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# On the solutions of a max-type system of difference equations of higher order

## Abstract

In this paper, we study the following max-type system of difference equations of higher order:

$$\textstyle\begin{cases} x_{n} = \max \{A ,\frac{y_{n-t}}{x_{n-s}} \}, \\ y_{n} = \max \{B ,\frac{x_{n-t}}{y_{n-s}} \},\end{cases}\displaystyle \quad n\in \{0,1,2,\ldots \},$$

where $$A,B\in (0, +\infty )$$, $$t,s\in \{1,2,\ldots \}$$ with $$\gcd (s,t)=1$$, the initial values $$x_{-d},y_{-d},x_{-d+1},y_{-d+1}, \ldots , x_{-1}, y_{-1}\in (0,+ \infty )$$ and $$d=\max \{t,s\}$$.

## Introduction

Concrete nonlinear difference equations and systems have attracted some recent attention (see, e.g., ). One of the classes of such equations/systems are max-type difference equations/systems. For some results of solutions of many max-type difference equations and systems, such as eventual periodicity, the boundedness character and attractivity, see, e.g. [15, 79, 1116, 1825, 2830, 3236, 38, 39] and the references therein. Our purpose in this paper is to study the eventual periodicity of the following max-type system of difference equation of higher order:

$$\textstyle\begin{cases} x_{n} = \max \{A ,\frac{y_{n-t}}{x_{n-s}} \}, \\ y_{n} = \max \{B ,\frac{x_{n-t}}{y_{n-s}} \},\end{cases}\displaystyle \quad n\in { \mathbf{{N}}}_{0}\equiv \{0,1,\ldots \},$$
(1.1)

where $$A,B\in {\mathbf{{R}}}_{+}\equiv (0,+\infty )$$, $$t,s\in {\mathbf{{N}\equiv }\{\mathbf{1},\mathbf{2},\ldots \}}$$ with $$\gcd (s,t)=1$$, the initial values $$x_{-d},y_{-d}, x_{-d+1},y_{-d+1}, \ldots , x_{-1}, y_{-1}\in \mathbf{{R}}_{+}$$ and $$d=\max \{t,s\}$$.

When $$t=1$$ and $$s=2$$, (1.1) reduces to the max-type system of difference equations

$$\textstyle\begin{cases} x_{n} = \max \{A ,\frac{y_{n-1}}{x_{n-2}} \}, \\ y_{n} = \max \{B ,\frac{x_{n-1}}{y_{n-2}} \},\end{cases}\displaystyle \quad n\in { \mathbf{{N}}}_{0}.$$
(1.2)

Fotiades and Papaschinopoulos in  showed that every positive solution of (1.2) is eventually periodic.

In 2012, Stević  obtained in an elegant way the general solution to the following max-type system of difference equations:

$$\textstyle\begin{cases} x_{n+1}=\max \{\frac{A}{x_{n}},\frac{y_{n}}{x_{n}} \}, \\ y_{n+1}=\max \{\frac{A}{y_{n}},\frac{x_{n}}{y_{n}} \},\end{cases}\displaystyle \quad n\in \mathbf{{N}}_{0},$$
(1.3)

for the case $$x_{0},y_{0}\geq A>0$$ and $$y_{0}/x_{0}\geq \max \{A,1/A\}$$.

In , Sun and Xi studied the following max-type system of difference equations:

$$\textstyle\begin{cases} x_{n} = \max \{\frac{1}{x_{n-m}} , \min \{1,\frac{A}{y_{n-r}} \} \}, \\ y_{n} = \max \{\frac{1}{y_{n-m}} , \min \{1,\frac{B}{x_{n-t}} \} \},\end{cases}\displaystyle \quad n\in { \mathbf{{N}}}_{0},$$
(1.4)

where $$A,B\in { \mathbf{{R}}}_{+}$$, $$m,r,t\in {\mathbf{{N}}}$$ and the initial values $$x_{-d},y_{-d},x_{-d+1},y_{-d+1}, \ldots , x_{-1}, y_{-1}\in {\mathbf{{R}}}_{+}$$ with $$d=\max \{m,r,t\}$$ and showed that every positive solution of (1.4) is eventually periodic with period 2m.

When $$m=r=t=1$$ and $$A=B$$, (1.4) reduces to the max-type system of difference equations

$$\textstyle\begin{cases} x_{n}=\max \{\frac{1}{x_{n-1}},\min \{1,\frac{A}{y_{n-1}} \} \}, \\ y_{n}=\max \{\frac{1}{y_{n-1}},\min \{1,\frac{A}{x_{n-1}} \} \}, \end{cases}\displaystyle \quad n\in \mathbf{{N}}_{0}.$$
(1.5)

Yazlik et al.  in 2015 obtained in an elegant way the general solution of (1.5).

In 2012, Stević  studied the following max-type system of difference equations:

$$\textstyle\begin{cases} y^{(1)}_{n}=\max_{1\leq i\leq m_{1}} \{f_{1i}(y^{(1)}_{n-k_{i,1}^{(1)}},y^{(2)}_{n-k_{i,2}^{(1)}}, \ldots ,y^{(l)}_{n-k_{i,l}^{(1)}},n),y_{n-s}^{(1)} \}, \\ y^{(2)}_{n}=\max_{1\leq i\leq m_{2}} \{f_{2i}(y^{(1)}_{n-k_{i,1}^{(2)}},y^{(2)}_{n-k_{i,2}^{(2)}}, \ldots ,y^{(l)}_{n-k_{i,l}^{(2)}},n),y_{n-s}^{(2)} \}, \\ \ldots , \\ y^{(l)}_{n}=\max_{1\leq i\leq m_{l}} \{f_{li}(y^{(1)}_{n-k_{i,1}^{(l)}},y^{(2)}_{n-k_{i,2}^{(l)}}, \ldots ,y^{(l)}_{n-k_{i,l}^{(l)}},n),y_{n-s}^{(l)} \},\end{cases}\displaystyle \quad n\in \mathbf{{N}}_{0},$$
(1.6)

where $$s,l,m_{j},k^{(j)}_{i,t}\in {\mathbf{{N}}}$$ ($$j,t\in \{1,2,\ldots ,l \}$$) and $$f_{ji}:\mathbf{{R}}_{+}^{l}\times {\mathbf{{ N}}}_{0}\longrightarrow \mathbf{{R}}_{+}$$ ($$j\in \{1,\ldots ,l\}$$ and $$i\in \{1,\ldots ,m_{j}\}$$), and showed that every positive solution of (1.6) is eventually periodic with (not necessarily prime) period s if $$f_{ji}$$ satisfy some conditions.

Moreover, Stević et al.  in 2014 investigated the following max-type system of difference equations:

$$\textstyle\begin{cases} y^{(1)}_{n}=\max_{1\leq i_{1}\leq m_{1}} \{f_{1i_{1}}(y^{(1)}_{n-k_{i_{1},1}^{(1)}},y^{(2)}_{n-k_{i_{1},2}^{(1)}}, \ldots ,y^{(l)}_{n-k_{i_{1},l}^{(1)}},n),y_{n-t_{1}s}^{(\sigma (1))} \}, \\ x^{(2)}_{n}=\max_{1\leq i_{2}\leq m_{2}} \{f_{2i_{2}}(y^{(1)}_{n-k_{i_{2},1}^{(2)}},y^{(2)}_{n-k_{i_{2},2}^{(2)}}, \ldots ,y^{(l)}_{n-k_{i_{2},l}^{(2)}},n),y_{n-t_{2}s}^{(\sigma (2))} \}, \\ \ldots, \\ y^{(l)}_{n}=\max_{1\leq i_{l}\leq m_{l}} \{f_{li_{l}}(y^{(1)}_{n-k_{i_{l},1}^{(l)}},y^{(2)}_{n-k_{i_{l},2}^{(l)}}, \ldots ,y^{(l)}_{n-k_{i_{l},l}^{(l)}},n),y_{n-t_{l}s}^{(\sigma (l))} \},\end{cases}\displaystyle \quad n\in \mathbf{{N}}_{0},$$
(1.7)

where $$s,l,m_{j},t_{j},k^{(j)}_{i_{j},h}\in { \mathbf{{N}}}$$ ($$j,h\in \{1,2,\ldots ,l\}$$), $$(\sigma (1),\ldots ,\sigma (l))$$ is a permutation of $$(1,\ldots ,l)$$ and $$f_{ji_{j}}:\mathbf{{R}}_{+}^{l}\times {\mathbf{{ N}}}_{0}\longrightarrow \mathbf{{R}}_{+}$$ ($$j\in \{1,\ldots ,l\}$$ and $$i_{j}\in \{1,\ldots ,m_{j}\}$$). They showed that every positive solution of (1.7) is eventually periodic with period sT for some $$T\in \mathbf{{N}}$$ if $$f_{ji_{j}}$$ satisfy some conditions.

## Main results and proofs

In this section, we study the eventual periodicity of positive solutions of system (1.1). Let $$\{(x_{n},y_{n})\}_{n\geq -d}$$ be a solution of (1.1) with the initial values $$x_{-d},y_{-d},x_{-d+1},y_{-d+1},\ldots , x_{-1}, y_{-1}\in { \mathbf{{R}}}_{+}$$.

### Lemma 2.1

If$$x_{n}=A$$eventually, then$$y_{n}$$is a periodic sequence with period 2seventually. If$$y_{n}=B$$eventually, then$$x_{n}$$is a periodic sequence with period 2seventually.

### Proof

Assume that $$x_{n}=A$$ eventually. By (1.1) we see

$$y_{n} = \max \biggl\{ B ,\frac{A}{y_{n-s}} \biggr\} \quad \mbox{eventually},$$
(2.1)

which implies $$y_{n}y_{n-s}\geq A$$ eventually and

\begin{aligned}[b] B&\leq y_{n} = \max \biggl\{ B , \frac{A}{y_{n-s}} \biggr\} \\ &= \max \biggl\{ B ,\frac{Ay_{n-2s}}{y_{n-s}y_{n-2s}} \biggr\} \\ &\leq \max \{B,y_{n-2s}\}\leq y_{n-2s} \quad \mbox{eventually}.\end{aligned}
(2.2)

Then, for any $$0\leq i\leq 2s-1$$, $$y_{2ns+i}$$ is eventually nonincreasing.

We claim that, for every $$0\leq i\leq 2s-1$$, $$y_{2ns+i}$$ is a constant sequence eventually. Assume on the contrary that, for some $$0\leq i\leq 2s-1$$, $$y_{2ns+i}$$ is not a constant sequence eventually. Then there exists a sequence of positive integers $$k_{1}< k_{2}<\cdots$$ such that, for any $$n\in { \mathbf{{N}}}$$, we have

\begin{aligned}[b] B&< y_{2sk_{n+1}+i}=\frac{A}{y_{2sk_{n+1}+i-s}} \\ &< y_{2sk_{n}+i}=\frac{A}{y_{2sk_{n}+i-s}},\end{aligned}
(2.3)

which implies $$y_{2sk_{n+1}+i-s}>y_{2sk_{n}+i-s}$$ for any $$n\in { \mathbf{{N}}}$$. This is a contradiction. Thus $$y_{n}$$ is a periodic sequence with period 2s eventually. The second case follows from the previously proved one by interchanging letters. The proof is complete. □

### Lemma 2.2

If$$A\geq B\geq 1/A$$, then$$x_{2(n+1)t+i}\leq x_{2nt+i}$$for any$$n\geq t+s$$and$$i\in { {\mathbf{{N}}}}_{0}$$. If$$B\geq A\geq 1/B$$, then$$y_{2(n+1)t+i}\leq y_{2nt+i}$$for any$$n\geq t+s$$and$$i\in { {\mathbf{{N}}}}_{0}$$.

### Proof

Assume that $$A\geq B\geq 1/A$$. By (1.1) we see that $$x_{n}\geq A$$ and $$y_{n}\geq B$$ for any $$n\in { \mathbf{{N}}}_{0}$$, and

$$x_{2(n+1)t+i} = \max \biggl\{ A,\frac{B}{x_{2(n+1)t+i-s}}, \frac{x_{2nt+i}}{x_{2(n+1)t+i-s}y_{2(n+1)t+i-t-s}} \biggr\} .$$
(2.4)

Since $$B/x_{2(n+1)t+i-s}\leq B/A\leq 1\leq A$$ and $$x_{2(n+1)t+i-s}y_{2(n+1)t+i-t-s}\geq AB\geq 1$$ for $$2(n+1)t+i\geq t+s$$, we obtain

$$x_{2(n+1)t+i}\leq \max \{A, x_{2nt+i}\}\leq x_{2nt+i}.$$
(2.5)

The second case follows from the previously proved one by interchanging letters. The proof is complete. □

### Theorem 2.1

Let$$AB>1$$. If$$A\geq B$$, then$$x_{n}=A$$eventually and$$y_{n}$$is a periodic sequence with period 2seventually. If$$B> A$$, then$$y_{n}=B$$eventually and$$x_{n}$$is a periodic sequence with period 2seventually.

### Proof

Assume that $$A\geq B$$. For any $$0\leq i\leq 2t-1$$ and $$n\in { \mathbf{{N}}}_{0}$$, we have

$$A\leq x_{2(n+1)t+i}= \max \biggl\{ A, \frac{x_{2nt+i}}{x_{2(n+1)t-s+i}y_{2(n+1)t-s-t+i}} \biggr\} .$$
(2.6)

By Lemma 2.2 we may let $$\lim_{n\longrightarrow \infty }x_{2nt+i}=A_{i}$$. Note that

$$\lim_{n\longrightarrow \infty } \frac{x_{2nt+i}}{x_{2(n+1)t-s+i}y_{2(n+1)t-s-t+i}}\leq \lim _{n \longrightarrow \infty }\frac{x_{2nt+i}}{AB}=\frac{A_{i}}{AB}< A_{i}$$
(2.7)

and

$$\lim_{n\longrightarrow \infty }x_{2(n+1)t+i}=A_{i}.$$
(2.8)

Thus we have $$x_{n}=A$$ eventually. By Lemma 2.1, we see that $$y_{n}$$ is a periodic sequence with period 2s eventually. The second case follows from the previously proved one by interchanging letters. The proof is complete. □

In the following, we assume $$AB=1$$. For any $$i\in { \mathbf{{N}}}_{0}$$, let

$$\lim_{n\longrightarrow \infty } x_{2nt+i}=A_{i} \quad \mbox{if } A \geq B$$
(2.9)

and

$$\lim_{n\longrightarrow \infty }y_{2nt+i}=B_{i}\quad \mbox{if } B \geq A.$$
(2.10)

Then $$A_{i}\geq A$$ and $$B_{i}\geq B$$.

### Lemma 2.3

If$$A\geq B=1/A$$and$$A_{i}>A$$for some$$i\in { {\mathbf{{N}}}}_{0}$$, then, for any$$k\in { {\mathbf{{N}}}}$$, $$x_{2nt+ks+i}$$and$$y_{2nt-t+ks+i}$$are constant sequences eventually. If$$B\geq A=1/B$$and$$B_{i}>B$$for some$$i\in { {\mathbf{{N}}}}_{0}$$, then, for any$$k\in { {\mathbf{{N}}}}$$, $$y_{2nt+ks+i}$$and$$x_{2nt-t+ks+i}$$are constant sequences eventually.

### Proof

Assume that $$A\geq B$$ and $$A_{i}>A$$ for some $$i\in { \mathbf{{N}}}_{0}$$. Since $$A_{i}>A$$, it follows from Lemma 2.2 and (1.1) that

\begin{aligned}[b] x_{2nt+i}& =\max \biggl\{ A, \frac{x_{2(n-1)t+i}}{x_{2nt-s+i}y_{2nt-t-s+i}} \biggr\} \\ &=\frac{x_{2(n-1)t+i}}{x_{2nt-s+i}y_{2nt-t-s+i}} \quad \mbox{eventually}.\end{aligned}
(2.11)

From this we have

\begin{aligned}[b] B&\leq \lim_{n\longrightarrow \infty }y_{2nt-t-s+i} \\ &=\lim_{n\longrightarrow \infty } \frac{x_{2(n-1)t+i}}{x_{2nt+i}x_{2nt-s+i}} \\ &=\frac{1}{A_{-s+i}}\leq \frac{1}{A}=B.\end{aligned}
(2.12)

This implies

$$\lim_{n\longrightarrow \infty }x_{2nt-s+i}=A$$
(2.13)

and

$$\lim_{n\longrightarrow \infty }y_{2nt-t-s+i}=B$$
(2.14)

and

$$\lim_{n\longrightarrow \infty }y_{2nt-t+i}=\lim_{n\longrightarrow \infty }x_{2nt+i}x_{2nt-s+i}=A_{i}A.$$
(2.15)

Since

$$x_{2nt+s+i} = \max \biggl\{ A, \frac{x_{2(n-1)t+s+i}}{x_{2nt+i}y_{2nt-t+i}} \biggr\}$$
(2.16)

and

$$\lim_{n\longrightarrow \infty } \frac{x_{2(n-1)t+s+i}}{x_{2nt+i}y_{2nt-t+i}}= \frac{A_{s+i}}{A_{i}^{2}A}< A_{s+i},$$
(2.17)

we see that $$x_{2nt+s+i} =A$$ eventually. Note that

$$\lim_{n\longrightarrow \infty }\frac{x_{2(n-1)t+s+i}}{y_{2nt-t+i}}= \frac{A}{AA_{i}}< B,$$
(2.18)

from which it follows that

\begin{aligned}[b] y_{2nt-t+s+i} &= \max \biggl\{ B, \frac{x_{2(n-1)t+s+i}}{y_{2nt-t+i}} \biggr\} \\ &=B \quad \mbox{eventually},\end{aligned}
(2.19)

and

\begin{aligned}[b] y_{2nt-t+2s+i} &= \max \biggl\{ B, \frac{x_{2(n-1)t+2s+i}}{y_{2nt-t+s+i}} \biggr\} \\ &=\frac{x_{2(n-1)t+2s+i}}{B}\quad \mbox{eventually}, \end{aligned}
(2.20)

and

\begin{aligned}[b] x_{2nt+2s+i}& = \max \biggl\{ A, \frac{y_{2nt-t+2s+i}}{x_{2nt+s+i}} \biggr\} \\ &=\max \{A,x_{2(n-1)t+2s+i}\} \\ &=x_{2(n-1)t+2s+i} \quad \mbox{eventually}.\end{aligned}
(2.21)

If $$x_{2nt+2s+i}>A$$ eventually, then, in a similar fashion, we obtain:

1. (1)

$$x_{2nt+3s+i} =A$$ eventually and $$y_{2nt-t+3s+i} =B$$ eventually.

2. (2)

$$x_{2nt+4s+i}$$ and $$y_{2nt-t+4s+i}$$ are constant sequences eventually.

If $$x_{2nt+2s+i}=A$$ eventually, then $$y_{2nt-t+2s+i} =A/B$$ eventually, and

\begin{aligned}[b] y_{2nt-t+3s+i}& = \max \biggl\{ B, \frac{x_{2(n-1)t+3s+i}}{y_{2nt-t+2s+i}} \biggr\} \\ &=\max \biggl\{ B,\frac{x_{2(n-1)t+3s+i}B}{A} \biggr\} \\ &=\frac{x_{2(n-1)t+3s+i}B}{A}\quad \mbox{eventually},\end{aligned}
(2.22)

and

\begin{aligned}[b] x_{2nt+3s+i}& = \max \biggl\{ A, \frac{y_{2nt-t+3s+i}}{x_{2nt+2s+i}} \biggr\} \\ &=\max \biggl\{ A,\frac{x_{2(n-1)t+3s+i}B}{A^{2}}\biggr\} \quad \mbox{eventually}.\end{aligned}
(2.23)

From this we see that if $$A=B$$, then

$$x_{2nt+3s+i}=x_{2(n-1)t+3s+i} \quad \mbox{eventually},$$
(2.24)

and if $$A>B$$, then

$$x_{2nt+3s+i}=A\quad \mbox{ eventually},$$
(2.25)

since

$$\lim_{n\longrightarrow \infty }\frac{x_{2(n-1)t+3s+i}B}{A^{2}}= \frac{A_{3s+i}B}{A^{2}}< A_{3s+i}.$$
(2.26)

Using induction and arguments similar to the ones developed in the above given proof, we can show that, for any $$k\in { \mathbf{{N}}}$$, $$x_{2nt+ks+i}$$ and $$y_{2nt-t+ks+i}$$ are constant sequences eventually. The second case follows from the previously proved one by interchanging letters. The proof is complete. □

### Lemma 2.4

If$$A=1/B>B$$and for some$$i\in { {\mathbf{{N}}}}_{0}$$, $$x_{2nt+i}>A$$eventually and$$A_{i}=A$$, then, for any$$k\in { {\mathbf{{N}}}}$$, $$x_{2nt+ks+i}$$and$$y_{2nt-t+ks+i}$$are constant sequences eventually. If$$B=1/A>A$$and for some$$i\in { {\mathbf{{N}}}}_{0}$$, $$y_{2nt+i}>B$$eventually and$$B_{i}=B$$, then, for any$$k\in { {\mathbf{{N}}}}$$, $$y_{2nt+ks+i}$$and$$x_{2nt-t+ks+i}$$are constant sequences eventually.

### Proof

Assume that $$A=1/B>B$$ and for some $$i\in { \mathbf{{N}}}_{0}$$, $$x_{2nt+i}>A$$ eventually and $$A_{i}=A$$. By (1.1) we have

\begin{aligned}[b] x_{2nt+i}& = \max \biggl\{ A, \frac{y_{2nt-t+i}}{x_{2nt-s+i}} \biggr\} \\ &=\frac{y_{2nt-t+i}}{x_{2nt-s+i}} \quad \mbox{eventually}, \end{aligned}
(2.27)

and

$$x_{2nt+s+i} = \max \biggl\{ A, \frac{x_{2(n-1)t+s+i}}{x_{2nt+i}y_{2nt-t+i}} \biggr\}$$
(2.28)

and

\begin{aligned}[b] \lim_{n\longrightarrow \infty } \frac{x_{2(n-1)t+s+i}}{x_{2nt+i}y_{2nt-t+i}} &=\lim_{n \longrightarrow \infty } \frac{x_{2(n-1)t+s+i}}{x^{2}_{2nt+i}x_{2nt-s+i}} \\ &\leq \frac{A_{s+i}}{A^{3}}< A_{s+i}. \end{aligned}
(2.29)

Then we see that $$x_{2nt+s+i} =A$$ eventually. From this and $$y_{2nt-t+i}\geq A^{2}$$ eventually it follows that

\begin{aligned}[b] y_{2nt-t+s+i}& = \max \biggl\{ B, \frac{x_{2(n-1)t+s+i}}{y_{2nt-t+i}} \biggr\} \\ &=\max \biggl\{ B,\frac{A}{y_{2nt-t+i}} \biggr\} \\ &=B \quad \mbox{eventually}, \end{aligned}
(2.30)

and

\begin{aligned}[b] y_{2nt-t+2s+i} &= \max \biggl\{ B, \frac{x_{2(n-1)t+2s+i}}{y_{2nt-t+s+i}} \biggr\} \\ &=\frac{x_{2(n-1)t+2s+i}}{B} \quad \mbox{eventually},\end{aligned}
(2.31)

and

\begin{aligned}[b] x_{2nt+2s+i}& = \max \biggl\{ A, \frac{y_{2nt-t+2s+i}}{x_{2nt+s+i}} \biggr\} \\ &=\max \{A,x_{2(n-1)t+2s+i}\} \\ &=x_{2(n-1)t+2s+i} \quad \mbox{eventually}.\end{aligned}
(2.32)

Thus $$x_{2nt+2s+i}$$ and $$y_{2nt-t+2s+i}$$ are constant sequences eventually. Using arguments similar to the ones developed in the proof of Lemma 2.3, we can show that, for any $$k\in { \mathbf{{N}}}$$, $$x_{2nt+ks+i}$$ and $$y_{2nt-t+ks+i}$$ are constant sequences eventually. The second case follows from the previously proved one by interchanging letters. The proof is complete. □

### Theorem 2.2

1. (1)

Assume$$A=1/B>B$$. Then one of the following statements holds.

1. (i)

$$x_{n}=A$$eventually and$$y_{n}$$is a periodic sequence with period 2seventually.

2. (ii)

Ifsis odd, then$$x_{n}$$, $$y_{n}$$are periodic sequences with period 2teventually.

3. (iii)

Ifsis even, then$$x_{n}$$is a periodic sequence with period 2teventually and$$y_{n}$$is a periodic sequence with period$$2st$$eventually.

2. (2)

Assume$$B=1/A>A$$. Then one of the following statements holds.

1. (i)

$$y_{n}=B$$eventually and$$x_{n}$$is a periodic sequence with period 2seventually.

2. (ii)

Ifsis odd, then$$x_{n}$$, $$y_{n}$$are periodic sequences with period 2teventually.

3. (iii)

Ifsis even, then$$y_{n}$$is a periodic sequence with period 2teventually and$$x_{n}$$is a periodic sequence with period$$2st$$eventually.

### Proof

Assume that $$A=1/B>B$$. If $$x_{n}=A$$ eventually, then by Lemma 2.1 we see that $$y_{n}$$ is a periodic sequence with period 2s eventually. Now we assume that $$x_{n}\neq A$$ eventually. Then we have $$A_{i}>A$$ (or $$x_{2nt+i}>A$$ eventually and $$A_{i}=A$$) for some $$0\leq i\leq 2t-1$$.

If s is odd, then $$\gcd (2t,s)=1$$. Thus, for every $$j\in \{0,1,2,\ldots ,2t-1\}$$, there exist some $$1\leq i_{j}\leq 2t$$ and integer $$\lambda _{j}$$ such that $$i_{j}s=\lambda _{j} 2t+j$$ since $$\{rs:0\leq r\leq 2t-1\}=\{0,1,2,\ldots ,2t-1\}$$ (mod 2t). By Lemma 2.3 and Lemma 2.4 we see that, for any $$k\in { \mathbf{{N}}}$$, $$x_{2nt+ks+i}$$ and $$y_{2nt-t+ks+i}$$ are constant sequences eventually. Thus, for any $$0\leq r\leq 2t-1$$, $$x_{2nt+r}$$ and $$y_{2nt+r}$$ are constant sequences eventually, which implies that $$x_{n}$$, $$y_{n}$$ are periodic sequences with period 2t eventually.

In the following, we assume that s is even with $$s=2s'$$. Then $$\gcd (t,s')=1$$ and t is odd. Thus, for every $$j\in \{0,1,2,\ldots ,t-1\}$$, there exist some $$1\leq i_{j}\leq t$$ and integer $$\lambda _{j}$$ such that $$i_{j}s'=\lambda _{j} t+j$$ and $$i_{j}s=\lambda _{j} 2t+2j$$.

If $$x_{2nt+i}\neq A$$ eventually for some $$i\in \{0,2,\ldots \}$$ and $$x_{2nt+l}\neq A$$ eventually for some $$l\in \{1,3,\ldots \}$$, then by Lemma 2.3 and Lemma 2.4 we see that, for any $$k\in { \mathbf{{N}}}$$, $$x_{2nt+ks+i}$$, $$y_{2nt-t+ks+i}$$, $$x_{2nt+ks+l}$$ and $$y_{2nt-t+ks+l}$$ are constant sequences eventually. Thus, for any $$0\leq r\leq 2t-1$$, $$x_{2nt+r}$$ and $$y_{2nt+r}$$ are constant sequences eventually, which implies that $$x_{n}$$, $$y_{n}$$ are periodic sequences with period 2t eventually.

If $$x_{2nt+i}\neq A$$ eventually for some $$i\in \{0,2,\ldots \}$$ and $$x_{2nt+l}=A$$ eventually for any $$l\in \{1,3,\ldots \}$$, then by Lemma 2.3 and Lemma 2.4 we see that, for any $$k\in { \mathbf{{N}}}$$, $$x_{2nt+ks+i}$$ and $$y_{2nt-t+ks+i}$$ are constant sequences eventually. This implies that, for every $$r\in \{0,1,2,\ldots ,2t-1\}$$, $$x_{2nt+r}$$ is constant sequence eventually and for every $$l\in \{1,3,\ldots \}$$, $$y_{2nst+l}$$ is constant sequence eventually. By (1.1) we see that there exists $$N\in { \mathbf{{N}}}$$ such that, for any $$n\geq N$$ and $$r\in \{0,2,\ldots \}$$,

$$y_{2nt+r} = \max \biggl\{ B ,\frac{A}{y_{2nt+r-s}} \biggr\} .$$
(2.33)

Then we have $$y_{2nt+r}y_{2nt+r-s}\geq A$$. Thus, for any $$n\geq N$$ and $$l\in \{1,3,\ldots \}$$ and $$k\in { \mathbf{{N}}}$$,

\begin{aligned} B&\leq y_{2nt+t+2ks+l} = \max \biggl\{ B , \frac{x_{2nt+2ks+l}}{y_{2nt+t+2ks+l-s}} \biggr\} \\ &= \max \biggl\{ B , \frac{Ay_{2nt+t+2ks+l-2s}}{y_{2nt+t+2ks+l-s}y_{2nt+t+2ks+l-2s}} \biggr\} \\ &\leq \max \{B,y_{2nt+t+2ks-2s+l}\} \\ &= y_{2nt+t+2ks-2s+l}\quad \mbox{eventually}. \end{aligned}
(2.34)

Then, for every $$n\geq N$$ and $$l\in \{1,3,\ldots \}$$, we have

$$B\leq \cdots \leq y_{2nt+t+2ks+l}\leq y_{2nt+t+2ks-2s+l}\leq y_{2nt+t+2s+l} \leq y_{2nt+t+l}.$$
(2.35)

We claim that, for every $$n\geq N$$ and $$l\in \{1,3,\ldots \}$$, $$\{y_{2nt+t+2ks+l}\}_{k\in { \mathbf{{N}}}}$$ is a constant sequence eventually. Assume on the contrary that, for some $$n\geq N$$ and some $$l\in \{1,3,\ldots \}$$, $$\{y_{2nt+t+2ks+l}\}_{k\in { \mathbf{{N}}}}$$ is not a constant sequence eventually. Then there exists a sequence of positive integers $$k_{1}< k_{2}<\cdots$$ such that, for any $$r\in { \mathbf{{N}}}$$, we have

\begin{aligned}[b] B&< y_{2nt+t+2k_{r}s+l} = \frac{A}{y_{2nt+t+2k_{r}s+l-s}} \\ &< y_{2nt+t+2k_{r-1}s+l}=\frac{A}{ y_{2nt+t+2k_{r-1}s+l-s}}, \end{aligned}
(2.36)

which implies $$y_{2nt+t+2k_{r}s+l-s}>y_{2nt+t+2k_{r-1}s+l-s}$$ for any $$r\in { \mathbf{{N}}}$$. This is a contradiction. Take $$ps>N$$. Then $$y_{2nst+t+l}=y_{2pst+t+2(nt-pt)s+l}$$ is a constant sequence eventually for any $$l\in \{1,3,\ldots \}$$. From the above we see that $$y_{n}$$ is a periodic sequence with period $$2st$$ eventually.

In a similar fashion, we can show that if $$x_{2nt+i}=A$$ eventually for any $$i\in \{0,2,\ldots \}$$ and $$x_{2nt+l}\neq A$$ eventually for some $$l\in \{1,3,\ldots \}$$, then also statement (1(iii)) holds.

The second case follows from the previously proved one by interchanging letters. The proof is complete. □

Now we assume that $$A=B=1$$. Then, for any $$0\leq i\leq 2t-1$$ and $$n\in { \mathbf{{N}}}_{0}$$, we have $$1\leq x_{2(n+1)t+i}\leq x_{2nt+i}$$ eventually and $$1\leq y_{2(n+1)t+i}\leq y_{2nt+i}$$ eventually.

### Lemma 2.5

Let$$A=B=1$$and$$s\geq t$$. Then the following statements hold.

1. (1)

If$$A_{i}=1$$, then$$B_{t+i}=1$$. If$$B_{i}=1$$, then$$A_{t+i}=1$$.

2. (2)

If$$x_{N}=1$$for some$$N\in { {\mathbf{{N}}}}$$and$$A_{2nt+N+ks}=1$$for any$$k,n\in { {\mathbf{{N}}}}$$, then$$x_{2nt+N+ks}=y_{2nt+t+ks+N}=1$$for any$$k,n\in { {\mathbf{{N}}}}$$. If$$y_{N}=1$$for some$$N\in { {\mathbf{{N}}}}$$and$$B_{2nt+N+ks}=1$$for any$$k,n\in { {\mathbf{{N}}}}$$, then$$y_{2nt+N+ks}=x_{2nt+t+ks+N}=1$$for any$$k,n\in { {\mathbf{{N}}}}$$.

3. (3)

Ifsis even and$$\gcd (s,t)=1$$, then$$1\in \{x_{n}:n\in \{0,2,\ldots \}\}\cup \{y_{t+n}:n\in \{0,2,\ldots \}\}$$and$$1\in \{x_{n}:n\in \{1,3,\ldots \}\}\cup \{y_{t+n}:n\in \{1,3,\ldots \}\}$$.

4. (4)

$$1\in \{x_{n}:n\in { {\mathbf{{N}}}}\}\cup \{y_{n}:n\in { {\mathbf{{N}}}}\}$$.

### Proof

(1) Assume that $$A_{i}=1$$. Assume on the contrary that $$B_{t+i}>1$$. It follows from (1.1) that

$$y_{2nt+t+i} =\frac{x_{2nt+i}}{y_{2nt+t-s+i}}.$$
(2.37)

This implies

$$1\leq \lim_{n\longrightarrow \infty }y_{2nt+t-s+i}=\frac{1}{B_{t+i}}< 1.$$
(2.38)

This is a contradiction. The second case follows from the previously proved one by interchanging letters.

(2) If $$x_{N}=1$$ for some $$N\in { \mathbf{{N}}}$$, then $$x_{2nt+N}=1$$ for any $$n\in { \mathbf{{N}}}$$. It follows from (1.1) that

\begin{aligned}[b] y_{2nt+t+N}& = \max \biggl\{ 1, \frac{x_{2nt+N}}{y_{2nt+t-s+N}} \biggr\} \\ &=\max \biggl\{ 1,\frac{1}{y_{2nt+t-s+N}}\biggr\} =1\end{aligned}
(2.39)

and

\begin{aligned}[b] y_{2nt+t+s+N}& = \max \biggl\{ 1, \frac{x_{2nt+s+N}}{y_{2nt+t+N}} \biggr\} \\ &=x_{2nt+s+N}\end{aligned}
(2.40)

and

\begin{aligned}[b] x_{2(n+1)t+N+s}& = \max \biggl\{ 1, \frac{y_{2nt+t+s+N}}{x_{2(n+1)t+N}} \biggr\} \\ &=\max \{1,y_{2nt+t+s+N}\} \\ &=y_{2nt+t+s+N} \\ &=x_{2nt+N+s}.\end{aligned}
(2.41)

Thus $$x_{2nt+N+s}=y_{2nt+t+s+N}=1$$ for any $$n\in { \mathbf{{N}}}$$. In a similar fashion, we can show that $$x_{2nt+N+ks}=y_{2nt+t+ks+N}=1$$ for any $$k,n\in { \mathbf{{N}}}$$. The second case follows from the previously proved one by interchanging letters.

(3) If s is even and $$\gcd (s,t)=1$$, then t is odd. Assume on the contrary that $$1\notin \{x_{n}:n\in \{0,2,\ldots \}\}\cup \{y_{t+n}:n\in \{0,2, \ldots \}\}$$. Then it follows from (1.1) that, for any $$n\in { \mathbf{{N}}}$$,

\begin{aligned}[b] y_{2nt+t}& = \max \biggl\{ 1, \frac{x_{2nt}}{y_{2nt+t-s}} \biggr\} \\ &=\frac{x_{2nt}}{y_{2nt+t-s}}>1 \end{aligned}
(2.42)

and

\begin{aligned}[b] x_{2nt+2t-s}& = \max \biggl\{ 1, \frac{y_{2nt+t-s}}{x_{2nt+2t-2s}} \biggr\} \\ &=\frac{y_{2nt+t-s}}{x_{2nt+2t-2s}}>1.\end{aligned}
(2.43)

Thus

\begin{aligned}[b] x_{2nt}& > x_{2nt-2(s-t)} \\ &>x_{2nt-4(s-t)} \\ &\quad \cdots \\ &>x_{2t(n-t+s)}. \end{aligned}
(2.44)

This is a contradiction.

(4) Case (4) is treated similarly to case (3). The proof is complete. □

### Theorem 2.3

Let$$A=B=1$$and$$s\geq t$$. Then one of the following statements holds.

1. (1)

$$x_{n}=1$$eventually and$$y_{n}$$is a periodic sequence with period 2seventually.

2. (2)

$$y_{n}=1$$eventually and$$x_{n}$$is a periodic sequence with period 2seventually.

3. (3)

$$x_{n}$$, $$y_{n}$$are periodic sequences with period 2teventually.

### Proof

If $$x_{n}=1$$ (or $$y_{n}=1$$) eventually, then by Lemma 2.1 we see that $$y_{n}$$ (or $$x_{n}$$) is a periodic sequence with period 2s eventually. Now we assume that $$x_{n}\neq1$$ eventually. Then we have $$A_{i}>1$$ for some $$0\leq i\leq 2t-1$$ or $$\lim_{n\longrightarrow \infty }x_{n}=1$$.

If s is odd, then $$\gcd (2t,s)=1$$. Thus, for every $$j\in \{0,1,2,\ldots ,2t-1\}$$, there exist some $$1\leq i_{j}\leq 2t$$ and integer $$\lambda _{j}$$ such that $$i_{j}s=\lambda _{j} 2t+j$$. By Lemma 2.3 and Lemma 2.5 we see that, for any $$k\in { \mathbf{{N}}}$$, $$x_{2nt+ks+i}$$ and $$y_{2nt-t+ks+i}$$ are constant sequences eventually, or for some $$N\in { \mathbf{{N}}}$$, $$x_{2nt+N+ks}=y_{2nt+t+ks+N}=1$$ for any $$k,n\in { \mathbf{{N}}}$$, or for some $$N\in { \mathbf{{N}}}$$, $$y_{2nt+N+ks}=x_{2nt+t+ks+N}=1$$ for any $$k,n\in { \mathbf{{N}}}$$. Thus, for any $$0\leq r\leq 2t-1$$, $$x_{2nt+r}$$ and $$y_{2nt+r}$$ are constant sequences eventually, which implies that $$x_{n}$$, $$y_{n}$$ are periodic sequences with period 2t eventually.

In the following, we assume that s is even with $$s=2s'$$, then $$\gcd (t,s')=1$$ and t is odd. Thus, for every $$j\in \{0,1,2,\ldots ,t-1\}$$, there exist some $$1\leq i_{j}\leq t$$ and integer $$\lambda _{j}$$ such that $$i_{j}s=\lambda _{j} 2t+2j$$.

If $$A_{i}>1$$ for some $$i\in \{0,2,\ldots \}$$, then by Lemma 2.3 we see that, for any $$k\in { \mathbf{{N}}}$$, $$x_{2nt+ks+i}$$ and $$y_{2nt-t+ks+i}$$ are constant sequences eventually. If $$A_{i}=1$$ for any $$i\in \{0,2,\ldots \}$$, then by Lemma 2.5 we have $$B_{t+i}=1$$ for any $$i\in \{0,2,\ldots \}$$ and $$x_{2nt+i+ks}=y_{2nt-t+ks+i}=1$$ for any $$k\in { \mathbf{{N}}}$$ eventually. In a similar fashion, also we can show that, for any $$i\in \{1,3,\ldots \}$$, $$x_{2nt+ks+i}$$ and $$y_{2nt-t+ks+i}$$ are constant sequences eventually for any $$k\in { \mathbf{{N}}}$$, or $$x_{2nt+i+ks}=y_{2nt-t+ks+i}=1$$ for any $$k\in { \mathbf{{N}}}$$ eventually for any $$i\in \{1,3,\ldots \}$$ and $$k\in { \mathbf{{N}}}$$. Thus, for any $$0\leq r\leq 2t-1$$, $$x_{2nt+r}$$ and $$y_{2nt+r}$$ are constant sequences eventually. This implies that $$x_{n}$$, $$y_{n}$$ are periodic sequences with period 2t eventually.

Using the previously proved one by interchanging letters, also we can show that if $$y_{n}\neq1$$ eventually, then $$x_{n}$$, $$y_{n}$$ are periodic sequences with period 2t eventually. The proof is complete. □

In Example 3.1 of , we showed that the equation

$$x_{n}=\frac{x_{n-t}}{x_{n-s}} (t>s)$$
(2.45)

has a positive solution $$z_{n}$$ ($$n\geq -t$$) with $$1< z_{n+1}< z_{n}$$ for any $$n\geq -t$$ and $$\lim_{n\longrightarrow \infty }z_{n}=1$$.

From Example 3.1 of , we obtain the following theorem.

### Theorem 2.4

Let$$A\leq 1$$and$$B\leq 1$$and$$s< t$$. Assume$$z_{n}$$ ($$n\geq -t$$) is a positive solution of (2.45) with$$1< z_{n+1}< z_{n}$$for any$$n\geq -t$$and$$\lim_{n\longrightarrow \infty }z_{n}=1$$. Then equation (1.1) have a solution$$(x_{n},y_{n})$$with$$1< x_{n+1}=y_{n+1}=z_{n+1}< x_{n}=y_{n}=z_{n}$$for any$$n\geq -t$$and$$\lim_{n\longrightarrow \infty }x_{n}=\lim_{n\longrightarrow \infty }y_{n}=1$$.

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### Acknowledgements

The authors would like to thank the referees for their valuable comments and suggestions.

None.

## Funding

The research was supported by NNSF of China (11761011, 71862003) and NSF of Guangxi (2018GXNSFAA294010) and SF of Guangxi University of Finance and Economics (2019QNB10).

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All authors participated in every phase of research conducted for this paper. All authors read and approved the final manuscript.

### Corresponding author

Correspondence to Caihong Han.

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