i,j) is a net occurrence if the corresponding string T⁢[i⁢…⁢j] is repeated, while both left extension T⁢[i−1⁢…⁢j] and right extension T⁢[i⁢…⁢j+1] are unique. When i=1, T⁢[i−1⁢…⁢j] is assumed to be unique; when j=|T|, T⁢[i⁢…⁢j+1] is assumed to be unique.

For an occurrence (i,j) in text T, we refer to T⁢[i−1] and T⁢[j+1] as the left and right extension characters of (i,j), respectively. For a string S occurring in T, we say x,y∈Σ are left and right extension characters of S if both strings x⁢S and S⁢y also occur in T.

Fibonacci words.

Let Fi denote the (finite) Fibonacci word of order i where F1:=b,F2:=a, and Fi:=Fi−1⁢Fi−2 for each i≥3. Let fi:=|Fi| be the length of the Fibonacci word of order i, which is also the ith Fibonacci number. We next review two useful results on Fi.

Lemma 2 ([12]).

Fi only occurs twice in Fi⁢Fi.

Lemma 3 ([32]).

The strings aaa and bb do not occur in Fi.

The following result can be readily derived by repeatedly applying the definition of Fi.

Observation 4.

For 1≤k≤i, there is a factorization of Fi where each factor is either Fk or Fk+1.

For example, for k=i−2⁢…⁢i−5, we have the following factorizations: Fi=Fi−1⁢Fi−2=Fi−2⁢Fi−3⁢Fi−2=Fi−3⁢Fi−4⁢Fi−3⁢Fi−3⁢Fi−4=Fi−4⁢Fi−5⁢Fi−4⁢Fi−4⁢Fi−5⁢Fi−4⁢Fi−5⁢Fi−4.

Thue-Morse words.

For a binary string S, let S¯ denote the string obtained by simultaneously replacing each a with b and each b with a. Let 𝒯i be the (finite) Thue-Morse word of order i where 𝒯1:=a and 𝒯i:=𝒯i−1⁢𝒯i−1¯ for each i≥2. Let τi:=|𝒯i|=2i−1 be the length of the Thue-Morse word of order i. We next review two properties of each 𝒯i.

Lemma 5 (Overlap-free [30]).

𝒯i has no overlapping occurrences of the same string.

Lemma 6 (Cube-free [30]).

𝒯i does not contain any string of the form x⁢x⁢x where x is a non-empty string.

The following result can be directly derived by repeatedly applying the definition of 𝒯i.

Observation 7.

For each i≥2 and 1≤j≤i, there is a factorization of 𝒯i where each factor is either 𝒯i−(j−1) or 𝒯i−(j−1)¯.

For example, for 2≤j≤3, 𝒯i=𝒯i−1⁢𝒯i−1¯=𝒯i−2⁢𝒯i−2¯⁢𝒯i−2¯⁢𝒯i−2. Figure 3 illustrates larger value of j. Also note that this result is analogous to Observation 4.

3 Overlapping Net Occurrence Cover

This section lays the foundation to prove the main results of this paper in the subsequent sections. Specifically, we aim to develop tools to show that certain net occurrences are the only ones in a text. To achieve this, we first provide two characteristics for non-net occurrences. The proofs in this section are presented in Appendix A.

Observation 8.

In a text T, if an occurrence (s,e) is a proper super-occurrence of a net occurrence, then (s,e) is not a net occurrence.

Observation 9.

In a text T, if an occurrence (s,e) is a proper sub-occurrence of a net occurrence, then (s,e) is not a net occurrence.

▶ Remark 10.

In a text, both a string and its substring can have positive NF, for example, in abaababaabaab, both abaaba and abaab have positive NF. However, this relationship does not hold for an occurrence and its sub-occurrence, as shown in the above two observations.

To show that a given set of net occurrences in T are the only ones in T, the above two observations allow us to ignore any occurrence that is either a sub-occurrence or a super-occurrence of a net occurrence. To fully use these two observations, we focus on the case when the given net occurrences “overlap” one another and collectively “cover” the text. Consequently, the only occurrences that need to be explicitly examined are the super-occurrences of those corresponding to the “overlapping regions” of these net occurrences. To formalize this, we introduce the following definition and lemma.

Definition 11 (ONOC and BNSO).

Consider a text T and a set of c net occurrences in T: 𝒞={(i1,j1),(i2,j2),…,(ic,jc)}. We say 𝒞 is an overlapping net occurrence cover (ONOC) of T if i1=1, ik+1≤jk for 1≤k≤c−1, and jc=n. Each occurrence in the set {(i2,j1),(i3,j2),…⁢(ic,jc−1)} is a bridging net sub-occurrence (BNSO) of 𝒞.

An example of Definition 11 is shown in Figure 1.

Figure 1: An example for Definition 11. The set {(1,6),(4,9),(9,14)} is an ONOC, with each of its net occurrences underlined in blue; {(4,6),(9,9)} is the corresponding set of BNSOs. Note that (2,7) is a net occurrence outside of this ONOC, underlined in orange.
Lemma 12.

For a text T, if there exists an ONOC 𝒞 of T such that 𝒞 does not contain all the net occurrences in T, then each net occurrence in T outside of 𝒞 must be a super-occurrence of (i−1,j+1), where (i,j) is a BNSO of 𝒞.

In the example in Figure 1, note that net occurrence (2,7) is indeed a super-occurrence of (4−1,6+1), where (4,6) is a BNSO.

In Section 6 and Section 7, we apply Lemma 12 in three steps. First, for a Fibonacci or Thue–Morse word, we show that an ONOC exists. Next, we examine the set of BNSOs of the ONOC. Finally, we prove that no super-occurrence of (i−1,j+1) (where (i,j) is a BNSO) is a net occurrence, thus concluding that the ONOC already contains all the net occurrences in the text.

4 Occurrences of Fibonacci Words of Smaller Order

We study the occurrences of Fi−j in Fi for appropriate i and j. These results will help us prove the only net occurrences in Fi in Section 6 and may also be of independent interest.

When j=1, with Fi=Fi−1⁢Fi−2, we have one occurrence of Fi−1 at position 1. The following result shows that this is the only one.

Lemma 13 ([32]).

Fi−1 only occurs at position 1 in Fi for i≥3.

The two factorizations in the following result reveal three occurrences of Fi−2 in Fi.

Observation 14 ([18]).

For each i≥6,

Fi =Fi−2⁢Fi−3⁢Fi−2 (1)
Fi =Fi−2⁢Fi−2⁢Fi−5⁢Fi−4. (2)

The following result confirms that these are the only three.

Lemma 15 ([32]).

Fi−2 only occurs at positions 1, fi−2+1, and fi−1+1 in Fi for i≥6.

We next provide the result when j=3 and i≥7.

Lemma 16.

Fi−3 only occurs at positions 1, fi−3+1, fi−2+1, and fi−1+1 in Fi.

Proof.

From Lemma 15, notice that the second occurrence of Fi−2 follows immediately after the first occurrence, and the second and the third occurrences of Fi−2 overlap. Then, based on Equations 1–2, we consider the following three cases.

  1. Case 1

    Fi−3 occurs within Fi−2. From Lemma 13, Fi−3 only occurs at position 1 in Fi−2. Thus, using Lemma 15, the only occurrences of Fi−3 within Fi−2 in Fi are at positions 1, fi−2+1, and fi−1+1.

  2. Case 2

    Fi−3 occurs across the boundary of Fi−2⁢Fi−3. Again from Lemma 15, the only occurrences of Fi−3 within Fi−2⁢Fi−3=Fi−1 are at positions 1, fi−3+1, and fi−2+1. Note that the occurrence at position fi−3+1 is the boundary-crossing one: we apply Equation 2 on Fi−1 and obtain Fi−2⁢Fi−3=Fi−3⁢Fi−3⁢Fi−6⁢Fi−5.

  3. Case 3

    Fi−3 occurs across the boundary of Fi−3⁢Fi−2. Note that Fi−3⁢Fi−2=Fi−3⁢Fi−3⁢Fi−4. Using Lemma 2, Fi−3 does not occur in Fi−3⁢Fi−3 and thus does not occur across the boundary of Fi−3⁢Fi−2.

◀

We now present the main result of the section, illustrated in Figure 2. Before that, we define the following. For a set of integers A and another integer i, A⊕i denotes the set {a+i:a∈A}. We write max⁡(A) for the maximum element of set A.

Figure 2: An illustration of Theorem 17 when j=4. Each row depicts a factorization of Fi with relevant factors highlighted in colors. The top two, middle two, and bottom two rows correspond to sets Θi,j−2, Θi,j−1 and Θi,j, respectively. Each green and blue occurrence of Fi−j is introduced by an occurrence of Fi−(j−2) and Fi−(j−1), respectively. The yellow occurrence is the rightmost one.
Theorem 17.

Let Θi,j denote the set of the starting positions of the occurrences of Fi−j in Fi. Then, Θi,0=Θi,1={1}, and for 2≤j≤i−4,

  • ■

    when j is even, Θi,j=Θi,j−1∪(Θi,j−2⊕fi−j)∪{fi−fi−j+1}, where the three sets in the union are mutually disjoint, and max⁡(Θi,j)=fi−fi−j+1;

  • ■

    when j is odd, Θi,j=Θi,j−1∪(Θi,j−2⊕fi−j) where the two sets in the union are disjoint, and max⁡(Θi,j)=fi−fi−(j−1)+1.

Proof.

We proceed by induction on j.

Base cases.

When j=0, we have Θi,0={1} trivially. When j=1, it follows from Lemma 13 that Θi,1={1}. When j=2, we obtain Θi,2={1,fi−2+1,fi−1+1} by Lemma 15. Thus, Θi,2=Θi,1∪(Θi,0⊕fi−2)∪{fi−fi−2+1}, where the three sets are mutually disjoint, and max⁡(Θi,2)=fi−1+1=fi−fi−2+1. Next, when j=3, we have Θi,3={1,fi−3+1,fi−2+1,fi−1+1} by Lemma 16. Hence, Θi,3=Θi,2∪(Θi,1⊕fi−3), Θi,2∩(Θi,1⊕fi−3)=∅, and max⁡(Θi,3)=fi−1+1=fi−fi−(3−1)+1.

Inductive step.

For each 4≤k≤i−4, assume the claim holds for j=k−2 and k−1, and we now prove the claim for j=k. We first prove the claim for even k. Define Λi,k:=Θi,k−1∪(Θi,k−2⊕fi−k)∪{fi−fi−k+1}. We aim to show that Θi,k=Λi,k by showing Λi,k⊂Θi,k and Θi,k⊂Λi,k. Before we proceed, note that Fi−(k+2),Fi−(k+1),Fi−k,Fi−(k−1),Fi−(k−2) are consecutive Fibonacci words of increasing orders.

To prove that Λi,k⊂Θi,k, we will show that each set in the union defining Λi,k is contained in Θi,k. Note that Θi,k−1⊂Θi,k because Fi−k is a prefix of Fi−(k−1)=Fi−k⁢Fi−(k+1). Next, we have (Θi,k−2⊕fi−k)⊂Θi,k because Fi−k occurs at position fi−k+1 in Fi−(k−2)=Fi−k⁢Fi−k⁢Fi−(k+3)⁢Fi−(k+2) where this factorization can be derived similarly to Equation 2. Lastly, by the induction hypothesis on j=k−2, we have fi−fi−(k−2)+1∈Θi,k−2 and it is the rightmost occurrence of Fi−(k−2) in Fi. Consider the factorization Fi−(k−2)=Fi−k⁢Fi−(k+1)⁢Fi−k, which can be derived similarly to Equation 1. Note that the rightmost occurrence of Fi−k in Fi−(k−2) is at position (fi−fi−(k−2)+1)+(fi−k+fi−(k+1))=(fi−(fi−k+fi−k+fi−(k+1))+1)+(fi−k+fi−(k+1))=fi−fi−k+1 in Fi. Thus, fi−fi−k+1∈Θi,k.

Next, we prove Θi,k⊂Λi,k by showing that each occurrence of Fi−k is in Λi,k. By Observation 4, there is a factorization of Fi where each factor is either Fi−(k−1) or Fi−k. We now examine the occurrences of Fi−k based on this factorization. First, when there is an occurrence of Fi−k within Fi−(k−1), this occurrence is in Θi,k−1⊂Λi,k. Next, when there is an occurrence of Fi−k (underlined) across the boundary of Fi−(k−1)⁢Fi−(k−1)=Fi−k⁢Fi−k¯⁢Fi−(k+3)⁢Fi−(k+2)⁢Fi−(k+1) or across the boundary of Fi−(k−1)⁢Fi−k=Fi−(k−2)=Fi−k⁢Fi−k¯⁢Fi−(k+3)⁢Fi−(k+2), then, by the fact that Fi−k only occurs at positions 1, fi−k+1, and fi−(k−1)+1 in Fi−(k−2) (a direct generalization of Lemma 15), this occurrence of Fi−k must be in (Θi,k−2⊕fi−k)⊂Λi,k. Finally, observe that there does not exist an occurrence of Fi−k across the boundary of Fi−k⁢Fi−(k−1)=Fi−k⁢Fi−k⁢Fi−(k−1) or across the boundary of Fi−k⁢Fi−k, because otherwise, this would contradict Lemma 2.

Now, consider a position x∈(Θi,k−2⊕fi−k). Assume, by contradiction, that x∈Θi,k−1, then, we have Fi−(k−2)=Fi−k⁢Fi−(k−1), which contradicts Fi−(k−2)=Fi−(k−1)⁢Fi−k≠Fi−k⁢Fi−(k−1). This is analogous to Fi−1⁢Fi−2≠Fi−2⁢Fi−1, which follows from the “near-commutative property” of Fibonacci words [26]. Thus, Θi,k−1∩(Θi,k−2⊕fi−k)=∅. Next, by the induction hypothesis, max⁡(Θi,k−1∪Θi,k−2)=fi−fi−(k−2)+1. Note that (fi−fi−(k−2)+1)+fi−k<fi−fi−k+1. Therefore, fi−fi−k+1∉Θi,k−1∪(Θi,k−2⊕fi−k) and max⁡(Θi,k)=fi−fi−k+1.

The proof for odd k is very similar to even k with the difference being that we do not need to consider fi−fi−k+1 for odd k. ◀

In the following result, the case where 0≤j≤i−4 has been addressed in [35], while the case where i−3≤j≤i−1 is straightforward. Our characterization in Theorem 17 can offer an alternative simpler proof for this result.

Corollary 18.

Consider Fi and 0≤j≤i−1. Let θi,j denote the number of occurrences of Fi−j in Fi, and define f−1:=1 and f0:=0 for convenience. Then,

θi,j={fj+2−(jmod2) if 0≤j≤i−4;fj+1 if i−3≤j≤i−2;fj−1 if j=i−1.

5 Occurrences of Thue-Morse Words of Smaller Order

Figure 3: An illustration of the occurrences of 𝒯i−j and 𝒯i−j¯ in 𝒯i for 1≤j≤6.

We study the occurrences of 𝒯i−j and 𝒯i−j¯ in each 𝒯i for appropriate i and j (the occurrences are shown in Figure 3 for 1≤j≤6). These results will help us identify the net occurrences in each 𝒯i in Section 6 and may also be of independent interest. We now present the main result of the section, illustrated in Figure 4.

Figure 4: An illustration of Theorem 19. Each dark blue, pink, and light blue occurrence of 𝒯i−j is introduced by an occurrence of 𝒯i−(j−1), 𝒯i−(j−1)¯, and 𝒯i−(j−2) respectively. Each occurrence of 𝒯i−j that is both dark blue and pink indicates that it is introduced by both an occurrence of 𝒯i−(j−1) and an occurrence of 𝒯i−(j−1)¯.
Theorem 19.

For each i≥2 and 0≤j≤i−1, let Ai,j and Bi,j denote the set of the starting positions of the occurrences of 𝒯i−j and 𝒯i−j¯ in 𝒯i, respectively. Then, Ai,0=Ai,1={1}. For each j≥2, we define

Bi,j−1′ :=Bi,j−1⊕τi−j,Ai,j−2′:=Ai,j−2⊕(τi−j+τi−(j+1)),
Ii,j−3 :={∅,j=2,Ai,j−3′∪Bi,j−3′,j≥3, where
Ai,j−3′ :=Ai,j−3⊕(τi−(j−1)+τi−j),Bi,j−3′:=Bi,j−3⊕τi−(j−2).

Then, Ai,j=Ai,j−1∪Bi,j−1′∪Ai,j−2′ with

Ai,j−1∩Bi,j−1′=Ii,j−3,Ai,j−1∩Ai,j−2′=∅,andBi,j−1′∩Ai,j−2′=∅.

Proof.

We proceed by induction on j.

Base cases.

When j=0, the claim holds trivially. When j=1, note that 𝒯i−1 only occurs at position 1 because it cannot occur at position τi−1+1 (where 𝒯i−1¯ occurs), and any other occurrences of 𝒯i−1 would overlap with its occurrence at position 1, contradicting Lemma 5. When j=2, observe that Ai,2−1={1}, Bi,2−1′={τi−1+τi−2+1} and Ai,2−2′={τi−2+τi−3+1} are mutually disjoint. Further, there are no occurrences of 𝒯i−2 outside of Ai,2=Ai,1∪Bi,1′∪Ai,0′ because any such occurrences would contradict Lemma 5.

Inductive step.

For each 3≤k≤i−1, assume the claim holds for j=k−3, k−2, and k−1 and we now prove the claim for j=k. Define Vi,k:=Ai,k−1∪Bi,k−1′∪Ai,k−2′. We prove Ai,k=Vi,k by showing Ai,k⊂Vi,k and Vi,k⊂Ai,k.

To prove Vi,k⊂Ai,k, we will show that each set in the union defining Vi,k is contained in Ai,k. Clearly, Ai,k−1⊂Ai,k because 𝒯i−j is a prefix of 𝒯i−(j−1)=𝒯i−j⁢𝒯i−j¯. Similarly, Bi,k−1′⊂Ai,k because 𝒯i−j is a suffix of 𝒯i−(j−1)¯=𝒯i−j¯⁢𝒯i−j. Lastly, Ai,k−2′⊂Ai,k because 𝒯i−j occurs at position τi−k+τi−(k+1) of

𝒯i−(k−2)=𝒯i−k⁢𝒯i−(k+1)¯⁢𝒯i−k⁢𝒯i−(k+1)⁢𝒯i−k (3)

Next, we prove Ai,k⊂Vi,k by showing that each occurrence of 𝒯i−k is in Vi,k. By Observation 7, there is a factorization of 𝒯i where each factor is either 𝒯i−(k−1) or 𝒯i−(k−1)¯. We thus consider the following four cases.

  1. Case 1

    When 𝒯i−k occurs within 𝒯i−(k−1)⁢𝒯i−(k−1)=𝒯i−k⁢𝒯i−k¯⁢𝒯i−k⁢𝒯i−k¯, by the overlap-free property (Lemma 5), positions 1 and τi−(k−1)+1 are the only two occurrences of 𝒯i−k. They are both contained in Ai,k−1, while the latter is also in Bi,k−1′. (The overlap-free property will be used similarly in the remaining three cases.)

  2. Case 2

    When 𝒯i−k occurs within 𝒯i−(k−1)¯⁢𝒯i−(k−1)¯=𝒯i−k¯⁢𝒯i−k⁢𝒯i−k¯⁢𝒯i−k, positions τi−k+1 and τi−(k−1)+τi−k+1 are the only two occurrences of 𝒯i−k. They are both contained in Bi,k−1′, while the former is also in Ai,k−1.

  3. Case 3

    When 𝒯i−k occurs within 𝒯i−(k−1)⁢𝒯i−(k−1)¯=𝒯i−(k−2), by Equation 3, positions 1, τi−k+τi−(k+1)+1 and τi−(k−1)+τi−k+1 are the only occurrences of 𝒯i−k: the first and third are both contained in Ai,k−1, the second is in Ai,k−2′, and the third is also in Bi,k−1′.

  4. Case 4

    When 𝒯i−k occurs within 𝒯i−(k−1)¯⁢𝒯i−(k−1)=𝒯i−k¯⁢𝒯i−k⁢𝒯i−k⁢𝒯i−k¯, position τi−k and τi−(k−1) are the only two occurrences of 𝒯i−k. The former is contained in Bi,k−1′, while the latter is in Ai,k−1.

After examining the above four cases, we conclude that Ai,k⊂Vi,k, and thus Ai,k=Vi,k. Next we will prove Ai,k−1∩Bi,k−1′=Ii,k−3 by showing Ai,k−1∩Bi,k−1′⊂Ii,k−3 and Ii,k−3⊂Ai,k−1∩Bi,k−1′. Recall that Ii,k−3:=Ai,k−3′∪Bi,k−3′.

First, we prove Ai,k−1∩Bi,k−1′⊂Ii,k−3 by establishing that if an occurrence of 𝒯i−k is in Ai,k−1∩Bi,k−1′, then this occurrence is in Ii,k−3. First observe that in Cases 1–2, some occurrences of 𝒯i−k are contained in both Ai,k−1 and Bi,k−1′. By Lemma 5, 𝒯i−(k−3)¯ and 𝒯i−(k−3) do not overlap in 𝒯i, it follow that, for each occurrence of 𝒯i−(k−3)¯ in 𝒯i, there is only one occurrence of 𝒯i−k contained in Bi,k−3′. Similarly, for each occurrence of 𝒯i−(k−3) in 𝒯i, there is only one occurrence of 𝒯i−k contained in Ai,k−3′. Now, consider the factorizations:

𝒯i−(k−3)¯ =𝒯i−(k−1)¯⁢𝒯i−(k−1)⁢𝒯i−(k−1)⁢𝒯i−(k−1)¯, and
𝒯i−(k−3) =𝒯i−(k−1)⁢𝒯i−(k−1)¯⁢𝒯i−(k−1)¯⁢𝒯i−(k−1).

Observe that the set of occurrences of 𝒯i−k in Case 1 is a subset of Bi,k−3′ since 𝒯i−(k−1)⁢𝒯i−(k−1) occurs at position τi−(k−1)+1 in 𝒯i−(k−3)¯. Similarly, the set of occurrences of 𝒯i−k in Case 2 is a subset of Ai,k−3′ since 𝒯i−(k−1)¯⁢𝒯i−(k−1)¯ occurs at position τi−(k−1)+1 in 𝒯i−(k−3).

Next, we prove Ii,k−3⊂Ai,k−1∩Bi,k−1′ by contraposition. Specifically, instead of directly showing that “if an occurrence of 𝒯i−k is in Ii,k−3, then this occurrence is in Ai,k−1∩Bi,k−1′”, we prove the equivalent contrapositive: “if an occurrence of 𝒯i−k is not in Ai,k−1∩Bi,k−1′, then this occurrence is not in Ii,k−3”. First observe that 𝒯i−k occurs at position 1 in 𝒯i−(k−1)=𝒯i−k⁢𝒯i−k¯ and occurs at position τi−k+1 in 𝒯i−(k−1)¯=𝒯i−k¯⁢𝒯i−k. Next, occurrences of 𝒯i−k⁢𝒯i−k¯ and 𝒯i−k⁢𝒯i−k¯ overlap in 𝒯i to form occurrences of 𝒯i−k¯⁢𝒯i−k⁢𝒯i−k¯ (see Cases 1–2). Hence, if an occurrence of 𝒯i−k is not in Ai,k−1∩Bi,k−1′, then this occurrence is not at position τi−k+1 in 𝒯i−k¯⁢𝒯i−k⁢𝒯i−k¯. Next, consider the factorizations:

𝒯i−(k−3)¯ =𝒯i−(k−1)¯⁢𝒯i−k⁢𝒯i−k¯⁢𝒯i−k⁢𝒯i−k¯⁢𝒯i−(k−1)¯, and
𝒯i−(k−3) =𝒯i−(k−1)⁢𝒯i−k¯⁢𝒯i−k⁢𝒯i−k¯⁢𝒯i−k⁢𝒯i−(k−1).

Since 𝒯i−(k−3)¯ and 𝒯i−(k−3) do not overlap in 𝒯i, we know that 𝒯i−k¯𝒯i−k⁢𝒯i−k¯ only occurs at position τi−(k−1)+τi−k+1 in 𝒯i−(k−3)¯ and at position τi−(k−1)+1 in 𝒯i−(k−3). Thus, if an occurrence of 𝒯i−k is not in Ai,k−1∩Bi,k−1′, then this occurrence is not in Ii,k−3. Therefore, we conclude that Ii,k−3⊂Ai,k−1∩Bi,k−1′. ◀

The analogous characterization of Bi,j is presented as follows, which can be proven in a way similar to the proof of Theorem 19.

Corollary 20.

For each i≥2 and 0≤j≤i−1, we have Bi,0=∅, Bi,1={τi−1+1}. For each j≥2, we define

Ai,j−1′′ :=Ai,j−1⊕τi−j,Bi,j−2′′:=Bi,j−2⊕(τi−j+τi−(j+1)),
Ii,j−3′ :={∅,j=2,Ai,j−3′′∪Bi,j−3′′,j≥3, where
Bi,j−3′′ :=Bi,j−3⊕(τi−(j−1)+τi−j),Ai,j−3′′:=Ai,j−3⊕τi−(j−2).

Then, Bi,j=Bi,j−1∪Ai,j−1′′∪Bi,j−2′′ with

Bi,j−1∩Ai,j−1′′=Ii,j−3′,Bi,j−1∩Bi,j−2′′=∅,andAi,j−1′′∩Bi,j−2′′=∅.⌟

We next use Theorem 19 and Corollary 20 to count the number of occurrences of 𝒯i−j and 𝒯i−j¯ in 𝒯i.

Corollary 21.

For i≥2, consider 𝒯i and 0≤j≤i−1. Let aj and bj denote the number of occurrences of 𝒯i−j and 𝒯i−j¯ in 𝒯i, respectively. Then,

  • ■

    a0=a1=1 and aj=aj−1+2⋅aj−2 for each j≥2;

  • ■

    b0=0 and bj=bj−1+aj−1 for each j≥1.

Proof.

We proceed by induction on j. For 1≤j≤2, the claim holds trivially. For j≥3, we have |Ii,j−3|=|Ai,j−3|+|Bi,j−3|=aj−3+bj−3=bj−2 and |Ai,j|=|Ai,j−1|+|Bi,j−1′|+|Ai,j−2′|−|Ii,j−3|=aj−1+bj−1+aj−2−bj−2=aj−1+(bj−1−bj−2)+aj−2=aj−1+2⋅aj−2. by induction hypothesis. We can prove bj similarly with Corollary 20. ◀

▶ Remark 22.

For each i≥2, aj is the (j+1)th Jacobsthal Number: Sequence A001045 of the On-Line Encyclopedia of Integer Sequences (https://oeis.org/A001045).

With the characterization of the occurrences of 𝒯i−j and 𝒯i−j¯ in 𝒯i, a natural next step is to investigate the structure of the strings that surround 𝒯i−j and 𝒯i−j¯, which correspond to the blank areas in each row in Figure 3. In Appendix D, we explore two smallest factorizations of 𝒯i, each containing all occurrences of 𝒯i−j and 𝒯i−j¯, respectively. The remaining factors in these factorizations represent the surrounding strings.

6 Net Occurrences in Fibonacci Words

In this section, we prove that there are only three net occurrences in each Fi, using the results on the occurrences of Fibonacci words of smaller order from Section 4, the notion of ONOC from Section 3, and new properties that we will develop in this section. We begin by reviewing the following results. Some of the proofs in this section are presented in Appendix B.

Lemma 23 ([18]).

For i≥7, let Qi:=Fi−5⁢Fi−6⁢⋯⁢F3⁢F2, Δ⁢(0):=ba, Δ⁢(1):=ab. Then,

Fi−4⁢Fi−5 =Qi⁢Δ⁢(1−(imod2))⁢ and (4)
Fi−5⁢Fi−4 =Qi⁢Δ⁢(imod2). (5)
Lemma 24 ([18]).

For each i≥7, the following are net occurrences in Fi:

  • ■

    one occurrence of Fi−2 at position fi−1+1;

  • ■

    two occurrences of Fi−2⁢Qi at positions 1 and fi−2+1.

Meanwhile, the two occurrences of Fi−2 at positions 1 and fi−2+1 are not net occurrences.

With Lemma 15, we know that these three are the only occurrences of Fi−2 in Fi. Similarly, we now strengthen Lemma 24 by showing the following result.

Lemma 25.

For each i≥7, Fi−2⁢Qi only occurs at positions 1 and fi−2+1 in Fi.

By combining Lemma 15, Lemma 24, and Lemma 25, we conclude that Fi−2 has only one net occurrence and Fi−2⁢Qi only has two net occurrences.

Lemma 26.

For each i≥7, the net occurrences identified in Lemma 24 are the only net occurrences of Fi−2 and Fi−2⁢Qi in Fi.

Figure 5: An illustration of several factorizations of Fi from Observation 14 and Lemma 23 where Δ:=Δ⁢(1−(imod2)) and Δ′:=Δ⁢(imod2). Net occurrences of Fi−2 and Fi−2⁢Qi are in yellow and green, respectively. Super-occurrences of the two BNSOs are shown as arrows.

It remains to show that there are no additional net occurrences in each Fi. To achieve this, we use the results from Section 3. First, observe that the three net occurrences in Lemma 24 form an ONOC of Fi. The two BNSOs of this ONOC correspond to an occurrence of Qi and an occurrence of Fi−4⁢Qi, respectively. See Figure 5 for an illustration. Next, we aim to show that no super-occurrences of these two occurrences can be a net occurrence. To establish this, we analyze the super-occurrences of the occurrences of Fi−3 in Lemma 32. This result covers the examination of the super-occurrences of the occurrence of Fi−4⁢Qi, since Fi−3 is a prefix of Fi−4⁢Qi=Fi−3⁢Qi−1. Furthermore, Lemma 32 helps examining super-occurrences of the occurrence of Qi in Lemma 33.

To prove these two lemmas, we introduce some properties of Fi−3 and Qi, which are proved in Appendix B.

Lemma 27.

|Qi|=fi−3−2.

Lemma 28.

For a substring S of Fi, if Fi−2⁢Qi is a proper substring of S, then S is unique.

Lemma 29.

Fi−3 is always followed by Qi−1 in Fi.

Lemma 30.

Fi−3⁢Fi−6⁢Fi−5 and its length-(fi−2−1) prefix are both unique in Fi.

Lemma 31.

The length-(fi−3−1) prefix of Fi−3 is always followed by Fi−3⁢[fi−3] in Fi.

Now, we introduce the two crucial lemmas motivated earlier.

(a) Case (1).
(b) Case (2).
(c) Case (3).
(d) Case (4).
(e) Case (5).
(f) Case (6).
(g) Case (7).
Figure 6: Illustration of the proof of Lemma 32. In each case, four factorizations of Fi are shown, each focusing on one occurrence of Fi−3, highlighted in green. For S=X⁢Fi−3⁢Y, each discussed X and Y is shown in pink and blue, respectively. Each discussed left or right extension character of S is shown in yellow. Recall that Δ:=Δ⁢(1−(imod2)) and Δ′:=Δ⁢(imod2).
Lemma 32.

Consider an occurrence (s,e) in Fi and let S:=Fi⁢[s⁢…⁢e]. If (s,e) is a super-occurrence of an occurrence of Fi−3, and S is neither Fi−2 nor Fi−2⁢Qi, then (s,e) is not a net occurrence.

Proof.

The proof is illustrated in Figure 6. Consider strings X and Y such that S=X⁢Fi−3⁢Y and X⁢Y≠ϵ. We examine the following cases depending on |Y|. Note that |Qi−1|=fi−4−2 and |Qi+1|=fi−2−2 from Lemma 27.

  1. (1)

    |Y|<|Qi−1|. Using Lemma 29, note that Y is a prefix of Qi−1 in this case. This means the right extension character of S is always Qi−1⁢[|Y|+1]. Thus, no occurrence of S is a net occurrence.

  2. (2)

    |Y|=|Qi−1|. Using Lemma 29, Fi−3⁢Y=Fi−3⁢Qi−1 always holds in this case. Next, if Fi−3⁢Y occurs at position 1, fi−2+1, or fi−1+1, then the right extension character is always Δ⁢(imod2)⁢[1]. On the other hand, if Fi−3⁢Y occurs at position fi−3+1, we examine the left extension character of S=X⁢Fi−3⁢Y. Notice that |X|≤fi−3 always holds in this case (and S becomes a prefix of Fi when |X|=fi−3). Now, observe that if Fi−3⁢Y occurs at position fi−3+1 or fi−1+1, the left extension character of S is always Fi−3⁢[fi−3−|X|−1]. Thus, this occurrence of S is also not a net occurrence.

  3. (3)

    |Qi−1|<|Y|<fi−4. If Fi−3⁢Y occurs at positions 1, fi−2+1, or fi−1+1, observe that occurrences of Fi−3 at these three positions are always followed by Fi−4. Thus, Y is a prefix of Fi−4 and the right extension character of S is always Fi−4⁢[|Y|+1]. So these three occurrences of S are not net occurrences. If Fi−3⁢Y occurs at position fi−3+1, since |Y|=|Qi−1|+1=fi−4−1, we have |Fi−3⁢Y|=fi−3+fi−4−1=fi−2−1. By Lemma 30, Fi−3⁢Y is unique, which means S is unique.

  4. (4)

    |Y|=fi−4. If Fi−3⁢Y occurs at position 1, then X is empty and S=Fi−3⁢Fi−4=Fi−2. If Fi−3⁢Y occurs at positions fi−3+1, then Fi−3⁢Y=Fi−3⁢Fi−6⁢Fi−5 is unique (Lemma 30) so S is also unique. If Fi−3⁢Y occurs at position fi−2+1, then X ends with Δ⁢(imod2). If Fi−3⁢Y occurs at position fi−1+1, then X ends with Δ⁢(1−(imod2)). Now, since Δ⁢(imod2)⁢Fi−2 and Δ⁢(1−(imod2))⁢Fi−2 are both unique by Lemma 15, S is also unique if Fi−3⁢Y occurs at these two positions.

  5. (5)

    fi−4<|Y|<|Qi+1|. First observe that the occurrences of Fi−3 at positions 1 and fi−2+1 are both followed by Fi−4⁢Qi=Qi+1. Thus, if Fi−3⁢Y occurs at these two positions, then Y is a prefix of Qi+1 and the right extension character of S is always Qi+1⁢[|Y|+1]. So these two occurrences of S are not net occurrences. Next, if Fi−3⁢Y occurs at position fi−3+1, then Fi−3⁢Y is unique because Fi−3⁢Fi−6⁢Fi−5 is a prefix of Fi−3⁢Y and the former is unique by Lemma 30. Thus, S is unique. Finally note that, Fi−3⁢Y cannot occur at position fi−1+1 because |Y|>|Fi−4| and Fi−3⁢Fi−4 is a suffix of Fi.

  6. (6)

    |Y|=|Qi+1|. Similar to the previous case, if Fi−3⁢Y occurs at position fi−3+1, then Fi−3⁢Y is unique, and Fi−3⁢Y cannot occur at position fi−1+1. If Fi−3⁢Y occurs at position 1, then X is empty and S=Fi−3⁢Y=Fi−3⁢Qi+1=Fi−2⁢Qi. If Fi−3⁢Y occurs at position fi−2+1, then S=X⁢Fi−2⁢Qi, which is unique by Lemma 28.

  7. (7)

    |Y|>|Qi+1|. Similar to the previous two cases, if Fi−3⁢Y occurs at position fi−3+1, then Fi−3⁢Y is unique, and Fi−3⁢Y cannot occur at position fi−1+1. If Fi−3⁢Y occurs at position fi−2+1, then Fi−3⁢Y is a prefix of Fi−3⁢Fi−2=Fi−2⁢Fi−5⁢Fi−4=Fi−2⁢Qi⁢Δ⁢(imod2), which is unique by Lemma 28. Thus, S is unique. Finally, since |Y|>|Qi+1|>fi−2−1, if Fi−3⁢Y occurs at positions 1, then Y begins with the length-(fi−2−1) prefix of Fi−3⁢Fi−6⁢Fi−5, which is unique by Lemma 30. Thus, S is also unique.

◀

Lemma 33.

Consider an occurrence (s,e) in Fi. If (s,e) is a proper super-occurrence of the occurrence of Qi at position fi−2+1, then (s,e) is not a net occurrence.

Proof.

Consider strings X and Y such that S=X⁢Qi⁢Y and X⁢Y≠ϵ. When |Y|≥2, notice that Fi−3 occurs in S. By Lemma 32, (s,e) is not a net occurrence. Now, we consider the case when |Y|=1. Note that Qi⁢Y is precisely the length-(fi−3−1) prefix of Fi−3. Thus, by Lemma 31, Qi⁢Y is always followed by the same right extension character, Fi−3⁢[fi−3], which means (s,e) is not a net occurrence. ◀

Finally, the main result follows from Lemma 24, Lemma 12, Lemma 32 and Lemma 33.

Theorem 34.

The three net occurrences identified in Lemma 24 are the only ones in Fi.

7 Net Occurrences in Thue-Morse Words

In this section, we prove the only nine net occurrences in each 𝒯i using the results on the occurrences of Thue-Morse words of smaller order from Section 5, the notion of ONOC from Section 3, and new results that we will introduce in this section. We will first show that each occurrence of each string in 𝒫i (defined below) is a net occurrence in 𝒯i, then show that they are the only ones.

Definition 35.

For each i≥5, define 𝒫i:={𝒯i−2,𝒯i−2¯,𝒯i−4⁢𝒯i−3¯,𝒯i−4¯⁢𝒯i−3}..

We next show several factorizations of 𝒯i, proved in Appendix C.

Lemma 36.

For each i≥5:

𝒯i =𝒯i−2⁢𝒯i−2¯⁢𝒯i−2¯⁢𝒯i−2 (6)
𝒯i =𝒯i−2⁢𝒯i−3¯⁢𝒯i−2⁢𝒯i−3⁢𝒯i−2 (7)
𝒯i =𝒯i−3⁢𝒯i−4¯⁢𝒯i−4⁢𝒯i−3¯⁢𝒯i−2⁢𝒯i−4⁢𝒯i−3¯⁢𝒯i−4¯⁢𝒯i−3¯ (8)
𝒯i =𝒯i−3⁢𝒯i−4¯⁢𝒯i−3⁢𝒯i−4⁢𝒯i−2⁢𝒯i−4⁢𝒯i−4¯⁢𝒯i−3⁢𝒯i−3¯ (9)
Figure 7: An illustration of several factorizations of 𝒯i from Lemma 36. Net occurrences of each string in Definition 35 are highlighted in a separate color. Super-occurrences of the eight BNSOs are shown as colored arrows, blue for 𝒯i−3 and orange for 𝒯i−3¯ (see Lemma 40). Each net occurrence is numbered in red at the top-right corner, and each arrow with label i⁢j corresponds to an overlap between the ith and jth net occurrences.

The following two results immediately follow from Theorem 19. Note that Corollary 37 also appears in [34].

Corollary 37.

For each i≥5:

  • ■

    𝒯i−2 only occurs at positions 1, τi−2+τi−3+1 and τi−1+τi−2+1 in 𝒯i.

  • ■

    𝒯i−2¯ only occurs at positions τi−2+1 and τi−1+1 in 𝒯i.

  • ■

    𝒯i−4⁢𝒯i−3¯ only occurs at positions τi−3+τi−4+1 and τi−1+τi−3+1 in 𝒯i.

  • ■

    𝒯i−4¯⁢𝒯i−3 only occurs at positions τi−3+1 and τi−1+τi−3+τi−4+1 in 𝒯i.

Corollary 38.

𝒯i−3 only occurs at positions 1,τi−3+τi−4+1,τi−2+τi−3+1,τi−1+τi−3+1, and ⁢τi−1+τi−2+1 in 𝒯i.

We now identify the nine net occurrences in 𝒯i.

Lemma 39.

Each occurrence of each string in 𝒫i is a net occurrence in 𝒯i.

Proof.

We proceed by examining the left and right extension characters of each occurrence of each string in 𝒫i.

Since 𝒯i−2 is a prefix and a suffix of 𝒯i, by the definition of occurrences, the occurrence of 𝒯i−2 at positions 1 has a unique left extension character, and the occurrence of 𝒯i−2 at positions τi−1+τi−2+1 has a unique right extension character. Next, note that the right extension character of the occurrence of 𝒯i−2 at position 1 differs from that of the occurrence at position τi−2+τi−3+1 because 𝒯i−3¯⁢[1]≠𝒯i−3⁢[1]. Similarly, the left extension character of the occurrence at position τi−2+τi−3+1 differs from that of the occurrence at position τi−1+τi−2+1, because 𝒯i−3¯⁢[τi−3]≠𝒯i−3⁢[τi−3]. Hence, all three occurrences of 𝒯i−2 are net occurrences.

For 𝒯i−2¯, a similar argument holds. the right extension characters satisfy 𝒯i−2¯⁢[1]≠𝒯i−2⁢[1] and the left extension characters satisfy 𝒯i−2⁢[τi−2]≠𝒯i−2¯⁢[τi−2]. Thus, both occurrences of 𝒯i−2¯ are net occurrences. For 𝒯i−4⁢𝒯i−3¯, similarly, the right extension characters satisfy 𝒯i−2⁢[1]≠𝒯i−4¯⁢[1] and the left extension characters satisfy 𝒯i−4¯⁢[τi−4]=𝒯i−2¯⁢[τi−2]≠𝒯i−2⁢[τi−2]. Thus, both occurrences of 𝒯i−4⁢𝒯i−3¯ are net occurrences. Finally, for 𝒯i−4¯⁢𝒯i−3, once again, the right extension characters satisfy 𝒯i−4⁢[1]≠𝒯i−3¯⁢[1] and the left extension characters satisfy 𝒯i−3⁢[τi−3]≠𝒯i−4⁢[τi−4]. Thus, both occurrences of 𝒯i−4¯⁢𝒯i−3 are net occurrences. ◀

To show that all other occurrences are not net occurrences, we follow Lemma 12. First note that the nine net occurrences we identified in Lemma 39 form an ONOC of 𝒯i. The eight BNSOs of this ONOC correspond to the occurrences of 𝒯i−3 and 𝒯i−3¯ shown in Figure 7. We next show that no super-occurrences of these occurrences are net occurrences to conclude that this ONOC already contains all the net occurrences in 𝒯i.

Lemma 40.

Consider an occurrence (s,e) in 𝒯i and let S:=𝒯i⁢[s⁢…⁢e]. If (s,e) is a proper super-occurrence of 𝒯i−3 or 𝒯i−3¯, and S∉𝒫i, then (s,e) is not a net occurrence.

Proof.

We first consider when (s,e) contains an occurrence of 𝒯i−3. Consider strings X and Y such that S=X⁢𝒯i−3⁢Y and X⁢Y≠ϵ. Let position u:=s+|X| be the starting position of this occurrence of 𝒯i−3. Let C:={1,τi−2+τi−3+1,τi−1+τi−2+1} and D:={τi−3+τi−4+1,τi−1+τi−3+1}. By Corollary 38, we have u∈C∪D. We next examine the following cases depending on which set u belongs to and how large |Y| is.

We first consider when u∈C.

  1. (a)

    |Y|<τi−3. By Corollary 37, note that Y is always a prefix of 𝒯i−3¯. This means the right extension character of S is always 𝒯i−3¯⁢[|Y|+1]. Thus, (s,e) is not a net occurrence.

  2. (b)

    |Y|=τi−3. By Corollary 37, S∈𝒫i in this case.

  3. (c)

    |Y|>τi−3. By Corollary 37, (s,e) contains a net occurrence of 𝒯i−3⁢𝒯i−3¯=𝒯i−2 as a proper sub-occurrence. By Observation 8 and Lemma 39, (s,e) is not a net occurrence.

We next consider when u∈D.

  1. (a)

    |Y|<τi−4. Recall that 𝒯i−4⁢𝒯i−3¯=𝒯i−4⁢𝒯i−4¯⁢𝒯i−4=𝒯i−3⁢𝒯i−4. By Corollary 37, note that Y is always a prefix of 𝒯i−4 in this case. This means the right extension character of S is always 𝒯i−4⁢[|Y|+1]. Thus, (s,e) is not a net occurrence.

  2. (b)

    |Y|=τi−4. Using Corollary 37, S∈𝒫i in this case.

  3. (c)

    |Y|>τi−4. By Corollary 37, in this case (s,e) contains a net occurrence of 𝒯i−3⁢𝒯i−4=𝒯i−4⁢𝒯i−4¯⁢𝒯i−4=𝒯i−4⁢𝒯i−3¯ as a proper sub-occurrence. Thus, by Observation 8 and Lemma 39, (s,e) is not a net occurrence.

We can prove the case when (s,e) contains an occurrence of 𝒯i−3¯ similarly. ◀

Finally, the main result follows from Lemma 12, Lemma 39 and Lemma 40.

Theorem 41.

The net occurrences in Lemma 39 are the only net occurrences in each 𝒯i.

8 Conclusion and Future Work

In this work, we investigate net occurrences in Fibonacci and Thue-Morse words, making two main contributions. First, we confirm the conjecture that each Fibonacci word contains exactly three net occurrences. Second, we establish that each Thue-Morse word contains exactly nine net occurrences. To achieve these results, we first introduce the notion of an overlapping net occurrence cover and show how it can be used to prove that certain net occurrences in a text are the only ones. We then develop recurrence relations that precisely characterize the occurrences of Fibonacci and Thue-Morse words of smaller order, which could be of independent interest. As an application, we illustrate how these results facilitate the counting of small-order occurrences.

An avenue of future work is to extend our findings to study the net occurrences in k-bonacci words [17, 25, 16, 15] and Thue-Morse-like words [1, 9]. Furthermore, since both Fibonacci and Thue-Morse words can be defined via morphisms, one could also explore net occurrences in other morphic words [14, 21, 8]. Finally, the net occurrences have been characterized in terms of minimal unique substrings [31]; this viewpoint may offer alternative and potentially simpler proofs than those presented in Sections 6–7.

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Appendix A Proofs Omitted from Section 3

See 8

Proof.

Let (s′,e′) be the net occurrence, then T⁢[s′−1⁢…⁢e′] and T⁢[s′⁢…⁢e′+1] are both unique. Since T⁢[s⁢…⁢e] contains at least one of these two strings as a substring, T⁢[s⁢…⁢e] is also unique. Thus, (s,e) is not a net occurrence. ◀

See 9

Proof.

Let (s′,e′) be the net occurrence, then T⁢[s′⁢…⁢e′] is repeated. Since (s,e) is a proper sub-occurrence of (s′,e′), T⁢[s−1⁢…⁢e] or T⁢[s⁢…⁢e+1] is also repeated. Thus, (s,e) is not a net occurrence. ◀

See 12

Proof.

Let (s,e) be a net occurrence in T that is outside of 𝒞. Assume, by contradiction, that (s,e) is not a super-occurrence of any occurrence (i−1,j+1), where (i,j) is an occurrence in the set of BNSOs, {(i2,j1),(i3,j2),…⁢(ic,jc−1)}. We consider the following two cases depending on the position of s.

First, when ik+1≤s<ik+2 for some 0≤k≤c−2. Given our assumption, (s,e) is not a super-occurrence of (ik+2−1,jk+1+1), where (ik+2,jk+1) is a BNSO. It follows that e satisfies e≤jk+1, implying that (s,e) must be a sub-occurrence of (ik+1,jk+1), which is a net occurrence in 𝒞. Now, if (s,e) is a proper sub-occurrence of (ik+1,jk+1), this this contradicts Observation 9; if (s,e) is (ik+1,jk+1), it contradicts the assumption that (s,e) is a net occurrence in T outside of 𝒞. Second, when s≥ic. In this case, (s,e) is a sub-occurrence of (ic,jc), the last net occurrence in 𝒞.

In both cases, (s,e) must be a sub-occurrence of some net occurrence in 𝒞, leading to a contradiction of either the assumption or Observation 9. Therefore, our initial assumption was false and we conclude that (s,e) is indeed a super-occurrence of (i−1,j+1), where (i,j) is a BNSO of 𝒞. ◀

Appendix B Proofs Omitted from Section 6

See 25

Proof.

By Lemma 15, there are only three positions where Fi−2⁢Qi could occur. First observe that Fi−2⁢Qi cannot occur at position fi−1+1. Next, by Observation 14 and Lemma 23, the occurrences of Fi−2 at positions 1 and fi−2+1 are both followed by Qi, thus Fi−2⁢Qi only occurs at these two positions. ◀

See 27

Proof.

Note that |Qi|=∑j=2i−5|Fj|=(∑j=1i−5fj)−f1=fi−3−1−f1=fi−3−2 where the third equality comes from the fact that ∑j=1kfj=fk+2−1. ◀

See 28

Proof.

By Lemma 25 and Lemma 24, Fi−2⁢Qi only occurs twice in Fi, and both are net occurrences, which means the extensions are unique. Thus, any string containing Fi−2⁢Qi as a substring is also unique. ◀

See 29

Proof.

Observe that the occurrence of Fi−3 at position fi−3+1 is followed by Fi−6⁢Fi−5=Qi−1⁢Δ⁢(1−(imod2)) while the other three occurrences of Fi−3 are all followed by Fi−4=Fi−5⁢Fi−6=Qi−1⁢Δ⁢(imod2). ◀

See 30

Proof.

From the proof of Lemma 29, Fi−3 is only followed by Fi−6⁢Fi−5 once and by Fi−4 three times, thus Fi−3⁢Fi−6⁢Fi−5 is unique. Next, by Lemma 27, |Fi−3⁢Qi−1|=fi−3+fi−4−2=fi−2−2. Also from the proof of Lemma 29, Fi−3⁢Qi−1 is only followed by Δ⁢(1−(imod2)) once and by Δ⁢(imod2) three times, thus, the length-(fi−2−1) prefix of Fi−3⁢Qi−1⁢Δ⁢(1−(imod2))=Fi−3⁢Fi−6⁢Fi−5 is also unique. ◀

See 31

Proof.

Let U be the length-(fi−3−1) prefix of Fi−3. By Equation 1, Fi−3=QiΔ(1−(imod2). Note that U⁢[|U|−1] is always a because Qi ends with F2=a. When U⁢[|U|]=a, the right extension character of U is always b. This is because, if it were not, aaa would occur in Fi, contradicting Lemma 3. On the other hand, when U⁢[|U|]=b, the right extension character of U is always a. This is because, similarly, an occurrence of bb would contradict Lemma 3. Finally, observe that, whether U⁢[|U|] is a or b, U⁢[|U|] concatenated with the right extension character of U is exactly Δ(1−(imod2). Therefore, the desired result follows. ◀

Appendix C Proof Omitted from Section 7

See 36

Proof.

The first two follow from Equation 10 and Equation 11. We next proceed by repeatedly applying the definition of Thue-Morse words. Substituting 𝒯i−2=𝒯i−3⁢𝒯i−3¯=𝒯i−3⁢𝒯i−4¯⁢𝒯i−4 and 𝒯i−3⁢𝒯i−2=𝒯i−4⁢𝒯i−4¯⁢𝒯i−3⁢𝒯i−3¯=𝒯i−4⁢𝒯i−4¯⁢𝒯i−4⁢𝒯i−4¯⁢𝒯i−3¯=𝒯i−4⁢𝒯i−3¯⁢𝒯i−4¯⁢𝒯i−3¯ to Equation 10, we have Equation 8. Finally, observe that 𝒯i−4⁢𝒯i−3¯=𝒯i−4⁢𝒯i−4¯⁢𝒯i−4=𝒯i−3⁢𝒯i−4 and similarly, 𝒯i−3¯⁢𝒯i−4¯=𝒯i−4¯⁢𝒯i−4⁢𝒯i−4¯=𝒯i−4¯⁢𝒯i−3. Substituting them to Equation 8, we derive Equation 9. ◀

Appendix D A Factorization of Thue-Morse Word

First, we define a smallest factorization of a string as one that contains the fewest number of factors while satisfying certain conditions. In this section, we explore two smallest factorizations of 𝒯i, each containing all occurrences of 𝒯i−j and 𝒯i−j¯, respectively. Observe that such factorizations exist due to the overlap-free property of each Thue-Morse word (Lemma 5).

Definition 42.

For each i≥2 and 0≤j≤i−1, we define the following.

  • ■

    Let ℱi,jA and ℱi,jB denote the smallest factorization of 𝒯i that contains all occurrences of 𝒯i−j and 𝒯i−j¯ in 𝒯i, respectively.

  • ■

    Define (Ti−j)−:=Ti−(j+1) and (Ti−j¯)−:=Ti−(j+1)¯.

  • ■

    Consider ℱi,j∈{ℱi,jA,ℱi,jB} and suppose ℱi,j=(xt)t=1m. We define two operators on ℱi,j:

    (ℱi,j)−:=((xt)−)t=1mandℱi,j¯:=(xt¯)t=1m.
  • ■

    Consider two factorizations 𝒳=(xk)k=1m and 𝒴=(yk)k=1ℓ. If xm=y1=𝒯i−j¯, then

    𝒳⊞𝒴:=(x1,x2,…,xm−1,𝒯i−(j+1)¯,𝒯i−j,𝒯i−(j+1),y2,y3,…,yℓ).⌟

For the definition of operator ⊞, note that |𝒳⊞𝒴|=|𝒳|+|𝒴|+1 and

𝒯i−j¯⁢𝒯i−j¯=𝒯i−(j+1)¯⁢𝒯i−(j+1)⁢𝒯i−(j+1)¯⁢𝒯i−(j+1)=𝒯i−(j+1)¯⁢𝒯i−j⁢𝒯i−(j+1).

We next introduce a simple characteristic of ℱi,jA and ℱi,jB.

Observation 43.

For each i≥2 and 0≤j≤i−1, consider a factorization 𝒳=(xk)k=1m of 𝒯i that contains all occurrences of 𝒯i−j (respectively, 𝒯i−j¯). If no two consecutive factors are both different from 𝒯i−j (respectively, 𝒯i−j¯), then 𝒳 is the smallest and thus 𝒳=ℱi,jA (respectively, 𝒳=ℱi,jB).

The observation holds because, otherwise, we could merge the two consecutive factors and obtain a smaller factorization.

Now, we present the main result of the section, illustrated in Figure 8.

Figure 8: An illustration of Theorem 44 for 1≤j≤4. Operators (⋅)− and (⋅)¯ are defined in Definition 42. Notice the green occurrences of 𝒯i−j are introduced from ⊞.
Theorem 44.

For i≥2 and 0≤j≤i−1, the following statements hold.

  1. (1)

    For each ℱi,j∈{ℱi,jA,ℱi,jB}, let ℱi,j=(xt)t=1m. Then each term xt is in the set ℬi,j:={𝒯i−j,𝒯i−j¯,𝒯i−(j+1),𝒯i−(j+1)¯}. Moreover, x1=𝒯i−j. If j is even, then xm=𝒯i−j; otherwise, xm=𝒯i−j¯.

  2. (2)

    ℱi,0A=(𝒯i), ℱi,1A=(𝒯i−1,𝒯i−1¯), and for each j≥2,

    ℱi,jA={(ℱi,j−1A)−⁢(ℱi,j−1B)−¯,j is odd,(ℱi,j−1A)−⊞(ℱi,j−1B)−¯,j is even.
  3. (3)

    ℱi,0B=(), ℱi,1B=(𝒯i−1,𝒯i−1¯), and for each j≥2, ℱi,jB=(ℱi,j−1B)−⁢(ℱi,j−1A)−¯.

Proof.

We proceed by induction on j.

Base cases.

The claim holds trivially for j=0. When j=1, (𝒯i−1,𝒯i−1¯) is the smallest factorization following Observation 43 and Statement 1 holds. Next, note that (ℱi,1A)−=(ℱi,1B)−=(𝒯i−2,𝒯i−2¯). When j=2, it follows that

ℱi,2A=(ℱi,1A)−⊞(ℱi,1B)−¯=(𝒯i−2,𝒯i−2¯)⊞(𝒯i−2¯,𝒯i−2)=(𝒯i−2,𝒯i−3¯,𝒯i−2,𝒯i−3,𝒯i−2) (10)

and

ℱi,2B=(ℱi,1B)−⁢(ℱi,1A)−¯=(𝒯i−2,𝒯i−2¯)⁢(𝒯i−2¯,𝒯i−2)=(𝒯i−2,𝒯i−2¯,𝒯i−2¯,𝒯i−2) (11)

Both factorizations are the smallest following Observation 43, and Statement 1 holds.

Inductive step.

Let k be an odd integer such that 3≤k≤i−1, and assume the claim holds for j=k−1. Specifically, assume ℱi,k−1A=(xt)t=1m, ℱi,k−1B=(yt)t=1l, x1=y1=𝒯i−(k−1), and xm=yl=𝒯i−(k−1). We now prove the result for j=k.

First, note that both ℱi,k−1A and ℱi,k−1B are factorizations of 𝒯i. Then, by the definition of operation (⋅)−, both (ℱi,k−1A)− and (ℱi,k−1B)− are factorizations of 𝒯i−1, and (ℱi,k−1B)−¯ is a factorization of 𝒯i−1¯. Now, since 𝒯i=𝒯i−1⁢𝒯i−1¯, it follows that 𝒴oddA:=(ℱi,k−1A)−⁢(ℱi,k−1B)−¯ is a factorization of 𝒯i. It remains to show that 𝒴oddA=ℱi,jA. Observe that

𝒴oddA=(ℱi,k−1A)−⁢(ℱi,k−1B)−¯=(x1)−⁢⋯⁢(xm−1)−⁢𝒯i−k⁢𝒯i−k¯⁢(y2)−¯⁢⋯⁢(yl)−¯. (12)

By the induction hypothesis on ℱi,k−1A and ℱi,k−1B and the definition of operation (⋅)−, we know that (ℱi,k−1A)− contains all the occurrences of 𝒯i−k in 𝒯i−1, and (ℱi,k−1B)−¯ contains all the occurrences of 𝒯i−k in 𝒯i−1¯. Since 𝒯i=𝒯i−1⁢𝒯i−1¯ and 𝒯i is overlap-free, it follows that 𝒴oddA contains all the occurrences of 𝒯i−k. Moreover, no two consecutive factors of 𝒴oddA are both different from 𝒯i−k, so by Observation 43, we conclude that 𝒴oddA=ℱi,jA.

Next we show that all factors of ℱi,jA are elements of ℬi,k. Since xt∈ℬi,k−1 for each 1≤t≤m and yt∈ℬi,k−1 for each 1≤t≤l, it follows from Equation 12 that all factors of ℱi,jA are elements of ℬi,k. Additionally, the first factor in ℱi,jA is (x1)−=𝒯i−k, and the last factor in ℱi,jA is (yl)−¯=𝒯i−k¯ since k−1 is even.

Similarly, we can show that 𝒴oddB:=(ℱi,k−1B)−⁢(ℱi,k−1A)−¯ is a factorization of 𝒯i, that 𝒴oddB=ℱi,jB, and that Statement 1 holds.

We can prove analogously when k is even. In this case, the operation ⊞ is used to ensure that ℱi,jA contains all occurrences of 𝒯i−j. Specifically, when k is even, we have (xm)−=(y1)−¯=𝒯i−k¯, and there is a occurrence of 𝒯i−k within 𝒯i−k¯⁢𝒯i−k¯=𝒯i−(k+1)¯⁢𝒯i−k⁢𝒯i−(k+1). ◀