Some Properties of the Quasi-Tribonacci Sequence

Abstract

The quasi-Tribonacci sequence $\mathbf{T}$, which is a transformation of the Tribonacci sequence, is the fixed point of morphism $\varphi :0\to 01$, $1\to 12$, $2\to 0$. In this paper, we study the properties of the factors in the quasi-Tribonacci sequence and give the singular factorization and the Lyndon factorization.

1. Introduction

Let $\mathbf{f}={\left\{f_n\right\}}_{n\ge 0}$ be the Fibonacci sequence that is the fixed point of the Fibonacci morphism ${\sigma }_f$ defined by $0\mapsto 01,1\mapsto 0$. The Fibonacci sequence has been extensively studied [1,2,3,4]. The combinatorial properties of the Fibonacci sequence are of great interest in number theory, fractal geometry, formal language, computational complexity and quasi-crystals [5,6,7,8,9]. There are several generalizations of the Fibonacci sequence on the alphabet of two letters. Ramírez, Rubiano and Castro [10,11,12] gave the generations of the Fibonacci sequence and proved them to be characteristic Sturmian words of irrational slopes. There are also extensions on alphabets containing at least three letters. A famous one is the Tribonacci sequence, which is the fixed point of the Rauzy substitution $0\mapsto 01,1\mapsto 02,2\mapsto 0$. Rauzy [13] studied the dynamical and geometrical aspects of the Tribonacci sequences; see also [4]. Huang and Wen [14] studied the occurrence of the arbitrary factors of $\mathbf{f}$ and gave a quite general decomposition of $\mathbf{f}$. Moreover, Huang and Wen [15,16] studied the kernel words, gap sequence and the numbers of repeated palindromes in the Tribonacci words. Kjos-Hanssen [17] studied the automatic complexity of Fibonacci and Tribonacci words. Břinda, Pelantová and Turek [18] gave the extension of the Fibonacci sequence to an alphabet of m letters, which is called m-bonacci sequence. Jahannia, Mohammad-Noori, Rampersad and Stipulanti [19,20] studied the Ziv–Lempel factorization of m-bonacci words. Recently, Zhang, Wen and Wu [21] gave the extension of the Fibonacci sequence to the infinite alphabet $\mathbb{N}$ and studied its combinatorial properties, including the growth order, digit sum and several decompositions. Ghareghani, Mohammad-Noori and Sharifani [22,23] gave the generations of the m-bonacci sequence to the infinite alphabet and studied the palindrome complexity, square factors and critical factors. Given the extensive research and significant importance of the Fibonacci and Tribonacci sequences, here, we study the transformation of the Tribonacci morphism, which is called quasi-Tribonacci morphism.

Consider the quasi-Tribonacci morphism $\varphi$ which is given by

$$\varphi :0\mapsto 01,1\mapsto 12\mathrm{and}2\mapsto 0.$$

The quasi-Tribonacci sequence $\mathbf{T}={\left\{T_i\right\}}_{i\ge 0}$ is the fixed point of $\varphi$ starting by 0. Wen and Wen [24] studied the properties of factors of the Fibonacci sequence $\mathbf{f}$ by analyzing its singular words. They proved that the adjacent singular words of the same order are positively separated, and gave the singular factorization of $\mathbf{f}$. Levé and Séébold [25] studied the singular factorization of the Fibonacci sequence and its conjugates. Melançon [26,27] studied the Lyndon factorization of the Fibonacci sequence and gave the link between the singular factorization and Lyndon factorization of the Fibonacci word. Tan and Wen [28] studied the factor properties of the Tribonacci sequence. Moreover, they gave the singular factorization and the Lyndon factorization of the Tribonacci sequences. Singular words have applications in the study of combinatorics on words [29,30,31,32,33], including the exploration the power of the factors, local isomorphism, special words and overlap of the subwords. Lyndon factorization has plenty of algorithmic applications and might be computed in an efficient way [34,35]. This factorization is also used in string processing algorithms (see [36]) and for the computation of runs in words (see, e.g., [37]).

In this paper, we mainly study the properties of the factors of the quasi-Tribonacci sequence. After giving notation and listing some known facts, we give the three following decompositions. The first one is a decomposition using its (slightly modified) prefixes (see Theorem 1). The second one is a decomposition using singular words (see Theorem 2). The third one is a decomposition using singular words and their combination (see Theorem 3). Moreover, the Lyndon factorization is also given (see Theorem 6).

This paper is organized as follows. In Section 2, we state some basic notation and study the fundamental properties of the quasi-Tribonacci sequence. In Section 3, we give three decompositions of $\mathbf{T}$. In Section 4, we study the Lyndon words and Lyndon decomposition of $\mathbf{T}$. In Section 5, we present the conclusion.

2. Preliminary

Let $\mathcal{A}=\{0,1,2\}$ be a three-letter alphabet. Let ${\mathcal{A}}^k$ denote the set of all words of length k on $\mathcal{A}$, and ${\mathcal{A}}^{*}$ denote the set of all finite words on $\mathcal{A}$. The length of a finite word $w\in {\mathcal{A}}^{*}$ is denoted by $\left|w\right|$. For any $V,W\in {\mathcal{A}}^{*}$, we write $V\prec W$ when the finite word V is a factor of the word W, that is, when there exist words $U,U^{\prime }\in {\mathcal{A}}^{*}$, such that $W=UV{U}^{\prime }$. We say that V is a prefix (resp. suffix) of a word W, and we write $V⊲W$ (resp. $V⊳W$) if there exists a word $U^{\prime }\in {\mathcal{A}}^{*}$, such that $W=V{U}^{\prime }$ (resp. $W=U^{\prime }V$).

Let $\mathbf{u}=u_0{u}_1{u}_2\cdots u_n$ be a finite word (or $\mathbf{u}=u_0{u}_1{u}_2\cdots u_n\cdots$ be a sequence). For any $i\le j\le n$, define $\mathbf{u}[i,j]:=u_i{u}_{i+1}\cdots u_{j-1}{u}_j$.

Denote by $W^{-1}$ the inverse of W, that is, $W^{-1}=w_p^{-1}\cdots w_2^{-1}{w}_1^{-1}$ where $W=w_1{w}_2\cdots w_p$. If V is a suffix of W, we can write $W{V}^{-1}=U$, with $W=UV$. This makes sense in ${\mathcal{A}}^{*}$, since the reduced word associated with $W{V}^{-1}$ belong to ${\mathcal{A}}^{*}$. This abuse of language will be very useful in what follows.

A morphism $\tau :{\mathcal{A}}^{*}\to {\mathcal{A}}^{*}$ is called a substitution of ${\mathcal{A}}^{*}$. We denote $\mathbf{F}$ by any one of the fixed points of $\tau$ (i.e., $\tau \left(\mathbf{F}\right)=\mathbf{F}$), if it exists. We define the nth iteration of $\tau$ by ${\tau }^n\left(a\right)=\tau \left({\tau }^{n-1}\left(a\right)\right)$, for $n\ge 2$ and $a\in \mathcal{A}$. By convention, we define ${\tau }^0\left(a\right)=a$, for all $a\in \mathcal{A}$.

The Tribonacci sequence $\mathbf{t}={\left\{t_n\right\}}_{n\ge 0}$ is generated by the Tribonacci morphism $\sigma$: $0\mapsto 01,1\mapsto 02,2\mapsto 0$, i.e., $\mathbf{t}={\sigma }^{\infty }\left(0\right)$. We know that $\left\{\right|{\sigma }^n\left(0\right){\left|\right\}}_{n\ge 0}$ are the Tribonacci numbers.

Consider the morphism $\varphi$ which is given by

$$\varphi :0\mapsto 01,1\mapsto 12,2\mapsto 0.$$

We know that $\varphi \left(0\right)=01,{\varphi }^2\left(0\right)=\varphi \left(\varphi \left(0\right)\right)=\varphi \left(01\right)=\varphi \left(0\right)\varphi \left(1\right)=0112,{\varphi }^3\left(0\right)=\varphi \left({\varphi }^2\left(0\right)\right)=0112120,\cdots$ Then, we observe that ${\varphi }^n\left(0\right)⊲{\varphi }^{n+1}\left(0\right)$. Thus, there exists a fixed point of $\varphi$ by iterating $\varphi$ with the letter 0. The quasi-Tribonacci sequence $\mathbf{T}={\left\{T_i\right\}}_{i\ge 0}$ is the fixed point of $\varphi$ starting by 0. We define $\tilde{\mathbf{T}}={\left\{\widetilde{T_i}\right\}}_{i\ge 0}$ as the fixed point of $\varphi$ starting by 1. The first several terms of $\mathbf{T}$, $\tilde{\mathbf{T}}$ and $\mathbf{t}$ are

$$\begin{array}{cc}\hfill \mathbf{T}& =011212012001120010112120\cdots \hfill \\ \hfill \tilde{\mathbf{T}}& =120010112011212001121201\cdots \hfill \\ \hfill \mathbf{t}& =010201001020101020100102\cdots \hfill \end{array}$$

For convenience, we define ${\varphi }^0\left(i\right)=i$, where $i\in \{0,1,2\}$. Let $L_n:=\left|{\varphi }^n\left(0\right)\right|$ and ${\mathsf{\ell }}_n:=\left|{\varphi }^n\left(1\right)\right|$, where $L_0=1$, $L_1=2$, ${\mathsf{\ell }}_0=1$ and ${\mathsf{\ell }}_1=2$. Then, we have the following properties:

Proposition 1. 

For all $n\ge 2$,

$$\left\{\begin{array}{cc}{L}_n=L_{n-1}+{\mathsf{\ell }}_{n-1},\hfill & \hfill \\ {\mathsf{\ell }}_n={\mathsf{\ell }}_{n-1}+L_{n-2},\hfill & \hfill \end{array}\right.$$

(1)

i.e.,

$$\left(\begin{array}{c}{L}_{n+1}\\ L_n\\ {\mathsf{\ell }}_{n+1}\\ {\mathsf{\ell }}_n\end{array}\right)=\left(\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 0& 1& 1& 0\\ 0& 0& 1& 0\end{array}\right)\left(\begin{array}{c}{L}_n\\ L_{n-1}\\ {\mathsf{\ell }}_n\\ {\mathsf{\ell }}_{n-1}\end{array}\right).$$

Proof. 

Since ${\varphi }^n\left(0\right)={\varphi }^{n-1}\left(01\right)={\varphi }^{n-1}\left(0\right){\varphi }^{n-1}\left(1\right)$, we have

$$L_n=|{\varphi }^n\left(0\right)|=|{\varphi }^{n-1}\left(0\right)|+|{\varphi }^{n-1}\left(1\right)|=L_{n-1}+{\mathsf{\ell }}_{n-1}.$$

By the definition of the morphism $\varphi$, ${\varphi }^n\left(1\right)={\varphi }^{n-1}\left(12\right)={\varphi }^{n-1}\left(1\right){\varphi }^{n-2}\left(0\right)$, so

$${\mathsf{\ell }}_n=|{\varphi }^n\left(1\right)|=|{\varphi }^{n-1}\left(1\right)|+|{\varphi }^{n-2}\left(0\right)|={\mathsf{\ell }}_{n-1}+L_{n-2}.$$

□

Proposition 2. 

For all $n\ge 2$, we have

$${\mathsf{\ell }}_n=2+\sum _{i=0}^{n-2}{L}_n.$$

(2)

Proof. 

Since

$$\begin{array}{ccc}\hfill {\varphi }^n\left(1\right)& =& {\varphi }^{n-1}\left(1\right){\varphi }^{n-1}\left(2\right)={\varphi }^{n-1}\left(1\right){\varphi }^{n-2}\left(0\right)\hfill \\ & =& {\varphi }^{n-2}\left(1\right){\varphi }^{n-2}\left(2\right){\varphi }^{n-2}\left(0\right)={\varphi }^{n-2}\left(1\right){\varphi }^{n-3}\left(0\right){\varphi }^{n-2}\left(0\right)\hfill \\ & =& \cdots \hfill \\ & =& \varphi \left(1\right){\varphi }^0\left(0\right){\varphi }^2\left(0\right)\cdots {\varphi }^{n-2}\left(0\right)\hfill \\ & =& 12{\varphi }^0\left(0\right){\varphi }^2\left(0\right)\cdots {\varphi }^{n-2}\left(0\right),\hfill \end{array}$$

according to the definition of ${\mathsf{\ell }}_n$, we have

$${\mathsf{\ell }}_n=2+\sum _{i=0}^{n-2}{L}_n.$$

□

Proposition 3. 

For all $n\ge 3$,

$$L_n=2+\sum _{i=0}^{n-3}{L}_n.$$

Proof. 

By Equations (1) and (2), we have

$$L_n=L_{n-1}+{\mathsf{\ell }}_{n-1}=L_{n-1}+2+\sum _{i=0}^{n-3}{L}_n=2+\sum _{i=0}^{n-3}{L}_n.$$

□

Here, we show that $L_n$ and ${\mathsf{\ell }}_n$ share the same recurrence relation with the A00521 in the OEIS [38], apart from the initial values. We shall prove it in the following.

Proposition 4. 

For all $n\ge 4$,

$$L_n=L_{n-1}+L_{n-2}+L_{n-4},$$

$${\mathsf{\ell }}_n={\mathsf{\ell }}_{n-1}+{\mathsf{\ell }}_{n-2}+{\mathsf{\ell }}_{n-4},$$

with $L_0=1,L_1=2,L_2=4,L_3=7,{\mathsf{\ell }}_0=1,{\mathsf{\ell }}_1=2,{\mathsf{\ell }}_2=3,{\mathsf{\ell }}_3=5$.

Proof. 

By Proposition 1, we know $L_4=L_3+{\mathsf{\ell }}_3=12$ and ${\mathsf{\ell }}_4={\mathsf{\ell }}_3+L_2=9$. It is easy to see that the result is true for $n=4$. Assuming that the result is true for all $m\le n$, then for $m=n+1$, we have

$$\begin{array}{ccc}\hfill L_{n+1}& =& L_n+{\mathsf{\ell }}_n\hfill \\ & =& L_{n-1}+L_{n-2}+L_{n-4}+{\mathsf{\ell }}_{n-1}+{\mathsf{\ell }}_{n-2}+{\mathsf{\ell }}_{n-4}\hfill \\ & =& \left(L_{n-1}+{\mathsf{\ell }}_{n-1}\right)+\left(L_{n-2}+{\mathsf{\ell }}_{n-2}\right)+\left(L_{n-4}+{\mathsf{\ell }}_{n-4}\right)\hfill \\ & =& L_n+L_{n-1}+L_{n-3},\hfill \end{array}$$

$$\begin{array}{ccc}\hfill {\mathsf{\ell }}_{n+1}& =& {\mathsf{\ell }}_n+L_{n-1}\hfill \\ & =& {\mathsf{\ell }}_{n-1}+{\mathsf{\ell }}_{n-2}+{\mathsf{\ell }}_{n-4}+L_{n-2}+L_{n-3}+L_{n-5}\hfill \\ & =& \left({\mathsf{\ell }}_{n-1}+L_{n-2}\right)+\left({\mathsf{\ell }}_{n-2}+L_{n-3}\right)+\left({\mathsf{\ell }}_{n-4}+L_{n-5}\right)\hfill \\ & =& {\mathsf{\ell }}_n+{\mathsf{\ell }}_{n-1}+{\mathsf{\ell }}_{n-3}.\hfill \end{array}$$

Hence, the result is true. □

3. Singular Words Decomposition

In this section, we present three decompositions (factorisations) of $\mathbf{T}$. These are given in Theorems 1–3. We will state the theorems firstly and introduce some notation, then give the three proofs.

Theorem 1. 

$$\mathbf{T}=\prod _{i=-1}^{\infty }{\varphi }^i\left(1\right),\tilde{\mathbf{T}}=12\prod _{i=0}^{\infty }{\varphi }^i\left(0\right),$$

where ${\varphi }^{-1}\left(1\right):=0$.

We observe that the following decomposition of the quasi-Tribonacci sequence is

$$\mathbf{T}=\underset{{\mathsf{\ell }}_0}{\underbrace{0}}\underset{{\mathsf{\ell }}_1}{\underbrace{11}}\underset{{\mathsf{\ell }}_2}{\underbrace{212}}\underset{{\mathsf{\ell }}_3}{\underbrace{01200}}\underset{{\mathsf{\ell }}_4}{\underbrace{112001011}}\underset{{\mathsf{\ell }}_5}{\underbrace{2120010112011212}}\cdots$$

That is, the length of the nth block in the decomposition is ${\mathsf{\ell }}_n$. Then, a question is posed naturally: What are these blocks? As we see, the second decomposition theorem will answer this question completely. To state the second decomposition theorem, we introduce the singular words. For all $n\ge 1$ and $i\in \{0,1,2\}$, the word

$$\alpha {\varphi }^n\left(i\right){\beta }^{-1}=:S_n^{\left(i\right)}$$

is called a singular word of order n (of letter i), where $\alpha$ and $\beta$ are two letters satisfying $\alpha \beta ⊳{\varphi }^n\left(i\right)$. In addition, we define $S_0^{\left(i\right)}:=0$. For example, since ${\varphi }^3\left(0\right)=0112120$, then $\alpha =2$, $\beta =0$ and $S_3^{\left(0\right)}=2011212$.

Theorem 2. 

$$\mathbf{T}=\prod _{i=0}^{\infty }{S}_i^{\left(1\right)}=\underline{0}\underline{11}\underline{212}\underline{01200}\underline{112001011}\cdots ,$$

$$\tilde{\mathbf{T}}=12\prod _{i=1}^{\infty }{S}_i^{\left(0\right)}=12\underline{00}\underline{1011}\underline{2011212}\underline{001121201200}\cdots .$$

We need the following notation to state the third theorem. Let $r_n\in \{0,1,2\}$ satisfy that $r_n\equiv n$$(mod3)$, for all $n\ge 0$.

Theorem 3. 

$$\mathbf{T}=011212012\prod _{i=1}^{\infty }\left(S_i^{\left(0\right)}{G}_i^{\left(0\right)}\right),\tilde{\mathbf{T}}=12\prod _{j=0}^{\infty }\left(r_j\prod _{i=0}^j{S}_i^{\left(1\right)}\right),$$

where $G_n^{\left(0\right)}=A_{r_n}{\prod }_{i=1}^n{S}_i^{\left(0\right)}(n\ge 1)$ and $A_0=012$, $A_1=112$, $A_2=212$.

Then, we will prove the above theorems.

Lemma 1. 

For all $n\ge 0$, ${\varphi }^n\left(0\right)={\prod }_{i=-1}^{n-1}{\varphi }^i\left(1\right)$.

Proof. 

We use induction on n to prove it. It is easy to see that the result is true for $n=0$, since ${\varphi }^0\left(0\right)=0$ and ${\varphi }^{-1}\left(1\right)=0$.

Assuming that the result is true for all $m\le n$, then for $m=n+1$, we have

$${\varphi }^{n+1}\left(0\right)={\varphi }^n\left(0\right){\varphi }^n\left(1\right)=\left(\prod _{i=-1}^{n-1}{\varphi }^i\left(1\right)\right){\varphi }^n\left(1\right)=\prod _{i=-1}^n{\varphi }^i\left(1\right).$$

Hence, the result is true. □

Lemma 2. 

For all $n\ge 0$, ${\varphi }^{n+2}\left(1\right)=\varphi \left(1\right){\prod }_{i=0}^n{\varphi }^i\left(0\right)$.

Proof. 

From the definition of the morphism $\varphi$, we have

$$\begin{array}{ccc}\hfill {\varphi }^{n+2}\left(1\right)& =& {\varphi }^{n+1}\left(1\right){\varphi }^{n+1}\left(2\right)\hfill \\ & =& {\varphi }^{n+1}\left(1\right){\varphi }^n\left(0\right)\hfill \\ & =& {\varphi }^n\left(1\right){\varphi }^{n-1}\left(0\right){\varphi }^n\left(0\right)\hfill \\ & =& \cdots \hfill \\ & =& \varphi \left(1\right){\varphi }^0\left(0\right)\varphi \left(0\right)\cdots {\varphi }^{n-1}\left(0\right){\varphi }^n\left(0\right)\hfill \\ & =& \varphi \left(1\right)\prod _{i=0}^n{\varphi }^i\left(0\right).\hfill \end{array}$$

□

The proof of Theorem 1 is direct from Lemmas 1 and 2.

To prove the second decomposition theorem, we introduce the following notation and properties. Let ${\alpha }_n,{\beta }_n$ be the last two letters in ${\varphi }^n\left(0\right)$, i.e., ${\alpha }_n{\beta }_n⊳{\varphi }^n\left(0\right)$, for all $n\ge 1$. For convenience, we denote ${\alpha }_0=2$ and ${\beta }_0=0$.

Proposition 5. 

For all $n\ge 0$, ${\alpha }_{3n}=2,{\alpha }_{3n+1}=0,{\alpha }_{3n+2}=1,{\beta }_{3n}=0,{\beta }_{3n+1}=1$ and ${\beta }_{3n+2}=2$. Moreover, ${\alpha }_n={\beta }_{n+2}$ for all $n\ge 0$.

Proof. 

By the definition of the morphism $\varphi$, we have

$$\begin{array}{ccc}\hfill {\varphi }^{3n}\left(0\right)& =& {\varphi }^{3n-1}\left(0\right){\varphi }^{3n-1}\left(1\right)\hfill \\ & =& {\varphi }^{3n-1}\left(0\right){\varphi }^{3n-2}\left(1\right){\varphi }^{3n-3}\left(0\right)\hfill \\ & =& {\varphi }^{3n-1}\left(0\right){\varphi }^{3n-2}\left(1\right){\varphi }^{3n-4}\left(0\right){\varphi }^{3n-5}\left(1\right){\varphi }^{3n-6}\left(0\right)\hfill \\ & =& \cdots \hfill \\ & =& {\varphi }^{3n-1}\left(0\right){\varphi }^{3n-2}\left(1\right){\varphi }^{3n-4}\left(0\right){\varphi }^{3n-5}\left(1\right)\cdots {\varphi }^2\left(0\right)\varphi \left(1\right){\varphi }^0\left(0\right),\hfill \end{array}$$

where $20⊳\varphi \left(1\right){\varphi }^0\left(0\right)=120$. So, ${\alpha }_{3n}=2,{\beta }_{3n}=0$. Since

$$\begin{array}{ccc}\hfill {\varphi }^{3n+1}\left(0\right)& =& {\varphi }^{3n}\left(0\right){\varphi }^{3n}\left(1\right)\hfill \\ & =& {\varphi }^{3n}\left(0\right){\varphi }^{3n-1}\left(1\right){\varphi }^{3n-2}\left(0\right)\hfill \\ & =& \cdots \hfill \\ & =& {\varphi }^{3n}\left(0\right){\varphi }^{3n-1}\left(1\right){\varphi }^{3n-3}\left(0\right){\varphi }^{3n-4}\left(1\right)\cdots {\varphi }^3\left(0\right){\varphi }^2\left(1\right)\varphi \left(0\right),\hfill \end{array}$$

where $01⊳\varphi \left(0\right)=01$, we have ${\alpha }_{3n+1}=0,{\beta }_{3n+1}=1$.

In addition,

$$\begin{array}{ccc}\hfill {\varphi }^{3n+2}\left(0\right)& =& {\varphi }^{3n+1}\left(0\right){\varphi }^{3n+1}\left(1\right)\hfill \\ & =& {\varphi }^{3n+1}\left(0\right){\varphi }^{3n}\left(1\right){\varphi }^{3n-1}\left(0\right)\hfill \\ & =& \cdots \hfill \\ & =& {\varphi }^{3n+1}\left(0\right){\varphi }^{3n}\left(1\right){\varphi }^{3n-2}\left(0\right){\varphi }^{3n-3}\left(1\right)\cdots {\varphi }^4\left(0\right){\varphi }^3\left(1\right){\varphi }^2\left(0\right),\hfill \end{array}$$

where $12⊳{\varphi }^2\left(0\right)=0112$. So, ${\alpha }_{3n+2}=1,{\beta }_{3n+2}=2$. Hence, the results are true. □

Remark 1. (1) For all $n\ge 0$, ${\alpha }_n{\beta }_n⊳{\varphi }^{n+2}\left(1\right)$ and ${\beta }_{n+2}{\beta }_{n+3}⊳{\varphi }^{n+2}\left(1\right)$, since ${\varphi }^{n+2}\left(1\right)={\varphi }^{n+1}\left(1\right){\varphi }^n\left(0\right)$, ${\alpha }_n={\beta }_{n+2}$ and ${\beta }_n={\beta }_{n+3}$.

(2) $S_n^{\left(1\right)}={\beta }_n{\varphi }^n\left(1\right){\beta }_{n+1}^{-1}$ for all $n\ge 0$.

(3) ${\alpha }_n=r_{n-1}={\beta }_{n+2}$ for all $n\ge 1$.

(4) $r_n={\beta }_n$ for all $n\ge 0$.

Remark 2. 

For all $n\ge 0$, ${\beta }_{n+4}{\beta }_{n+3}^{-1}{A}_{r_n}=A_{r_{n+1}}$.

Proof. 

When $n=3k$, then $r_n=0$, ${\beta }_{n+4}=1$, ${\beta }_{n+3}=0$. So,

$${\beta }_{n+4}{\beta }_{n+3}^{-1}{A}_{r_n}=10^{-1}012=112=A_1=A_{r_{n+1}}.$$

When $n=3k+1$, then $r_n=1$, ${\beta }_{n+4}=2$, ${\beta }_{n+3}=1$. So,

$${\beta }_{n+4}{\beta }_{n+3}^{-1}{A}_{r_n}=21^{-1}112=212=A_2=A_{r_{n+1}}.$$

When $n=3k+2$, then $r_n=2$, ${\beta }_{n+4}=0$, ${\beta }_{n+3}=2$. So,

$${\beta }_{n+4}{\beta }_{n+3}^{-1}{A}_{r_n}=02^{-1}212=012=A_0=A_{r_{n+1}}.$$

Hence, the result is true. □

Lemma 3. 

For all $n\ge 1$, ${\varphi }^n\left(0\right)=\left({\prod }_{i=0}^{n-1}{S}_i^{\left(1\right)}\right){\beta }_n$.

Proof. 

We use induction on n to prove it. It is easy to see that the result is true for $n=1$, since ${\beta }_1=1$, $S_0^{\left(1\right)}=0$ and $\varphi \left(0\right)=01=S_0^{\left(1\right)}{\beta }_1$.

Assuming that the result is true for all $m<n$, then for $m=n$, we have

$$\begin{array}{ccc}\hfill {\varphi }^n\left(0\right)& =& {\varphi }^{n-1}\left(0\right){\varphi }^{n-1}\left(1\right)\hfill \\ & =& \left(\prod _{i=0}^{n-2}{S}_i^{\left(1\right)}\right){\beta }_{n-1}{\varphi }^{n-1}\left(1\right)\hfill \\ & =& \left(\prod _{i=0}^{n-2}{S}_i^{\left(1\right)}\right)\left({\beta }_{n-1}{\varphi }^{n-1}\left(1\right){\beta }_n^{-1}\right){\beta }_n\hfill \\ & =& \left(\prod _{i=0}^{n-2}{S}_i^{\left(1\right)}\right)S_{n-1}^{\left(1\right)}{\beta }_n\hfill \\ & =& \left(\prod _{i=0}^{n-1}{S}_i^{\left(1\right)}\right){\beta }_n.\hfill \end{array}$$

Hence, the result is true. □

Lemma 4. 

For all $n\ge 3$, ${\varphi }^n\left(1\right)=12\left({\prod }_{i=1}^{n-2}{S}_i^{\left(0\right)}\right){\alpha }_{n-1}$.

Proof. 

We use induction on n to prove it. It is easy to see that the result is true for $n=3$, since ${\alpha }_2=1$, $S_1^{\left(0\right)}=00$ and ${\varphi }^3\left(1\right)=12001=12S_1^{\left(0\right)}{\alpha }_2$.

Assuming that the result is true for all $m\le n$, then for $m=n+1$, we have

$$\begin{array}{ccc}\hfill {\varphi }^{n+1}\left(1\right)& =& {\varphi }^n\left(1\right){\varphi }^{n-1}\left(0\right)\hfill \\ & =& 12\left(\prod _{i=1}^{n-2}{S}_i^{\left(0\right)}\right){\alpha }_{n-1}{\varphi }^{n-1}\left(0\right)\hfill \\ & =& 12\left(\prod _{i=1}^{n-2}{S}_i^{\left(0\right)}\right)\left({\alpha }_{n-1}{\varphi }^{n-1}\left(0\right){\beta }_{n-1}^{-1}\right){\beta }_{n-1}\hfill \\ & =& 12\left(\prod _{i=1}^{n-2}{S}_i^{\left(0\right)}\right)S_{n-1}^{\left(0\right)}{\alpha }_n(\mathrm{by} \mathrm{Proposition} 5)\hfill \\ & =& 12\left(\prod _{i=1}^{n-1}{S}_i^{\left(0\right)}\right){\alpha }_n.\hfill \end{array}$$

Hence, the result is true. □

Proof of Theorem 2 

According to the definition of ${\alpha }_n$, we know ${\alpha }_n⊲S_n^{\left(0\right)}$. From Remark 1(2), we have ${\beta }_n⊲S_n^{\left(1\right)}$. Thus, the results are true from Lemmas 3 and 4. □

The following are the relationship between the singular words $S_n^{\left(0\right)}$ and $S_n^{\left(1\right)}$, for all $n\ge 0$.

Proposition 6. 

For all $n\ge 0$, $S_{n+1}^{\left(0\right)}=r_n\left({\prod }_{i=0}^n{S}_i^{\left(1\right)}\right)$.

Proof. 

We use induction on n to prove it. It is easy to see that the result is true for $n=0$, since $S_1^{\left(0\right)}=00$, $S_0^{\left(1\right)}=0$, $r_0=0$ and $S_1^{\left(0\right)}=r_0{S}_0^{\left(1\right)}$.

Assuming that the result is true for all $m\le n$, then for $m=n+1$, we have

$$\begin{array}{ccc}\hfill S_{n+2}^{\left(0\right)}& =& {\alpha }_{n+2}{\varphi }^{n+2}\left(0\right){\beta }_{n+2}^{-1}\hfill \\ & =& {\alpha }_{n+2}{\varphi }^{n+1}\left(0\right){\varphi }^{n+1}\left(1\right){\beta }_{n+2}^{-1}\hfill \\ & =& {\alpha }_{n+2}{\alpha }_{n+1}^{-1}\left({\alpha }_{n+1}{\varphi }^{n+1}\left(0\right){\beta }_{n+1}^{-1}\right)\left({\beta }_{n+1}{\varphi }^{n+1}\left(1\right){\beta }_{n+2}^{-1}\right)\hfill \\ & =& {\alpha }_{n+2}{\alpha }_{n+1}^{-1}{S}_{n+1}^{\left(0\right)}{S}_{n+1}^{\left(1\right)}(\mathrm{by} \mathrm{Remark} 1\left(2\right))\hfill \\ & =& {\alpha }_{n+2}{\alpha }_{n+1}^{-1}{r}_n\left(\prod _{i=0}^n{S}_i^{\left(1\right)}\right)S_{n+1}^{\left(1\right)}\hfill \\ & =& r_{n+1}\left(r_n^{-1}{r}_n\right)\left(\prod _{i=0}^n{S}_i^{\left(1\right)}\right)S_{n+1}^{\left(1\right)}(\mathrm{by} \mathrm{Remark} 1\left(3\right))\hfill \\ & =& r_{n+1}\left(\prod _{i=0}^{n+1}{S}_i^{\left(1\right)}\right).\hfill \end{array}$$

Hence, the result is true. □

Proposition 7. 

For all $n\ge 1$,

$$S_{n+2}^{\left(1\right)}=A_{r_{n-1}}\left(\prod _{i=1}^n{S}_i^{\left(0\right)}\right).$$

Moreover, $S_{n+2}^{\left(1\right)}=r_{n-1}12\left({\prod }_{i=1}^n{S}_i^{\left(0\right)}\right)$.

Proof. 

We use induction on n to prove it. It is easy to see that the result is true for $n=1$, since $S_1^{\left(0\right)}=00$, $S_3^{\left(1\right)}=01200$, $r_0=0$, $A_{r_0}=012$ and $S_3^{\left(1\right)}=A_{r_0}{S}_1^{\left(0\right)}$.

Assuming that the result is true for all $m<n$, then for $m=n$, we have

$$\begin{array}{ccc}\hfill S_{n+2}^{\left(1\right)}& =& {\beta }_{n+2}{\varphi }^{n+2}\left(1\right){\beta }_{n+3}^{-1}(\mathrm{by} \mathrm{Remark} 1\left(2\right))\hfill \\ & =& {\beta }_{n+2}{\varphi }^{n+1}\left(1\right){\varphi }^n\left(0\right){\beta }_{n+3}^{-1}\hfill \\ & =& {\beta }_{n+2}{\beta }_{n+1}^{-1}\left({\beta }_{n+1}{\varphi }^{n+1}\left(1\right){\beta }_{n+2}^{-1}\right){\beta }_{n+2}{\varphi }^n\left(0\right){\beta }_{n+3}^{-1}\hfill \\ & =& {\beta }_{n+2}{\beta }_{n+1}^{-1}\left({\beta }_{n+1}{\varphi }^{n+1}\left(1\right){\beta }_{n+2}^{-1}\right)\left(alph{a}_n{\varphi }^n\left(0\right){\beta }_n^{-1}\right)(\mathrm{by} \mathrm{Proposition} 5)\hfill \\ & =& {\beta }_{n+2}{\beta }_{n+1}^{-1}{S}_{n+1}^{\left(1\right)}{S}_n^{\left(0\right)}\hfill \\ & =& {\beta }_{n+2}{\beta }_{n+1}^{-1}{A}_{r_{n-2}}\left(\prod _{i=1}^{n-1}{S}_i^{\left(0\right)}\right)S_n^{\left(0\right)}\hfill \\ & =& A_{r_{n-1}}\left(\prod _{i=1}^n{S}_i^{\left(0\right)}\right)(\mathrm{by} \mathrm{Remark} 2).\hfill \end{array}$$

Hence, the result is true. □

Proposition 8. 

$$\prod _{i=1}^{\infty }{S}_i^{\left(0\right)}\prec \mathbf{T},\prod _{i=1}^{\infty }{S}_i^{\left(1\right)}\prec \tilde{\mathbf{T}}.$$

Proof. 

From Proposition 7 and Theorem 2, we have

$$S_{n+2}^{\left(1\right)}=r_{n}12\left(\prod _{i=1}^n{S}_i^{\left(0\right)}\right)$$

and $\mathbf{T}={\prod }_{i=0}^{\infty }{S}_i^{\left(1\right)}$, so ${\prod }_{i=1}^{\infty }{S}_i^{\left(0\right)}\prec \mathbf{T}$. Moreover, from Proposition 6 and Theorem 2, we have

$$S_{n+1}^{\left(0\right)}=r_n\left(\prod _{i=0}^n{S}_i^{\left(1\right)}\right)$$

and $\tilde{\mathbf{T}}=12{\prod }_{i=1}^{\infty }{S}_i^{\left(0\right)}$, so ${\prod }_{i=1}^{\infty }{S}_i^{\left(1\right)}\prec \tilde{\mathbf{T}}$. □

To prove the third decomposition theorem, Theorem 3, we introduce the following lemmas.

Lemma 5. 

For all $n\ge 5$, ${\beta }_n{\varphi }^n\left(1\right)=G_{n-3}^{\left(0\right)}{S}_{n-2}^{\left(0\right)}{\beta }_{n+1}$.

Proof. 

We use induction on n to prove it. For $n=5$, it is easy to see that ${\beta }_5=2$, ${\beta }_6=0$, $G_2^{\left(0\right)}=212001011$ and $S_3^{\left(0\right)}=2011212$. So,

$$\begin{array}{ccc}\hfill {\beta }_5{\varphi }^5\left(1\right)& =& \underline{212001011}\underline{2011212}0\hfill \\ & =& G_2^{\left(0\right)}{S}_3^{\left(0\right)}{\beta }_6.\hfill \end{array}$$

The result is true for $n=5$. By the definition of $G_n^0$, we have $G_n^0=A_{r_n}{\prod }_{i=1}^n{S}_i^{\left(0\right)}$. Assuming that the result is true for all $m\le n$, then for $m=n+1$, we have

$$\begin{array}{ccc}\hfill {\beta }_{n+1}{\varphi }^{n+1}\left(1\right)& =& {\beta }_{n+1}{\varphi }^n\left(1\right){\varphi }^n\left(2\right)\hfill \\ & =& {\beta }_{n+1}{\varphi }^n\left(1\right){\varphi }^{n-1}\left(0\right)\hfill \\ & =& {\beta }_{n+1}{\beta }_n^{-1}\left({\beta }_n{\varphi }^n\left(1\right)\right){\varphi }^{n-1}\left(0\right)\hfill \\ & =& {\beta }_{n+1}{\beta }_n^{-1}\left(G_{n-3}^{\left(0\right)}{S}_{n-2}^{\left(0\right)}{\beta }_{n+1}\right){\varphi }^{n-1}\left(0\right)\hfill \\ & =& {\beta }_{n+1}{\beta }_n^{-1}{A}_{r_{n-3}}\left(\prod _{i=1}^{n-3}{S}_i^{\left(0\right)}\right)S_{n-2}^{\left(0\right)}{\beta }_{n+1}{\varphi }^{n-1}\left(0\right)\hfill \\ & =& A_{r_{n-2}}\left(\prod _{i=1}^{n-3}{S}_i^{\left(0\right)}\right)S_{n-2}^{\left(0\right)}\left({\alpha }_{n-1}{\varphi }^{n-1}\left(0\right){\beta }_{n-1}^{-1}\right){\beta }_{n-1}(\mathrm{by} \mathrm{Remark} 2)\hfill \\ & =& A_{r_{n-2}}\left(\prod _{i=1}^{n-2}{S}_i^{\left(0\right)}\right)S_{n-1}^{\left(0\right)}{\beta }_{n+2}\hfill \\ & =& G_{n-2}{S}_{n-1}^{\left(0\right)}{\beta }_{n+2}.\hfill \end{array}$$

Hence, the result is true. □

Lemma 6. 

For all $n\ge 5$,

$${\varphi }^n\left(0\right)=011212012\prod _{i=1}^{n-4}\left(S_i^{\left(0\right)}{G}_i^{\left(0\right)}\right)S_{n-3}^{\left(0\right)}{\beta }_n.$$

Proof. 

We use induction on n to prove it. For $n=5$, it is easy to see that ${\beta }_5=2$, $S_1^{\left(0\right)}=00$, $S_2^{\left(0\right)}=1011$ and $G_1^{\left(0\right)}=11200$. Thus,

$$\begin{array}{ccc}\hfill {\varphi }^5\left(0\right)& =& 011212012\underline{0011200}\underline{1011}2\hfill \\ & =& 011212012S_1^{\left(0\right)}{G}_1^{\left(0\right)}{S}_2^{\left(0\right)}{\beta }_5.\hfill \end{array}$$

The result is true for $n=5$. Assuming that the result is true for all $m\le n$, then for $m=n+1$, we have

$$\begin{array}{ccc}\hfill {\varphi }^{n+1}\left(0\right)& =& {\varphi }^n\left(0\right){\varphi }^n\left(1\right)\hfill \\ & =& 011212012\prod _{i=1}^{n-4}\left(S_i^{\left(0\right)}{G}_i^{\left(0\right)}\right)S_{n-3}^{\left(0\right)}{\beta }_n{\varphi }^n\left(1\right)\hfill \\ & =& 011212012\prod _{i=1}^{n-4}\left(S_i^{\left(0\right)}{G}_i^{\left(0\right)}\right)S_{n-3}^{\left(0\right)}{G}_{n-3}^{\left(0\right)}{S}_{n-2}^{\left(0\right)}{\beta }_{n+1}(\mathrm{by} \mathrm{Lemma} 5)\hfill \\ & =& 011212012\prod _{i=1}^{n-3}\left(S_i^{\left(0\right)}{G}_i^{\left(0\right)}\right)S_{n-2}^{\left(0\right)}{\beta }_{n+1}.\hfill \end{array}$$

Hence, the result is true. □

Lemma 7. 

For all $n\ge 0$, ${\beta }_n{\varphi }^{n+1}\left(0\right)=r_n\left({\prod }_{i=0}^n{S}_i^{\left(1\right)}\right){\beta }_{n+1}$.

Proof. 

We use induction on n to prove it. For $n=0$, it is easy to see that ${\beta }_0=0$, ${\beta }_1=1$, $r_0=0$ and $S_0^{\left(1\right)}=0$. Thus, ${\beta }_0\varphi \left(0\right)=001=r_0{S}_0^{\left(1\right)}{\beta }_1$. Thus, the result is true for $n=0$. Assuming that the result is true for all $m\le n$, then for $m=n+1$, we have

$$\begin{array}{ccc}\hfill {\beta }_{n+1}{\varphi }^{n+2}\left(0\right)& =& {\beta }_{n+1}{\varphi }^{n+1}\left(0\right){\varphi }^{n+1}\left(1\right)\hfill \\ & =& {\beta }_{n+1}{\beta }_n^{-1}\left({\beta }_n{\varphi }^{n+1}\left(0\right)\right){\varphi }^{n+1}\left(1\right)\hfill \\ & =& {\beta }_{n+1}{\beta }_n^{-1}\left(r_n\left(\prod _{i=0}^n{S}_i^{\left(1\right)}\right){\beta }_{n+1}\right){\varphi }^{n+1}\left(1\right)\hfill \\ & =& \left({\beta }_{n+1}{\beta }_n^{-1}{r}_n\right)\left(\prod _{i=0}^n{S}_i^{\left(1\right)}\right)\left({\beta }_{n+1}{\varphi }^{n+1}\left(1\right){\beta }_{n+2}^{-1}\right){\beta }_{n+2}\hfill \\ & =& r_{n+1}\left(\prod _{i=0}^n{S}_i^{\left(1\right)}\right)S_{n+1}^{\left(1\right)}{\beta }_{n+2}(\mathrm{by} \mathrm{Remark} 1\left(2\right))\hfill \\ & =& r_{n+1}\left(\prod _{i=0}^{n+1}{S}_i^{\left(1\right)}\right){\beta }_{n+2}.\hfill \end{array}$$

Hence, the result is true. □

Lemma 8. 

For all $n\ge 3$,

$${\varphi }^n\left(1\right)=12\left(\prod _{j=0}^{n-3}\left(r_j\prod _{i=0}^j{S}_i^{\left(1\right)}\right)\right){\beta }_{n-2}.$$

Proof. 

We use induction on n to prove it. For $n=3$, it is easy to see that $r_0=0$, ${\beta }_1=1$, $S_0^{\left(1\right)}=0$ and $S_1^{\left(1\right)}=11$. Thus,

$$\begin{array}{ccc}\hfill {\varphi }^3\left(1\right)& =& 12\underline{00}\underline{1}\hfill \\ & =& 12r_0{S}_0^{\left(1\right)}{\beta }_1.\hfill \end{array}$$

The result is true for $n=3$. Assuming that the result is true for all $m\le n$, then for $m=n+1$, we have

$$\begin{array}{ccc}\hfill {\varphi }^{n+1}\left(1\right)& =& {\varphi }^n\left(1\right){\varphi }^{n-1}\left(0\right)\hfill \\ & =& 12\left(\prod _{j=0}^{n-3}\left(r_j\prod _{i=0}^j{S}_i^{\left(1\right)}\right)\right){\beta }_{n-2}{\varphi }^{n-1}\left(0\right)\hfill \\ & =& 12\left(\prod _{j=0}^{n-3}\left(r_j\prod _{i=0}^j{S}_i^{\left(1\right)}\right)\right)\left(r_{n-2}\prod _{i=0}^{n-2}{S}_i^{\left(1\right)}\right){\beta }_{n-1}(\mathrm{by} \mathrm{Lemma} 7)\hfill \\ & =& 12\left(\prod _{j=0}^{n-2}\left(r_j\prod _{i=0}^j{S}_i^{\left(1\right)}\right)\right){\beta }_{n-1}.\hfill \end{array}$$

Hence, the result is true. □

Proof of Theorem 3 

For all $n\ge 5$, we have $G_{n-3}^{\left(0\right)}=A_{r_{n-3}}{\prod }_{i=1}^{n-3}{S}_i^{\left(0\right)}$. Since $A_0=012,A_1=112and{A}_2=212$, by Proposition 5, we know ${\beta }_n⊲G_{n-3}^{\left(0\right)}$. Moreover, from Remark 1$\left(4\right)$, we have ${\beta }_n=r_n$. We have known that ${\varphi }^n\left(0\right)⊲\mathbf{T}$ and ${\varphi }^n\left(1\right)⊲\tilde{\mathbf{T}}$, so the result is true from Lemmas 6 and 8. □

4. Lyndon Decomposition

In this section, we mainly give the Lyndon factorization of $\mathbf{T}$. We totally order $\mathcal{A}$ by $0<1<2$ and extend this order to the set ${\mathcal{A}}^{*}$ of all words. Lyndon words are defined as primitive words, which are minimal in the class of all their conjugates [26,27]. We list two main conclusions about Lyndon words in the following.

Theorem 4 

(Lyndon Theorem [39]). Any non-empty word is a unique product of non-increasing Lyndon words.

In [40], the authors defined the Lyndon sequences as those sequences which have infinitely many prefixes being Lyndon words. Moreover, they proved the generalization of the Lyndon theorem as follows:

Theorem 5. 

Any sequence s can be uniquely factorized in one of the following forms:

(1) $s=s_0{s}_1{s}_2\cdots$, with $s_0\ge s_1\ge s_2\ge \cdots$, are Lyndon words;

(2) $s=s_0{s}_1{s}_2\cdots s_k{s}_{k+1}$, with $s_0\ge s_1\ge s_2\ge \cdots \ge s_k$, are Lyndon words, and $s_{k+1}<s_k$ is a Lyndon sequence.

The following theorem gives the Lyndon factorization of $\mathbf{T}$.

Theorem 6. 

$$\mathbf{T}={\varphi }^3\left(0\right)\varphi \left(1\right)\prod _{j=1}^{\infty }\left(\left(\prod _{i=0}^j{\varphi }^i\left(0\right)\right)\varphi \left(1\right)\right)$$

is the Lyndon factorization of the sequence $\mathbf{T}$. That is,

(1) For any $n\ge 0$, ${\varphi }^3\left(0\right)\varphi \left(1\right)$ and $\left({\prod }_{i=0}^{n+1}{\varphi }^i\left(0\right)\right)\varphi \left(1\right)$ are Lyndon words, and

$$\begin{array}{ccc}\hfill {\varphi }^3\left(0\right)\varphi \left(1\right)& >& {\varphi }^0\left(0\right)\varphi \left(0\right)\varphi \left(1\right)\hfill \\ & >& {\varphi }^0\left(0\right)\varphi \left(0\right){\varphi }^2\left(0\right)\varphi \left(1\right)\hfill \\ & >& {\varphi }^0\left(0\right)\varphi \left(0\right){\varphi }^2\left(0\right){\varphi }^3\left(0\right)\varphi \left(1\right)>\cdots .\hfill \end{array}$$

(2)

$$\begin{array}{ccc}\hfill \mathbf{T}& =& \left[{\varphi }^3\left(0\right)\varphi \left(1\right)\right]\left[{\varphi }^0\left(0\right)\varphi \left(0\right)\varphi \left(1\right)\right]\left[{\varphi }^0\left(0\right)\varphi \left(0\right){\varphi }^2\left(0\right)\varphi \left(1\right)\right]\hfill \\ & & \left[{\varphi }^0\left(0\right)\varphi \left(0\right){\varphi }^2\left(0\right){\varphi }^3\left(0\right)\varphi \left(1\right)\right]\cdots .\hfill \end{array}$$

Proof. 

(1) Since ${\varphi }^3\left(0\right)\varphi \left(1\right)=011212012$ and ${\varphi }^0\left(0\right)\varphi \left(0\right)\varphi \left(1\right)=00112$. It is obvious that $00112<011212012$. From Theorem 5, we only need to show that

$$\left(\prod _{i=0}^{n+1}{\varphi }^i\left(0\right)\right)\varphi \left(1\right)>\left(\prod _{i=0}^{n+2}{\varphi }^i\left(0\right)\right)\varphi \left(1\right).$$

(3)

Since

$$\left(\prod _{i=0}^{n+2}{\varphi }^i\left(0\right)\right)\varphi \left(1\right)=\left(\prod _{i=0}^{n+1}{\varphi }^i\left(0\right)\right){\varphi }^{n+2}\left(0\right)\varphi \left(1\right),$$

and $0⊲{\varphi }^{n+2}\left(0\right)$. However, $1⊲\varphi \left(1\right)$. Thus, (3) holds.

(2) We define

$$\begin{array}{ccc}\hfill \xi & :=& \left[{\varphi }^3\left(0\right)\varphi \left(1\right)\right]\left[{\varphi }^0\left(0\right)\varphi \left(0\right)\varphi \left(1\right)\right]\left[{\varphi }^0\left(0\right)\varphi \left(0\right){\varphi }^2\left(0\right)\varphi \left(1\right)\right]\hfill \\ & & \left[{\varphi }^0\left(0\right)\varphi \left(0\right){\varphi }^2\left(0\right){\varphi }^3\left(0\right)\varphi \left(1\right)\right]\cdots .\hfill \end{array}$$

It is easy to see $0⊲\xi$. We only need to show $\varphi \left(\xi \right)=\xi$, then $\mathbf{T}=\xi$.

Since

$$\begin{array}{ccc}& & \varphi \left(\xi \right)\hfill \\ & =& \left[\varphi \left({\varphi }^3\left(0\right)\varphi \left(1\right)\right)\right]\left[\varphi \left({\varphi }^0\left(0\right)\varphi \left(0\right)\varphi \left(1\right)\right)\right]\left[\varphi \left({\varphi }^0\left(0\right)\varphi \left(0\right){\varphi }^2\left(0\right)\varphi \left(1\right)\right)\right]\hfill \\ & & \left[\varphi \left({\varphi }^0\left(0\right)\varphi \left(0\right){\varphi }^2\left(0\right){\varphi }^3\left(0\right)\varphi \left(1\right)\right)\right]\cdots \hfill \\ & =& \left[{\varphi }^4\left(0\right){\varphi }^2\left(1\right)\right]\left[\varphi \left(0\right){\varphi }^2\left(0\right){\varphi }^2\left(1\right)\right]\left[\varphi \left(0\right){\varphi }^2\left(0\right){\varphi }^3\left(0\right){\varphi }^2\left(1\right)\right]\hfill \\ & & \left[\varphi \left(0\right){\varphi }^2\left(0\right){\varphi }^3\left(0\right){\varphi }^4\left(0\right){\varphi }^2\left(1\right)\right]\cdots \hfill \\ & =& \left[{\varphi }^3\left(0\right)\varphi \left(1\right)\right]\left[{\varphi }^0\left(0\right)\varphi \left(0\right)\varphi \left(1\right)\right]\left[{\varphi }^0\left(0\right)\left(\varphi \left(0\right){\varphi }^2\left(0\right)\varphi \left(1\right){\varphi }^0\left(0\right)\right)\right]\hfill \\ & & \left[\left(\varphi \left(0\right){\varphi }^2\left(0\right){\varphi }^3\left(0\right)\varphi \left(1\right){\varphi }^0\left(0\right)\right)\right]\cdots \hfill \\ & =& \left[{\varphi }^3\left(0\right)\varphi \left(1\right)\right]\left[{\varphi }^0\left(0\right)\varphi \left(0\right)\varphi \left(1\right)\right]\left[{\varphi }^0\left(0\right)\varphi \left(0\right){\varphi }^2\left(0\right)\varphi \left(1\right)\right]\hfill \\ & & \left[{\varphi }^0\left(0\right)\varphi \left(0\right){\varphi }^2\left(0\right){\varphi }^3\left(0\right)\varphi \left(1\right)\right]\cdots \hfill \\ & =& \xi ,\hfill \end{array}$$

where ${\varphi }^4\left(0\right)={\varphi }^3\left(0\right){\varphi }^3\left(1\right)={\varphi }^3\left(0\right){\varphi }^2\left(1\right)\varphi \left(0\right)={\varphi }^3\left(0\right)\varphi \left(1\right){\varphi }^0\left(0\right)\varphi \left(0\right)$. Thus, the result is true. □

5. Conclusions

We studied the combinatorial properties of the factors in the quasi-Tribonacci sequence. Using these factors, we could rebuild the quasi-Tribonacci sequence. We also established two decompositions of the quasi-Tribonacci sequence in singular words and their consequences. These results could be used to discuss the local isomorphism, the overlap properties of the factors and the power of the factors in the quasi-Tribonacci sequence. Meanwhile, we gave the Lyndon factorization of the quasi-Tribonacci sequence. The Lyndon factorization of the quasi-Tribonacci sequence helps us analyze the properties of the sequence, such as its repetitive patterns and structural characteristics. It provides a way to understand the sequence in terms of its minimal components and can be useful in various applications, including combinatorics, string algorithms and data compression. Studying the singular factorization and Lyndon factorization of the quasi-Tribonacci sequence can also contribute to a deeper understanding of its structure and potential connections to other mathematical concepts, which we shall discuss in our following work.