On the Recursive Sequence 𝑥_𝑛+𝟏=max{𝑥_𝑛,𝐴}/𝑥^𝟐𝑛𝑥_𝑛−𝟏

Abstract

We investigate the periodic nature of the solution of the max-type difference equation 𝑥_𝑛+1=max{𝑥_𝑛,𝐴}/𝑥²_𝑛𝑥_𝑛−1, 𝑛=0,1,2,…, where the initial conditions are 𝑥₋₁=𝐴^𝑟1 and 𝑥₀=𝐴^𝑟2 for 𝐴∈(0,∞), and that 𝑟₁ and 𝑟₂ are positive rational numbers. The results in this paper solve the Open Problem proposed by Grove and Ladas (2005).

1. Introduction

Max-type difference equations stem from, for example, certain models in automatic control theory (see [1, 2]). Although max-type difference equations are relatively simple in form, it is, unfortunately, extremely difficult to understand thoroughly the behavior of their solutions, see, for example, [1–41] and the relevant references cited therein. Furthermore, difference equation appear naturally as a discrete analogues, and as a numerical solution of differential and delay differential equations having applications various scientific branches, such as in ecology, economy, physics, technics, sociology, and biology. For some papers on periodicity of difference equation see, for example, [8, 9, 11, 12, 15] and the relevant references cited therein.

In [22], the following open problem was posed.

Open Problem 1 (see [22, page 218, Open Problem 6.4]). Assume that 𝐴∈(0,∞), and that 𝑟₁ and 𝑟₂ are positive rational numbers. Investigate the periodic nature of the solution of the difference equation: 𝑥_𝑛+1=max{𝑥_𝑛,𝐴}𝑥²_𝑛𝑥_𝑛−1,𝑛=0,1,…, (1.1) where the initial conditions are 𝑥₋₁=𝐴^𝑟1 and 𝑥₀=𝐴^𝑟2.

Now, in this paper we give answer to the open problem 6.4.

2. Main Results

2.1. The Case 𝐴>1

We consider (1.1) where 𝐴>1. It is clear that the change of variables𝑥_𝑛=𝐴^𝑟𝑛for𝑛≥−1 (2.1) reduces (1.1) to the difference equation:𝑟_𝑛+1=max{𝑟_𝑛,1}−2𝑟_𝑛−𝑟_𝑛−1,𝑛=0,1,2,…, (2.2) where the initial conditions 𝑟₋₁ and 𝑟₀ are positive rational numbers.

In this section we consider the behavior of the solutions of (2.2) (or equivalently of (1.1)) in this case 𝐴>1. We give the following lemmas which give us explicit solutions for some consecutive terms and show us the behavior of the solutions of (2.2) (or equivalently of (1.1)).

Lemma 2.1. Assume that {𝑟_𝑛}^∞_𝑛=−1 is a solution of (2.2). If 2<𝑟₀+𝑟₋₁=𝑟, then the following statements are true for some integer 𝑁 such that 𝑁>0.(i)If 𝑟_𝑁=−𝑟, then 𝑟_𝑁−1>1 and 𝑟_𝑁+1≥1.(ii)If 𝑟_𝑁=−𝑟, then 𝑟_𝑁=𝑟_𝑁+3 or 𝑟_𝑁=𝑟_𝑁+5.(iii)If 𝑟_𝑁=−𝑟 and 𝑟_𝑁+2>1, then 𝑟_𝑁+3=−𝑟 and 𝑟_𝑁+2=𝑟_𝑁−1+𝑟_𝑁−1.(iv)If 𝑟_𝑁=−𝑟 and 𝑟_𝑁+2≤1, then 𝑟_𝑁+5=−𝑟 and 𝑟_𝑁+4=𝑟_𝑁−1−𝑟_𝑁−2.

Proof. (i) Assume that 2<𝑟₀+𝑟₋₁=𝑟. If 𝑟₀>1, then from (2.2) we get 𝑟₁=−𝑟 and 𝑟₂=1+𝑟₀+2𝑟₋₁>1.If 𝑟₀≤1, then we get 𝑟₁=1−2𝑟₀−𝑟₋₁<1,𝑟₂=3𝑟₀+2𝑟₋₁−1>1,𝑟₃=−𝑟,𝑟₄=2−𝑟₀≥1and𝑟₅=2𝑟₀+𝑟₋₁−2.If 𝑟₅>1, then 𝑟₆=−𝑟 and 𝑟₇=𝑟₋₁+3>1.If 𝑟₅≤1, then 𝑟₆=3−3𝑟₀−2𝑟₋₁<1,𝑟₇=4𝑟₀+3𝑟₋₁−3>1,𝑟₈=−𝑟 and 𝑟₉=4−2𝑟₀−𝑟₋₁≥1.
Working inductively we have 𝑟_𝑁−1>1 and 𝑟_𝑁+1≥1 for 𝑟_𝑁=−𝑟. So, the proof of (i) is complete.
(ii) Assume that 𝑟_𝑁=−𝑟. From (i) and (2.2) we get that𝑟_𝑁+1=1−2𝑟_𝑁−𝑟_𝑁−1𝑟≥1,_𝑁+2=𝑟_𝑁+𝑟_𝑁−1−1. (2.3)
If 𝑟_𝑁+2>1, then 𝑟_𝑁+3=𝑟_𝑁=−𝑟.
If 𝑟_𝑁+2≤1, then 𝑟_𝑁+3=2−𝑟_𝑁−1<1,𝑟_𝑁+4=𝑟_𝑁−1−𝑟_𝑁−2>1 and 𝑟_𝑁+5=𝑟_𝑁=−𝑟. So, the proof of (ii) is complete.
As for (iii) and (iv), they are immediately obtained from (ii) and (2.2).

Lemma 2.2. Assume that {𝑟_𝑛}^∞_𝑛=−1 is a solution of (2.2). If 2<𝑟₀+𝑟₋₁=𝑘/𝑚,GCD(𝑘,𝑚)=1, then the following statements are true.(i)If (8𝑘−𝑚)≠0(mod3) for 0<𝑁<(8𝑘−𝑚+2), then the number of integers 𝑁 for which Lemma 2.1(iii) holds is (𝑘−2𝑚) and the number of integers 𝑁 for which Lemma 2.1(iv) holds is (𝑘+𝑚).(ii)If (8𝑘−𝑚)=0(mod3) for 0<𝑁<((8𝑘−𝑚)/3+2), then the number of integers 𝑁 for which Lemma 2.1(iii) holds is ((𝑘−2𝑚)/3) and the number of integers 𝑁 for which Lemma 2.1(iv) holds is ((𝑘+𝑚)/3).

Proof. (i) Assume that 2<𝑟₀+𝑟₋₁=𝑟=𝑘/𝑚,GCD(𝑘,𝑚)=1,(8𝑘−𝑚)≠0(mod3) and 0<𝑁<(8𝑘−𝑚+2).
Assume that number of integers 𝑁 satisfying Lemma 2.1(iii) is (𝑘−2𝑚+1).This assumption made Lemma 2.1(iii) be applied consecutively for (𝑘−2𝑚+1) times such that;𝑟_𝑁=𝑟_𝑁+3=𝑟_𝑁+6=⋯=𝑟_{𝑁+3(𝑘−2𝑚)}=𝑟_{𝑁+3(𝑘−2𝑚+1)}𝑟=−𝑟,_𝑁+2=𝑟_𝑁−1+𝑟_𝑁𝑟−1>1,_𝑁+5=𝑟_𝑁−1+2(𝑟_𝑁𝑟−1)>1,_𝑁+8=𝑟_𝑁−1+3(𝑟_𝑁⋮𝑟−1)>1,_{𝑁+3(𝑘−2𝑚)+2}=𝑟_𝑁−1+(𝑘−2𝑚+1)(𝑟_𝑁−1)>1. (2.4)
Thus, 𝑟_𝑁−1>1+(𝑘−2𝑚+1)(−𝑟_𝑁+1)>−2𝑟_𝑁. But, from Lemma 2.1(i) we have 𝑟_𝑁+1=1−2𝑟_𝑁−𝑟_𝑁−1≥1 and (−2𝑟_𝑁)≥𝑟_𝑁−1. This means that Lemma 2.1(iii) cannot be applied consecutively for (𝑘−2𝑚+1) times. So, the number of integers 𝑁 satisfying Lemma 2.1(iii) is not more than (𝑘−2𝑚).
Similarly, assume that the number of integers 𝑁 satisfying Lemma 2.1(iv) is (𝑘+𝑚+1). So, we can apply Lemma 2.1(iv) consecutively for (𝑘+𝑚+1) times such that𝑟_𝑁=𝑟_𝑁+5=𝑟_𝑁+10=⋯=𝑟_{𝑁+5(𝑘+𝑚)}=𝑟_{𝑁+5(𝑘+𝑚+1)}𝑟=−𝑟,_𝑁+4=𝑟_𝑁−1−𝑟_𝑁𝑟−2>1,_𝑁+9=𝑟_𝑁−1+2(−𝑟_𝑁𝑟−2)>1,_𝑁+14=𝑟_𝑁−1+3(−𝑟_𝑁⋮𝑟−2)>1,_{𝑁+5(𝑘+𝑚−1)+4}=𝑟_𝑁−1+(𝑘+𝑚)(−𝑟_𝑁−2)>1. (2.5)
Thus, we have 𝑟_{𝑁+5(𝑘+𝑚)+2}=𝑟_𝑁−1+(𝑘+𝑚)(−𝑟_𝑁−2)+𝑟_𝑁−1≤1 and 𝑟_𝑁−1≤1+(𝑘+𝑚)(𝑟_𝑁+2)−𝑟_𝑁+1<1. But, it contradicts Lemma 2.1(i). So, the number of integers 𝑁 satisfying Lemma 2.1(iv) is not more than (𝑘+𝑚).
Now, assume that the number of integers 𝑁 satisfying Lemma 2.1(iii) is (𝑘−2𝑚−1). We have just had the number of integers 𝑁 satisfying Lemma 2.1(iii) is less than(𝑘−2𝑚+1). From the proof of Lemma 2.1(i), if 𝑗 is the smallest integer 𝑁 satisfying Lemma 2.1(i), then we have 𝑟_𝑗=−𝑟 for 𝑗∈{1,3}. We apply Lemma 2.1(iii) for (𝑘−2𝑚−1) times such that𝑟_𝑗=𝑟_𝑗+3=𝑟_𝑗+6=⋯=𝑟_{𝑗+3(𝑘−2𝑚−2)}=𝑟_{𝑗+3(𝑘−2𝑚−1)}=−𝑟 (2.6) and from Lemma 2.1(ii) 𝑟_{𝑗+3(𝑘−2𝑚−1)}=𝑟_{𝑗+3(𝑘−2𝑚−1)+5}=⋯=𝑟_{𝑗+3(𝑘−2𝑚−1)+5(𝑘+𝑚)}=𝑟_{𝑗+3(𝑘−2𝑚−1)+5(𝑘+𝑚+1)}=−𝑟. (2.7)
Thus, the number of integers 𝑁 satisfying Lemma 2.1(iv) is (𝑘+𝑚+1). But it is not possible. So, the number of integers 𝑁 satisfying Lemma 2.1(iii) is (𝑘−2𝑚).
Similarly, assume that number of integers 𝑁 satisfying Lemma 2.1(iv) is (𝑘+𝑚−1). From Lemma 2.1(ii)–(iv), we have𝑟_𝑗=𝑟_𝑗+5=𝑟_𝑗+10=⋯=𝑟_{𝑗+5(𝑘+𝑚−2)}=𝑟_{𝑗+5(𝑘+𝑚−1)}𝑟=−𝑟,_{𝑗+5(𝑘+𝑚−1)}=𝑟_{𝑗+5(𝑘+𝑚−1)+3}=⋯=𝑟_{𝑗+5(𝑘+𝑚−1)+3(𝑘−2𝑚+1)}=𝑟_{𝑗+5(𝑘+𝑚−1)+3(𝑘−2𝑚+2)}=−𝑟. (2.8)
Thus, the number of integers 𝑁 satisfying Lemma 2.1(iii) is (𝑘−2𝑚+2). But it is not possible. So, the number of integers 𝑁 satisfying Lemma 2.1(iv) is (𝑘+𝑚). So, the proof of (i) is completed.
(ii) Proof of (ii) is similar to the proof of (i). So, it is omitted.

We omit the proofs of Lemmas 2.3 and 2.4 since they can easily be obtained in a way similar to the proofs of Lemmas 2.1 and 2.2.

Lemma 2.3. Assume that {𝑟_𝑛}^∞_𝑛=−1 is a solution of (2.2). If 1<𝑟₀+𝑟₋₁=𝑟<2, then the following statements are true for some integer 𝑁 such that 𝑁>0.(i)If 𝑟_𝑁=−𝑟, then 𝑟_𝑁−1>1 and 𝑟_𝑁+1≥1.(ii)If 𝑟_𝑁=−𝑟, then 𝑟_𝑁=𝑟_𝑁+5 or 𝑟_𝑁=𝑟_𝑁+7.(iii)If 𝑟_𝑁=−𝑟 and 𝑟_𝑁+4>1, then 𝑟_𝑁+5=−𝑟 and 𝑟_𝑁+4=𝑟_𝑁−1−𝑟_𝑁−2.(iv)If 𝑟_𝑁=−𝑟 and 𝑟_𝑁+4≤1, then 𝑟_𝑁+7=−𝑟 and 𝑟_𝑁+6=𝑟_𝑁−1−3𝑟_𝑁−3.

Lemma 2.4. Assume that {𝑟_𝑛}^∞_𝑛=−1 is a solution of (2.2). If 1<𝑟₀+𝑟₋₁=𝑘/𝑚<2,GCD(𝑘,𝑚)=1, then the following statements are true.(i)If (8𝑘−𝑚)≠0(mod3) for 0<𝑁<(8𝑘−𝑚−1), then number of integers 𝑁 for which Lemma 2.3(iii) holds is 3(𝑘−𝑚) and the number of integers 𝑁 for which Lemma 2.3(iv) holds is (2𝑚−𝑘).(ii)If (8𝑘−𝑚)=0(mod3) for 0<𝑁<((8𝑘−𝑚)/3−1), then the number of integers 𝑁 for which Lemma 2.3(iii) holds is (𝑘−𝑚) and the number of integers 𝑁 for which Lemma 2.3(iv) holds is ((2𝑚−𝑘)/3).

Lemma 2.5. Assume that {𝑟_𝑛}^∞_𝑛=−1 is a solution of (2.2). If𝑟₀+𝑟₋₁<1 and 𝑟₀+𝑟₋₁≠1/2, then the following statements are true for some integer 𝑁 such that 𝑁≥−1.(i)Assume that 𝑟_𝑁>1, and 𝑟_𝑁+2≥1. If 𝑟₀+𝑟₋₁<1/2,then𝑟_𝑁+1=𝑟₀+𝑟₋₁−1 or if 𝑟₀+𝑟₋₁>1/2, then 𝑟_𝑁+1=−𝑟₀−𝑟₋₁. (ii)If 𝑟_𝑁>1 and 𝑟_𝑁+2≥1, then 𝑟_𝑁+3<1,𝑟_𝑁+4<1,𝑟_𝑁+5<1,𝑟_𝑁+6=3+2𝑟_𝑁+1−𝑟_𝑁<1 and 𝑟_𝑁+7=𝑟_𝑁−3(𝑟_𝑁+1+1),(iii)If 𝑟_𝑁∉(𝑟_𝑁𝑖,𝑟_𝑁𝑖+1,𝑟_𝑁𝑖+2,𝑟_𝑁𝑖+3,𝑟_𝑁𝑖+4,𝑟_𝑁𝑖+5,𝑟_𝑁𝑖+6) such that 𝑟_𝑁𝑖>1 and 𝑟_𝑁𝑖+2≥1, then 𝑟_𝑁≤1.

Proof. (i) Assume that 𝑟₀+𝑟₋₁<1/2. From (2.2), we get that 𝑟₁=1−2𝑟₀−𝑟₋₁<1,𝑟₂=3𝑟₀+2𝑟₋₁−1<1 and 𝑟₃=2−4𝑟₀−3𝑟₋₁.If 𝑟₃>1, then 𝑟₄=𝑟₀+𝑟₋₁−1,𝑟₅=1+2𝑟₀+𝑟₋₁>1.If 𝑟₃≤1, then 𝑟₄=5𝑟₀+4𝑟₋₁−2<1, 𝑟₅=3−6𝑟₀−5𝑟₋₁ and 𝑟₅>1 or 𝑟₅≤1.If 𝑟₅>1, then 𝑟₆=𝑟₀+𝑟₋₁−1 and 𝑟₇=4𝑟₀+3𝑟₋₁≥1.If 𝑟₅≤1, then 𝑟₆=7𝑟₀+6𝑟₋₁−3<1, 𝑟₇=4−8𝑟₀−7𝑟₋₁ and 𝑟₇>1 or𝑟₇≤1.
Working inductively, we have 𝑟_𝑁+1=𝑟₀+𝑟₋₁−1 for 𝑟₀+𝑟₋₁<1/2,𝑟_𝑁>1 and 𝑟_𝑁+2≥1.
Assume that 𝑟₀+𝑟₋₁>1/2. From (2.2), we get that 𝑟₁=1−2𝑟₀−𝑟₋₁<1 and 𝑟₂=3𝑟₀+2𝑟₋₁−1.
If 𝑟₂>1, then 𝑟₃=−𝑟₀−𝑟₋₁ and 𝑟₄=2−𝑟₀>1.If 𝑟₂≤1, then 𝑟₃=2−4𝑟₀−3𝑟₋₁<1 and 𝑟₄=5𝑟₀+4𝑟₋₁−2.If 𝑟₄>1, then 𝑟₅=−𝑟₀−𝑟₋₁ and 𝑟₆=3−3𝑟₀−2𝑟₋₁>1.
Working inductively, we have 𝑟_𝑁+1=−𝑟₀−𝑟₋₁ for 1/2<𝑟₀+𝑟₋₁<1,𝑟_𝑁>1 and 𝑟_𝑁+2≥1. So, the proof of (i) is completed.
(ii)Assume that 𝑟₀+𝑟₋₁<1 and that 𝑟_𝑁>1,𝑟_𝑁+2≥1. From (2.2) and (i), we get that 𝑟_𝑁+2=max{𝑟_𝑁+1,1}−2𝑟_𝑁+1−𝑟_𝑁=1−2𝑟_𝑁+1−𝑟_𝑁,𝑟_𝑁≤−2𝑟_𝑁+1. (2.9) Then,𝑟_𝑁+3=max{𝑟_𝑁+2,1}−2𝑟_𝑁+2−𝑟_𝑁+1=𝑟_𝑁+𝑟_𝑁+1𝑟−1<1,_𝑁+4=max{𝑟_𝑁+3,1}−2𝑟_𝑁+3−𝑟_𝑁+2=2−𝑟_𝑁𝑟<1,_𝑁+5=max{𝑟_𝑁+4,1}−2𝑟_𝑁+4−𝑟_𝑁+3=𝑟_𝑁−𝑟_𝑁+1𝑟−2<1,_𝑁+6=max{𝑟_𝑁+5,1}−2𝑟_𝑁+5−𝑟_𝑁+4=−𝑟_𝑁+2𝑟_𝑁+1𝑟+3<1,_𝑁+7=max{𝑟_𝑁+6,1}−2𝑟_𝑁+6−𝑟_𝑁+5=𝑟_𝑁−3(𝑟_𝑁+1+1). (2.10)
So, the proof of (ii) is completed.
(iii)Assume that the smallest integer of integers 𝑁 satisfying (i) is 𝑁₁. From the proof of (i), we have 𝑟_𝑗<1 for 𝑗<𝑁₁. Also, from this assumption we have the subsequence (𝑟_𝑁1,𝑟_𝑁1+1,𝑟_𝑁1+2,𝑟_𝑁1+3,𝑟_𝑁1+4,𝑟_𝑁1+5,𝑟_𝑁1+6) such that 𝑟_𝑁1>1 and 𝑟_𝑁1+2≥1. Then, from (ii) we get that 𝑟_𝑁1+7=𝑟_𝑁1−3(𝑟_𝑁1+1+1) and 𝑟_𝑁1+7≤1 or 𝑟_𝑁1+7>1. If 𝑟_𝑁1+7≤1, we get that 𝑟_𝑁1+8=−𝑟_𝑁1+4(𝑟_𝑁1+1+1)<1. It means that 𝑟_𝑁1+7 and𝑟_𝑁1+8 are not the element previous and later subsequences satisfying (ii). If 𝑟_𝑁1+7>1, then we get that 𝑟_𝑁1+9≥1 and that 𝑟_𝑁1+7 is a element of the subsequence (𝑟_𝑁1+7,𝑟_𝑁1+8,𝑟_𝑁1+9,𝑟_𝑁1+10,𝑟_𝑁1+11,𝑟_𝑁1+12,𝑟_𝑁1+13) such that𝑟_𝑁1+7>1 and 𝑟_𝑁1+9≥1. If this proceeds, we have 𝑟_𝑁≤1, for 𝑟_𝑁∉(𝑟_𝑁𝑖,𝑟_𝑁𝑖+1,𝑟_𝑁𝑖+2,𝑟_𝑁𝑖+3,𝑟_𝑁𝑖+4,𝑟_𝑁𝑖+5,𝑟_𝑁𝑖+6) such that 𝑟_𝑁𝑖>1 and 𝑟_𝑁𝑖+2≥1.

Lemma 2.6. Assume that {𝑟_𝑛}^∞_𝑛=−1 is a solution of (2.2). If 𝑟₀+𝑟₋₁=𝑘/𝑚<1/2,GCD(𝑘,𝑚)=1, then the following statements are true.(i)If (7𝑚−8𝑘)≠0(mod3) for 𝑁<(7𝑚−8𝑘−1), then number of integers 𝑁 for which Lemma 2.5(ii) holds is (𝑚−2𝑘) and the number of integers 𝑁 for which Lemma 2.5(iii) holds is 6𝑘.(ii)If (7𝑚−8𝑘)=0(mod3) for 𝑁<((7𝑚−8𝑘)/3−1), then number of integers 𝑁 for which Lemma 2.5(ii) holds is ((𝑚−2𝑘)/3) and the number of integers 𝑁 for which Lemma 2.5(iii) holds is 2𝑘.

Lemma 2.7. Assume that {𝑟_𝑛}^∞_𝑛=−1 is a solution of (2.2). If 1/2<𝑟₀+𝑟₋₁=𝑘/𝑚<1,GCD(𝑘,𝑚)=1, then the following statements are true.(i)If (8𝑘−𝑚)≠0(mod3) for 𝑁<(8𝑘−𝑚−1), then number of integers 𝑁 for which Lemma 2.5(ii) holds is (2𝑘−𝑚) and the number of integers 𝑁 for which Lemma 2.5(iii) holds is 6(𝑚−𝑘).(ii)If (8𝑘−𝑚)=0(mod3) for 𝑁<((8𝑘−𝑚)/3−1), then number of integers 𝑁 for which Lemma 2.5(ii) holds is ((2𝑘−𝑚)/3) and the number of integers 𝑁 for which Lemma 2.5(iii) holds is 2(𝑚−𝑘).

Theorem 2.8. Consider (2.2). If 𝑟₋₁+𝑟₀=𝑘/𝑚≤1/2 and GCD(𝑘,𝑚)=1, then the following statements are true.(i)If (7𝑚−8𝑘)≠0(mod3), then {𝑟_𝑛}^∞_𝑛=−1 is periodic with prime period (7𝑚−8𝑘),(ii)If (7𝑚−8𝑘)=0(mod3), then {𝑟_𝑛}^∞_𝑛=−1 is periodic with prime period ((7𝑚−8𝑘)/3).

Proof. (i) Assume that 𝑟₋₁+𝑟₀=𝑘/𝑚<1/2,GCD(𝑘,𝑚)=1 and (7𝑚−8𝑘)≠0(mod3). It suffices to prove 𝑟_𝑛=𝑟_{𝑛+7𝑚−8𝑘} for 𝑛≥−1. We must show that 𝑟₋₁=𝑟_{7𝑚−8𝑘−1} and 𝑟₀=𝑟_{7𝑚−8𝑘}. From Lemmas 2.5 and 2.6, for getting the last two terms we need, we can assume that 𝑟_𝑛≤1for𝑛=−1,0,…,(6𝑘−2) (2.11) and that 𝑟_{6𝑘−1+7𝑙}>1,𝑟_{6𝑘+1+7𝑙}≥1for𝑙=0,1,…,(𝑚−2𝑘−1). (2.12)
From this assumption and (2.2), we get that for 𝑛=1,2,…,(3𝑘−1)𝑟_2𝑛−1=𝑟₋₁+^𝑛_𝑖=1[1−2(𝑟₋₁+𝑟₀𝑟)]≤1_2𝑛=𝑟₀+^𝑛_𝑖=1[2(𝑟₋₁+𝑟₀)−1]≤1 (2.13) and that 𝑟_6𝑘−1=𝑟₋₁+^3𝑘_𝑖=1[1−2(𝑟₋₁+𝑟₀)]>1,𝑟_6𝑘=(𝑟₋₁+𝑟₀)−1,𝑟_6𝑘+1=2−𝑟₋₁+^3𝑘−1_𝑖=1[2(𝑟₋₁+𝑟₀)−1]≥1,⋮𝑟_6𝑘+5=3𝑟₋₁+4𝑟₀+^3𝑘−1_𝑖=1[2(𝑟₋₁+𝑟₀)−1]≤1,𝑟_6𝑘+6=1−(4𝑟₋₁+5𝑟₀)−^3𝑘−1_𝑖=1[2(𝑟₋₁+𝑟₀)−1]>1,𝑟_6𝑘+7=(𝑟₋₁+𝑟₀)−1,⋮𝑟_6𝑘+12=6𝑟₋₁+7𝑟₀+^3𝑘−1_𝑖=1[2(𝑟₋₁+𝑟₀)−1]≤1,𝑟_6𝑘+13=1−(7𝑟₋₁+8𝑟₀)−^3𝑘−1_𝑖=1[2(𝑟₋₁+𝑟₀)−1]>1,⋮𝑟_{6𝑘−1+7(𝑚−2𝑘−1)}=1−𝑟₋₁−2𝑟₀−3(𝑚−2𝑘−1)(𝑟₋₁+𝑟₀)−^3𝑘−1_𝑖=1[2(𝑟₋₁+𝑟₀)−1]>1,⋮𝑟_{6𝑘−1+7(𝑚−2𝑘−1)+6}=𝑟₀+3(𝑚−2𝑘)(𝑟₋₁+𝑟₀)+^3𝑘−1_𝑖=1[2(𝑟₋₁+𝑟₀)−1]≤1,𝑟_{6𝑘−1+7(𝑚−2𝑘)}=1−[−𝑟₋₁+(3𝑚−6𝑘+2)(𝑟₋₁+𝑟₀)]−^3𝑘−1_𝑖=1[2(𝑟₋₁+𝑟₀)−1]. (2.14)(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)(14)(15)
Then we get 𝑟_{7𝑚−8𝑘−1}=𝑟₋₁ and that𝑟_{7𝑚−8𝑘}=1−2𝑟_{7𝑚−8𝑘−1}−𝑟_{7𝑚−8𝑘−2}=1−2𝑟₋₁−𝑟_{7𝑚−8𝑘−2}−[𝑟₀+3(𝑚−2𝑘)(𝑟₋₁+𝑟₀)+^3𝑘−1_𝑖=1[2(𝑟₋₁+𝑟₀)−1]]=𝑟₀. (2.15)
So, we have 𝑟_𝑛=𝑟_{𝑛+7𝑚−8𝑘} for 𝑛≥−1.
(ii) Assume that 𝑟₋₁+𝑟₀=1/2. From (2.2), we get immediately 𝑟₁=𝑟₋₁ and 𝑟₂=𝑟₀. So, we have 𝑟_𝑛=𝑟_{𝑛+(7𝑚−8𝑘)/3} for 𝑛≥−1. If 𝑟₋₁+𝑟₀<1/2,GCD(𝑘,𝑚)=1 and (7𝑚−8𝑘)=0(mod3), then the proof of (ii) is similar to the proof of (i). So, it is omitted.

Theorem 2.9. Consider (2.2). If 𝑟=𝑟₋₁+𝑟₀=𝑘/𝑚>1/2 and GCD(𝑘,𝑚)=1, then the following statements are true.(i)If (8𝑘−𝑚)≠0(mod3), then {𝑟_𝑛}^∞_𝑛=−1 is periodic with prime period (8𝑘−𝑚).(ii)If (8𝑘−𝑚)=0(mod3), then {𝑟_𝑛}^∞_𝑛=−1 is periodic with prime period ((8𝑘−𝑚)/3).

Proof. (i) If 𝑟₋₁+𝑟₀=1, then from (2.2) we get that 𝑟₁=−𝑟₀,𝑟₂=1+𝑟₀,𝑟₃=−1,𝑟₄=2−𝑟₀,𝑟₅=𝑟₀−1,𝑟₆=𝑟₋₁ and 𝑟₇=𝑟₀. So, we have 𝑟_𝑛=𝑟_{𝑛+8𝑘−𝑚} for 𝑛≥−1.
In the case 1/2<𝑟₋₁+𝑟₀<1, the proof is similar to the proof of Theorem 2.8(i) such that; from Lemmas 2.5 and 2.7, we can assume that𝑟_𝑛≤1for𝑛=−1,0,…,[6(𝑚−𝑘)−3] (2.16) and that 𝑟_{6(𝑚−𝑘)−2+7𝑙}>1,𝑟_{6(𝑚−𝑘)+7𝑙}≥1,𝑟_{6(𝑚−𝑘)−2+7(2𝑘−𝑚)}≤1 (2.17) for 𝑙=0,1,…,(2𝑘−𝑚−1). Thus, we get that 𝑟_2𝑛=𝑟₀+^𝑛_𝑖=1[2(𝑟₋₁+𝑟₀𝑟)−1]≤1,_2𝑛+1=𝑟₋₁+^𝑛+1_𝑖=1[1−2(𝑟₋₁+𝑟₀)]≤1 (2.18) for 𝑛=1,2,…,[3(𝑚−𝑘)−2] and that 𝑟_{6(𝑚−𝑘)−2}=𝑟₀+^{3(𝑚−𝑘)−1}_𝑖=1[2(𝑟₋₁+𝑟₀𝑟)−1]>1,_{6(𝑚−𝑘)−1}=−(𝑟₋₁+𝑟₀𝑟),_{6(𝑚−𝑘)}=1+2𝑟₋₁+𝑟₀+^{3(𝑚−𝑘)−1}_𝑖=1[1−2(𝑟₋₁+𝑟₀⋮𝑟)]≥1,_{6(𝑚−𝑘)−2+7(2𝑘−𝑚)−1}=𝑟₋₁+3(2𝑘−𝑚)(1−𝑟₋₁−𝑟₀)−^{3(𝑚−𝑘)−1}_𝑖=1[2(𝑟₋₁+𝑟₀𝑟)−1]≤1,_{6(𝑚−𝑘)−2+7(2𝑘−𝑚)}=𝑟₀+3(2𝑘−𝑚)(𝑟₋₁+𝑟₀−1)+^{3(𝑚−𝑘)−1}_𝑖=1[2(𝑟₋₁+𝑟₀𝑟)−1]≤1,_{6(𝑚−𝑘)+7(2𝑘−𝑚)−1}=1−2𝑟_{6(𝑚−𝑘)+7(2𝑘−𝑚)−2}−𝑟_{6(𝑚−𝑘)+7(2𝑘−𝑚)−3}=𝑟₋₁,𝑟_{6(𝑚−𝑘)+7(2𝑘−𝑚)}=1−2𝑟_{6(𝑚−𝑘)+7(2𝑘−𝑚)−1}−𝑟_{6(𝑚−𝑘)+7(2𝑘−𝑚)−2}=𝑟₀ (2.19)
So, we have 𝑟_𝑛=𝑟_{𝑛+8𝑘−𝑚} for 𝑛≥−1.
Assume that 1<𝑟₋₁+𝑟₀<2. From Lemmas 2.3 and 2.4, for getting the last two terms, we can assume that𝑟_𝑁+5𝑙1=𝑟_{𝑁+5(𝑙1+1)}𝑟=−𝑟,_{𝑁+5[3(𝑘−𝑚)]+7𝑙2}=𝑟_{𝑁+5[3(𝑘−𝑚)]+7(𝑙2+1)}=−𝑟 (2.20) for 𝑙₁=0,1,…,[3(𝑘−𝑚)−1] and 𝑙₂=0,1,…,(2𝑚−𝑘−1). From this assumption and Lemma 2.4(iii)-(iv), we get that 𝑟_𝑁+4=𝑟_𝑁−1−𝑟_𝑁𝑟−2,_𝑁+9=𝑟_𝑁−1+2(−𝑟_𝑁⋮𝑟−2),_{𝑁+5[3(𝑘−𝑚)−1]+4}=𝑟_𝑁−1+3(𝑘−𝑚)(−𝑟_𝑁−2), (2.21) and that 𝑟_{𝑁+5[3(𝑘−𝑚)]+6}=𝑟_{𝑁+5[3(𝑘−𝑚)−1]+4}−3𝑟_𝑁𝑟−3,_{𝑁+5[3(𝑘−𝑚)]+13}=𝑟_{𝑁+5[3(𝑘−𝑚)−1]+4}+2(−3𝑟_𝑁⋮𝑟−3),_{𝑁+5[3(𝑘−𝑚)]+7(2𝑚−𝑘−1)+6}=𝑟_{𝑁+5[3(𝑘−𝑚)−1]+4}+(2𝑚−𝑘)(−3𝑟_𝑁−3)>1 (2.22)
Thus, we get 𝑟_{𝑁+8𝑘−𝑚−1}=𝑟_𝑁−1 and 𝑟_{𝑁+8𝑘−𝑚}=𝑟_𝑁. From 𝑟_𝑛−1=max{𝑟_𝑛,1}−2𝑟_𝑛−𝑟_𝑛+1, we have 𝑟_𝑛=𝑟_{𝑛+8𝑘−𝑚} for 𝑛≥−1. Also, it is easy to see that 𝑟_𝑁−1≠𝑟_{𝑁+5𝑙1−1} and 𝑟_𝑁−1≠𝑟_{𝑁+5[3(𝑘−𝑚)]+7𝑙2−1} for 𝑙₁=1,2,…,[3(𝑘−𝑚)] and 𝑙₂=1,2,…,(2𝑚−𝑘) which imply that 8𝑘−𝑚 is the smallest period.
In the case 2<𝑟₋₁+𝑟₀, the proof is similar. So, it is omitted.
(ii) Assume that 𝑟₋₁+𝑟₀=2. If 𝑟₀>1, then from (2.2) we get that 𝑟₁=−2,𝑟₂=5−𝑟₀,𝑟₃=𝑟₀−3,𝑟₄=𝑟₋₁ and 𝑟₅=𝑟₀. If 𝑟₀≤1, then 𝑟₁=−1−𝑟₀,𝑟₂=3+𝑟₀,𝑟₃=−2,𝑟₄=𝑟₋₁ and 𝑟₅=𝑟₀. So, we have 𝑟_𝑛=𝑟_{𝑛+(8𝑘−𝑚)/3} for 𝑛≥−1. The rest of proof is similar to the proof of (i) and is omitted.

2.2. The Case 𝐴<1

We consider (1.1) where 𝐴<1. It is clear that the change of variables 𝑥_𝑛=𝐴^𝑟𝑛for𝑛≥−1 (2.23) reduces (1.1) to the difference equation:𝑟_𝑛+1=min{𝑟_𝑛,1}−2𝑟_𝑛−𝑟_𝑛−1,𝑛=0,1,2,…, (2.24) where the initial conditions 𝑟₋₁ and 𝑟₀ are positive rational numbers.

In this section we consider the behavior of the solutions of (2.24) (or equivalently of (1.1)) in this case 𝐴<1. We omit the proofs of the following results since they can easily be obtained in a way similar to the proofs of the lemmas and theorems in the previous section.

Lemma 2.10. Assume that {𝑟_𝑛}^∞_𝑛=−1 is a solution of (2.24). If 𝑟₋₁,𝑟₀>1 and 𝑟=𝑟₀+𝑟₋₁, then the following statements are true for some integer 𝑁>0.(i)If 𝑟_𝑁=𝑟−1, then 𝑟_𝑁−1≤1 and 𝑟_𝑁+1≤1.(ii)If 𝑟_𝑁=𝑟−1, then 𝑟_𝑁=𝑟_𝑁+3 or 𝑟_𝑁=𝑟_𝑁+5.(iii)If 𝑟_𝑁=𝑟−1 and 𝑟_𝑁+2<1, then 𝑟_𝑁+3=𝑟−1 and 𝑟_𝑁+2=𝑟_𝑁−1+𝑟_𝑁−1.(iv)If 𝑟_𝑁=𝑟−1 and 𝑟_𝑁+2≥1, then 𝑟_𝑁+5=𝑟−1 and 𝑟_𝑁+4=𝑟_𝑁−1−𝑟_𝑁−2.

Lemma 2.11. Assume that {𝑟_𝑛}^∞_𝑛=−1 is a solution of (2.24) for which Lemma 2.10 holds. If 𝑟=𝑟₀+𝑟₋₁=𝑘/𝑚, GCD(𝑘,𝑚)=1, then the following statements are true.(i)If (8𝑘−7𝑚)≠0(mod3) for 0<𝑁<(8𝑘−7𝑚−1), then number of integers 𝑁 for which Lemma 2.10(iii) holds is (𝑘+𝑚) and the number of integers 𝑁 for which Lemma 2.10(iv) holds is (𝑘−2𝑚).(ii)If (8𝑘−7𝑚)=0(mod3) for 0<𝑁<((8𝑘−7𝑚)/3−1), then number of integers 𝑁 for which Lemma 2.10(iii) holds is ((𝑘+𝑚)/3) and the number of integers 𝑁 for which Lemma 2.10(iv) holds is ((𝑘−2𝑚)/3).

Lemma 2.12. Assume that {𝑟_𝑛}^∞_𝑛=−1 is a solution of (2.24). If max{𝑟₋₁,𝑟₀}=𝑟>1 and one of the initial contitions 𝑟₋₁,𝑟₀ is less than or equal to one, then the following statements are true for some integer 𝑁>0.(i)If 𝑟_𝑁=𝑟, then 𝑟_𝑁−1≤1 and 𝑟_𝑁+1≤1.(ii)If 𝑟_𝑁=𝑟, then 𝑟_𝑁=𝑟_𝑁+3 or 𝑟_𝑁=𝑟_𝑁+5.(iii)If 𝑟_𝑁=𝑟 and 𝑟_𝑁+2<1, then 𝑟_𝑁+3=𝑟 and 𝑟_𝑁+2=𝑟_𝑁−1+𝑟_𝑁−1.(iv)If 𝑟_𝑁=𝑟 and 𝑟_𝑁+2≥1, then 𝑟_𝑁+5=𝑟 and 𝑟_𝑁+4=𝑟_𝑁−1−𝑟_𝑁−2.

Lemma 2.13. Assume that {𝑟_𝑛}^∞_𝑛=−1 is a solution of (2.24) for which Lemma 2.12 holds. If 𝑟=max{𝑟₋₁,𝑟₀}=𝑘/𝑚, GCD(𝑘,𝑚)=1, then the following statements are true.(i)If (8𝑘+𝑚)≠0(mod3) for 0<𝑁<(8𝑘+𝑚−1), then number of integers 𝑁 for which Lemma 2.12(iii) holds is (2𝑚+𝑘) and the number of integers 𝑁 for which Lemma 2.12(iv) holds is (𝑘−𝑚).(ii)If (8𝑘+𝑚)=0(mod3) for 0<𝑁<((8𝑘+𝑚)/3−1), then number of integers 𝑁 for which Lemma 2.12(iii) holds is ((2𝑚+𝑘)/3) and the number of integers 𝑁 for which Lemma 2.12(iv) holds is ((𝑘−𝑚)/3).

Theorem 2.14. Consider (2.24). If 𝑟₋₁,𝑟₀>1,𝑟=(𝑟₀+𝑟₋₁)=𝑘/𝑚 and GCD(𝑘,𝑚)=1, then the following statements are true.(i)If (8𝑘−7𝑚)≠0(mod3), then {𝑟_𝑛}^∞_𝑛=−1 is periodic with prime period (8𝑘−7𝑚).(ii)If (8𝑘−7𝑚)=0(mod3), then {𝑟_𝑛}^∞_𝑛=−1 is periodic with prime period ((8𝑘−7𝑚)/3).

Theorem 2.15. Consider (2.24). If at least one of the initial conditions of (2.24) is less than or equal to one, max{𝑟₋₁,𝑟₀,1}=𝑘/𝑚 and GCD(𝑘,𝑚)=1, then the following statements are true.(i)If (8𝑘+𝑚)≠0(mod3), then {𝑟_𝑛}^∞_𝑛=−1 is periodic with prime period (8𝑘+𝑚).(ii)If (8𝑘+𝑚)=0(mod3), then {𝑟_𝑛}^∞_𝑛=−1 is periodic with prime period ((8𝑘+𝑚)/3).

3. Conclusion

Difference equation appears naturally as a discrete analogue and as a numerical solutions of differential and delay differential equations having applications various scientific branches, such as in ecology, economy, physics, technics, sociology, and biology. Specially, max-type difference equations stem from, for example, certain models in automatic control theory. In this paper, periodic nature of the solution of (1.1) which was open problem proposed by Grove and Ladas [22] was investigated. We describe a new method in investigating periodic character of max-type difference equations. It is expected that after some modifications our method will be applicable to Open Problem 6.3 in [22].