# On the Recursive Sequencexn+1=max{xn,A}/xn2xn−1

Authors: Ibrahim Yalcinkaya; Cengiz Cinar; Ali Gelisken
Journal: Discrete Dynamics in Nature and Society
 (2010)

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## Abstract

We investigate the periodic nature of the solution of the max-type difference equationxn+1=max{xn,A}/xn2xn-1, n=0,1,2,…, where the initial conditions are x-1=Ar1 and x0=Ar2 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).


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## Body

## 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 thatA∈(0,∞), and that r1 and r2 are positive rational numbers. Investigate the periodic nature of the solution of the difference equation:
(1.1)xn+1=max{xn,A}xn2xn-1,n=0,1,…,
where the initial conditions are x-1=Ar1 and x0=Ar2.Now, in this paper we give answer to the open problem6.4.


## 2. Main Results

### 2.1. The CaseA>1

We consider (1.1) where A>1. It is clear that the change of variables(2.1)xn=Arnforn≥-1
reduces (1.1) to the difference equation:(2.2)rn+1=max{rn,1}-2rn-rn-1,n=0,1,2,…,
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{rn}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)
IfrN=-r, then rN-1>1 and rN+1≥1.(ii)
IfrN=-r, then rN=rN+3 or rN=rN+5.(iii)
IfrN=-r and rN+2>1, then rN+3=-r and rN+2=rN-1+rN-1.(iv)
IfrN=-r and rN+2≤1, then rN+5=-r and rN+4=rN-1-rN-2.Proof.
(i) Assume that2<r0+r-1=r. If r0>1, then from (2.2) we get r1=-r and r2=1+r0+2r-1>1.Ifr0≤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.Ifr5>1, then r6=-r and r7=r-1+3>1.Ifr5≤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 haverN-1>1 and rN+1≥1 for rN=-r. So, the proof of (i) is complete.
(ii) Assume thatrN=-r. From (i) and (2.2) we get that(2.3)rN+1=1-2rN-rN-1≥1,rN+2=rN+rN-1-1.
IfrN+2>1, then rN+3=rN=-r.
IfrN+2≤1, then rN+3=2-rN-1<1,rN+4=rN-1-rN-2>1 and rN+5=rN=-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{rn}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 that2<r0+r-1=r=k/m,GCD(k,m)=1,(8k-m)≠0(mod3) and 0<N<(8k-m+2).
Assume  that number of integersN 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;(2.4)rN=rN+3=rN+6=⋯=rN+3(k-2m)=rN+3(k-2m+1)=-r,rN+2=rN-1+rN-1>1,rN+5=rN-1+2(rN-1)>1,rN+8=rN-1+3(rN-1)>1,⋮rN+3(k-2m)+2=rN-1+(k-2m+1)(rN-1)>1.
Thus,rN-1>1+(k-2m+1)(-rN+1)>-2rN. But, from Lemma 2.1(i) we have rN+1=1-2rN-rN-1≥1 and (-2rN)≥rN-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 integersN satisfying Lemma 2.1(iv) is (k+m+1). So, we can apply Lemma 2.1(iv) consecutively for (k+m+1) times such that(2.5)rN=rN+5=rN+10=⋯=rN+5(k+m)=rN+5(k+m+1)=-r,rN+4=rN-1-rN-2>1,rN+9=rN-1+2(-rN-2)>1,rN+14=rN-1+3(-rN-2)>1,⋮rN+5(k+m-1)+4=rN-1+(k+m)(-rN-2)>1.
Thus, we haverN+5(k+m)+2=rN-1+(k+m)(-rN-2)+rN-1≤1 and rN-1≤1+(k+m)(rN+2)-rN+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 integersN 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 rj=-r for j∈{1,3}. We apply Lemma 2.1(iii) for (k-2m-1) times such that(2.6)rj=rj+3=rj+6=⋯=rj+3(k-2m-2)=rj+3(k-2m-1)=-r
and from Lemma 2.1(ii)
(2.7)rj+3(k-2m-1)=rj+3(k-2m-1)+5=⋯=rj+3(k-2m-1)+5(k+m)=rj+3(k-2m-1)+5(k+m+1)=-r.
Thus, the number of integersN 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 integersN satisfying Lemma 2.1(iv) is (k+m-1). From Lemma 2.1(ii)–(iv), we have(2.8)rj=rj+5=rj+10=⋯=rj+5(k+m-2)=rj+5(k+m-1)=-r,rj+5(k+m-1)=rj+5(k+m-1)+3=⋯=rj+5(k+m-1)+3(k-2m+1)=rj+5(k+m-1)+3(k-2m+2)=-r.
Thus, the number of integersN 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 Lemmas2.3 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{rn}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)
IfrN=-r, then rN-1>1 and rN+1≥1.(ii)
IfrN=-r, then rN=rN+5 or rN=rN+7.(iii)
IfrN=-r and rN+4>1, then rN+5=-r and rN+4=rN-1-rN-2.(iv)
IfrN=-r and rN+4≤1, then rN+7=-r and rN+6=rN-1-3rN-3.Lemma 2.4.
Assume that{rn}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{rn}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 thatrN>1, and rN+2≥1. If r0+r-1<1/2,thenrN+1=r0+r-1-1 or if r0+r-1>1/2, then rN+1=-r0-r-1.(ii)
IfrN>1 and rN+2≥1, then rN+3<1,rN+4<1,rN+5<1,rN+6=3+2rN+1-rN<1 and rN+7=rN-3(rN+1+1),(iii)
IfrN∉(rNi,rNi+1,rNi+2,rNi+3,rNi+4,rNi+5,rNi+6) such that rNi>1 and rNi+2≥1, then rN≤1.Proof.
(i) Assume thatr0+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.Ifr3>1, then r4=r0+r-1-1,r5=1+2r0+r-1>1.Ifr3≤1, then r4=5r0+4r-1-2<1, r5=3-6r0-5r-1 and r5>1 or r5≤1.Ifr5>1, then r6=r0+r-1-1 and r7=4r0+3r-1≥1.Ifr5≤1, then r6=7r0+6r-1-3<1, r7=4-8r0-7r-1 and r7>1 orr7≤1.
Working inductively, we haverN+1=r0+r-1-1 for r0+r-1<1/2,rN>1 and rN+2≥1.
Assume thatr0+r-1>1/2. From (2.2), we get that r1=1-2r0-r-1<1 and r2=3r0+2r-1-1.
Ifr2>1, then r3=-r0-r-1 and r4=2-r0>1.Ifr2≤1, then r3=2-4r0-3r-1<1 and r4=5r0+4r-1-2.Ifr4>1, then r5=-r0-r-1 and r6=3-3r0-2r-1>1.
Working inductively, we haverN+1=-r0-r-1 for 1/2<r0+r-1<1,rN>1 and rN+2≥1. So, the proof of (i) is completed.
(ii)
Assume thatr0+r-1<1 and that rN>1,rN+2≥1. From (2.2) and (i), we get that
(2.9)rN+2=max{rN+1,1}-2rN+1-rN=1-2rN+1-rN,rN≤-2rN+1.
Then,(2.10)rN+3=max{rN+2,1}-2rN+2-rN+1=rN+rN+1-1<1,rN+4=max{rN+3,1}-2rN+3-rN+2=2-rN<1,rN+5=max{rN+4,1}-2rN+4-rN+3=rN-rN+1-2<1,rN+6=max{rN+5,1}-2rN+5-rN+4=-rN+2rN+1+3<1,rN+7=max{rN+6,1}-2rN+6-rN+5=rN-3(rN+1+1).
So, the proof of (ii) is completed.
(iii)
Assume that the smallest integer of integersN satisfying (i) is N1. From the proof of (i), we have rj<1 for j<N1. Also, from this assumption we have the subsequence (rN1,rN1+1,rN1+2,rN1+3,rN1+4,rN1+5,rN1+6) such that rN1>1 and rN1+2≥1. Then, from (ii) we get that rN1+7=rN1-3(rN1+1+1) and rN1+7≤1 or rN1+7>1. If rN1+7≤1, we get that rN1+8=-rN1+4(rN1+1+1)<1. It means that rN1+7 andrN1+8 are not the element previous and later subsequences satisfying (ii). If rN1+7>1, then we get that rN1+9≥1 and that rN1+7 is a element of the subsequence (rN1+7,rN1+8,rN1+9,rN1+10,rN1+11,rN1+12,rN1+13) such thatrN1+7>1 and rN1+9≥1. If this proceeds, we have rN≤1, for rN∉(rNi,rNi+1,rNi+2,rNi+3,rNi+4,rNi+5,rNi+6) such that rNi>1 and rNi+2≥1.Lemma 2.6.
Assume that{rn}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{rn}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 {rn}n=-1∞ is periodic with prime period (7m-8k),(ii)
If(7m-8k)=0(mod3), then {rn}n=-1∞ is periodic with prime period ((7m-8k)/3).Proof.
(i) Assume thatr-1+r0=k/m<1/2,GCD(k,m)=1 and (7m-8k)≠0(mod3). It suffices to prove rn=rn+7m-8k for n≥-1. We must show that r-1=r7m-8k-1 and r0=r7m-8k. From Lemmas 2.5 and 2.6, for getting the last two terms we need, we can assume that
(2.11)rn≤1forn=-1,0,…,(6k-2)
and that
(2.12)r6k-1+7l>1,r6k+1+7l≥1forl=0,1,…,(m-2k-1).
From this assumption and (2.2), we get that for n=1,2,…,(3k-1)(2.13)r2n-1=r-1+∑i=1n[1-2(r-1+r0)]≤1r2n=r0+∑i=1n[2(r-1+r0)-1]≤1
and that
(2.14)r6k-1=r-1+∑i=13k[1-2(r-1+r0)]>1,r6k=(r-1+r0)-1,r6k+1=2-r-1+∑i=13k-1[2(r-1+r0)-1]≥1,⋮r6k+5=3r-1+4r0+∑i=13k-1[2(r-1+r0)-1]≤1,r6k+6=1-(4r-1+5r0)-∑i=13k-1[2(r-1+r0)-1]>1,r6k+7=(r-1+r0)-1,⋮r6k+12=6r-1+7r0+∑i=13k-1[2(r-1+r0)-1]≤1,r6k+13=1-(7r-1+8r0)-∑i=13k-1[2(r-1+r0)-1]>1,⋮r6k-1+7(m-2k-1)=1-r-1-2r0-3(m-2k-1)(r-1+r0)-∑i=13k-1[2(r-1+r0)-1]>1,⋮r6k-1+7(m-2k-1)+6=r0+3(m-2k)(r-1+r0)+∑i=13k-1[2(r-1+r0)-1]≤1,r6k-1+7(m-2k)=1-[-r-1+(3m-6k+2)(r-1+r0)]-∑i=13k-1[2(r-1+r0)-1].
Then we getr7m-8k-1=r-1 and that(2.15)r7m-8k=1-2r7m-8k-1-r7m-8k-2=1-2r-1-r7m-8k-2-[r0+3(m-2k)(r-1+r0)+∑i=13k-1[2(r-1+r0)-1]]=r0.
So, we havern=rn+7m-8k for n≥-1.
(ii)
Assume thatr-1+r0=1/2. From (2.2), we get immediately r1=r-1 and r2=r0. So, we have rn=rn+(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 {rn}n=-1∞ is periodic with prime period (8k-m).(ii)
If(8k-m)=0(mod3), then {rn}n=-1∞ is periodic with prime period ((8k-m)/3).Proof.
(i) Ifr-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 rn=rn+8k-m for n≥-1.
In the case1/2<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 that(2.16)rn≤1forn=-1,0,…,[6(m-k)-3]
and that
(2.17)r6(m-k)-2+7l>1,r6(m-k)+7l≥1,r6(m-k)-2+7(2k-m)≤1
for l=0,1,…,(2k-m-1). Thus, we get that
(2.18)r2n=r0+∑i=1n[2(r-1+r0)-1]≤1,r2n+1=r-1+∑i=1n+1[1-2(r-1+r0)]≤1
for n=1,2,…,[3(m-k)-2] and that
(2.19)r6(m-k)-2=r0+∑i=13(m-k)-1[2(r-1+r0)-1]>1,r6(m-k)-1=-(r-1+r0),r6(m-k)=1+2r-1+r0+∑i=13(m-k)-1[1-2(r-1+r0)]≥1,⋮r6(m-k)-2+7(2k-m)-1=r-1+3(2k-m)(1-r-1-r0)-∑i=13(m-k)-1[2(r-1+r0)-1]≤1,r6(m-k)-2+7(2k-m)=r0+3(2k-m)(r-1+r0-1)+∑i=13(m-k)-1[2(r-1+r0)-1]≤1,r6(m-k)+7(2k-m)-1=1-2r6(m-k)+7(2k-m)-2-r6(m-k)+7(2k-m)-3=r-1,r6(m-k)+7(2k-m)=1-2r6(m-k)+7(2k-m)-1-r6(m-k)+7(2k-m)-2=r0
So, we havern=rn+8k-m for n≥-1.
Assume that1<r-1+r0<2. From Lemmas 2.3 and 2.4, for getting the last two terms, we can assume that(2.20)rN+5l1=rN+5(l1+1)=-r,rN+5[3(k-m)]+7l2=rN+5[3(k-m)]+7(l2+1)=-r
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
(2.21)rN+4=rN-1-rN-2,rN+9=rN-1+2(-rN-2),⋮rN+5[3(k-m)-1]+4=rN-1+3(k-m)(-rN-2),
and that
(2.22)rN+5[3(k-m)]+6=rN+5[3(k-m)-1]+4-3rN-3,rN+5[3(k-m)]+13=rN+5[3(k-m)-1]+4+2(-3rN-3),⋮rN+5[3(k-m)]+7(2m-k-1)+6=rN+5[3(k-m)-1]+4+(2m-k)(-3rN-3)>1
Thus, we getrN+8k-m-1=rN-1 and rN+8k-m=rN. From rn-1=max{rn,1}-2rn-rn+1, we have rn=rn+8k-m for n≥-1. Also, it is easy to see that rN-1≠rN+5l1-1 and rN-1≠rN+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 case2<r-1+r0, the proof is similar. So, it is omitted.
(ii)
Assume thatr-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 rn=rn+(8k-m)/3 for n≥-1. The rest of proof is similar to the proof of (i) and is omitted.


### 2.2. The CaseA<1

We consider (1.1) where A<1. It is clear that the change of variables
(2.23)xn=Arnforn≥-1
reduces (1.1) to the difference equation:(2.24)rn+1=min{rn,1}-2rn-rn-1,n=0,1,2,…,
where the initial conditions r-1 and r0 are positive rational numbers.In this section we consider the behavior of the solutions of (2.24) (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{rn}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)
IfrN=r-1, then rN-1≤1 and rN+1≤1.(ii)
IfrN=r-1, then rN=rN+3 or rN=rN+5.(iii)
IfrN=r-1 and rN+2<1, then rN+3=r-1 and rN+2=rN-1+rN-1.(iv)
IfrN=r-1 and rN+2≥1, then rN+5=r-1 and rN+4=rN-1-rN-2.Lemma 2.11.
Assume that{rn}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{rn}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)
IfrN=r, then rN-1≤1 and rN+1≤1.(ii)
IfrN=r, then rN=rN+3 or rN=rN+5.(iii)
IfrN=r and rN+2<1, then rN+3=r and rN+2=rN-1+rN-1.(iv)
IfrN=r and rN+2≥1, then rN+5=r and rN+4=rN-1-rN-2.Lemma 2.13.
Assume that{rn}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 {rn}n=-1∞ is periodic with prime period (8k-7m).(ii)
If(8k-7m)=0(mod3), then {rn}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 {rn}n=-1∞ is periodic with prime period (8k+m).(ii)
If(8k+m)=0(mod3), then {rn}n=-1∞ is periodic with prime period ((8k+m)/3).


## 2.1. The CaseA>1

We consider (1.1) where A>1. It is clear that the change of variables(2.1)xn=Arnforn≥-1
reduces (1.1) to the difference equation:(2.2)rn+1=max{rn,1}-2rn-rn-1,n=0,1,2,…,
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{rn}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)
IfrN=-r, then rN-1>1 and rN+1≥1.(ii)
IfrN=-r, then rN=rN+3 or rN=rN+5.(iii)
IfrN=-r and rN+2>1, then rN+3=-r and rN+2=rN-1+rN-1.(iv)
IfrN=-r and rN+2≤1, then rN+5=-r and rN+4=rN-1-rN-2.Proof.
(i) Assume that2<r0+r-1=r. If r0>1, then from (2.2) we get r1=-r and r2=1+r0+2r-1>1.Ifr0≤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.Ifr5>1, then r6=-r and r7=r-1+3>1.Ifr5≤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 haverN-1>1 and rN+1≥1 for rN=-r. So, the proof of (i) is complete.
(ii) Assume thatrN=-r. From (i) and (2.2) we get that(2.3)rN+1=1-2rN-rN-1≥1,rN+2=rN+rN-1-1.
IfrN+2>1, then rN+3=rN=-r.
IfrN+2≤1, then rN+3=2-rN-1<1,rN+4=rN-1-rN-2>1 and rN+5=rN=-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{rn}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 that2<r0+r-1=r=k/m,GCD(k,m)=1,(8k-m)≠0(mod3) and 0<N<(8k-m+2).
Assume  that number of integersN 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;(2.4)rN=rN+3=rN+6=⋯=rN+3(k-2m)=rN+3(k-2m+1)=-r,rN+2=rN-1+rN-1>1,rN+5=rN-1+2(rN-1)>1,rN+8=rN-1+3(rN-1)>1,⋮rN+3(k-2m)+2=rN-1+(k-2m+1)(rN-1)>1.
Thus,rN-1>1+(k-2m+1)(-rN+1)>-2rN. But, from Lemma 2.1(i) we have rN+1=1-2rN-rN-1≥1 and (-2rN)≥rN-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 integersN satisfying Lemma 2.1(iv) is (k+m+1). So, we can apply Lemma 2.1(iv) consecutively for (k+m+1) times such that(2.5)rN=rN+5=rN+10=⋯=rN+5(k+m)=rN+5(k+m+1)=-r,rN+4=rN-1-rN-2>1,rN+9=rN-1+2(-rN-2)>1,rN+14=rN-1+3(-rN-2)>1,⋮rN+5(k+m-1)+4=rN-1+(k+m)(-rN-2)>1.
Thus, we haverN+5(k+m)+2=rN-1+(k+m)(-rN-2)+rN-1≤1 and rN-1≤1+(k+m)(rN+2)-rN+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 integersN 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 rj=-r for j∈{1,3}. We apply Lemma 2.1(iii) for (k-2m-1) times such that(2.6)rj=rj+3=rj+6=⋯=rj+3(k-2m-2)=rj+3(k-2m-1)=-r
and from Lemma 2.1(ii)
(2.7)rj+3(k-2m-1)=rj+3(k-2m-1)+5=⋯=rj+3(k-2m-1)+5(k+m)=rj+3(k-2m-1)+5(k+m+1)=-r.
Thus, the number of integersN 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 integersN satisfying Lemma 2.1(iv) is (k+m-1). From Lemma 2.1(ii)–(iv), we have(2.8)rj=rj+5=rj+10=⋯=rj+5(k+m-2)=rj+5(k+m-1)=-r,rj+5(k+m-1)=rj+5(k+m-1)+3=⋯=rj+5(k+m-1)+3(k-2m+1)=rj+5(k+m-1)+3(k-2m+2)=-r.
Thus, the number of integersN 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 Lemmas2.3 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{rn}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)
IfrN=-r, then rN-1>1 and rN+1≥1.(ii)
IfrN=-r, then rN=rN+5 or rN=rN+7.(iii)
IfrN=-r and rN+4>1, then rN+5=-r and rN+4=rN-1-rN-2.(iv)
IfrN=-r and rN+4≤1, then rN+7=-r and rN+6=rN-1-3rN-3.Lemma 2.4.
Assume that{rn}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{rn}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 thatrN>1, and rN+2≥1. If r0+r-1<1/2,thenrN+1=r0+r-1-1 or if r0+r-1>1/2, then rN+1=-r0-r-1.(ii)
IfrN>1 and rN+2≥1, then rN+3<1,rN+4<1,rN+5<1,rN+6=3+2rN+1-rN<1 and rN+7=rN-3(rN+1+1),(iii)
IfrN∉(rNi,rNi+1,rNi+2,rNi+3,rNi+4,rNi+5,rNi+6) such that rNi>1 and rNi+2≥1, then rN≤1.Proof.
(i) Assume thatr0+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.Ifr3>1, then r4=r0+r-1-1,r5=1+2r0+r-1>1.Ifr3≤1, then r4=5r0+4r-1-2<1, r5=3-6r0-5r-1 and r5>1 or r5≤1.Ifr5>1, then r6=r0+r-1-1 and r7=4r0+3r-1≥1.Ifr5≤1, then r6=7r0+6r-1-3<1, r7=4-8r0-7r-1 and r7>1 orr7≤1.
Working inductively, we haverN+1=r0+r-1-1 for r0+r-1<1/2,rN>1 and rN+2≥1.
Assume thatr0+r-1>1/2. From (2.2), we get that r1=1-2r0-r-1<1 and r2=3r0+2r-1-1.
Ifr2>1, then r3=-r0-r-1 and r4=2-r0>1.Ifr2≤1, then r3=2-4r0-3r-1<1 and r4=5r0+4r-1-2.Ifr4>1, then r5=-r0-r-1 and r6=3-3r0-2r-1>1.
Working inductively, we haverN+1=-r0-r-1 for 1/2<r0+r-1<1,rN>1 and rN+2≥1. So, the proof of (i) is completed.
(ii)
Assume thatr0+r-1<1 and that rN>1,rN+2≥1. From (2.2) and (i), we get that
(2.9)rN+2=max{rN+1,1}-2rN+1-rN=1-2rN+1-rN,rN≤-2rN+1.
Then,(2.10)rN+3=max{rN+2,1}-2rN+2-rN+1=rN+rN+1-1<1,rN+4=max{rN+3,1}-2rN+3-rN+2=2-rN<1,rN+5=max{rN+4,1}-2rN+4-rN+3=rN-rN+1-2<1,rN+6=max{rN+5,1}-2rN+5-rN+4=-rN+2rN+1+3<1,rN+7=max{rN+6,1}-2rN+6-rN+5=rN-3(rN+1+1).
So, the proof of (ii) is completed.
(iii)
Assume that the smallest integer of integersN satisfying (i) is N1. From the proof of (i), we have rj<1 for j<N1. Also, from this assumption we have the subsequence (rN1,rN1+1,rN1+2,rN1+3,rN1+4,rN1+5,rN1+6) such that rN1>1 and rN1+2≥1. Then, from (ii) we get that rN1+7=rN1-3(rN1+1+1) and rN1+7≤1 or rN1+7>1. If rN1+7≤1, we get that rN1+8=-rN1+4(rN1+1+1)<1. It means that rN1+7 andrN1+8 are not the element previous and later subsequences satisfying (ii). If rN1+7>1, then we get that rN1+9≥1 and that rN1+7 is a element of the subsequence (rN1+7,rN1+8,rN1+9,rN1+10,rN1+11,rN1+12,rN1+13) such thatrN1+7>1 and rN1+9≥1. If this proceeds, we have rN≤1, for rN∉(rNi,rNi+1,rNi+2,rNi+3,rNi+4,rNi+5,rNi+6) such that rNi>1 and rNi+2≥1.Lemma 2.6.
Assume that{rn}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{rn}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 {rn}n=-1∞ is periodic with prime period (7m-8k),(ii)
If(7m-8k)=0(mod3), then {rn}n=-1∞ is periodic with prime period ((7m-8k)/3).Proof.
(i) Assume thatr-1+r0=k/m<1/2,GCD(k,m)=1 and (7m-8k)≠0(mod3). It suffices to prove rn=rn+7m-8k for n≥-1. We must show that r-1=r7m-8k-1 and r0=r7m-8k. From Lemmas 2.5 and 2.6, for getting the last two terms we need, we can assume that
(2.11)rn≤1forn=-1,0,…,(6k-2)
and that
(2.12)r6k-1+7l>1,r6k+1+7l≥1forl=0,1,…,(m-2k-1).
From this assumption and (2.2), we get that for n=1,2,…,(3k-1)(2.13)r2n-1=r-1+∑i=1n[1-2(r-1+r0)]≤1r2n=r0+∑i=1n[2(r-1+r0)-1]≤1
and that
(2.14)r6k-1=r-1+∑i=13k[1-2(r-1+r0)]>1,r6k=(r-1+r0)-1,r6k+1=2-r-1+∑i=13k-1[2(r-1+r0)-1]≥1,⋮r6k+5=3r-1+4r0+∑i=13k-1[2(r-1+r0)-1]≤1,r6k+6=1-(4r-1+5r0)-∑i=13k-1[2(r-1+r0)-1]>1,r6k+7=(r-1+r0)-1,⋮r6k+12=6r-1+7r0+∑i=13k-1[2(r-1+r0)-1]≤1,r6k+13=1-(7r-1+8r0)-∑i=13k-1[2(r-1+r0)-1]>1,⋮r6k-1+7(m-2k-1)=1-r-1-2r0-3(m-2k-1)(r-1+r0)-∑i=13k-1[2(r-1+r0)-1]>1,⋮r6k-1+7(m-2k-1)+6=r0+3(m-2k)(r-1+r0)+∑i=13k-1[2(r-1+r0)-1]≤1,r6k-1+7(m-2k)=1-[-r-1+(3m-6k+2)(r-1+r0)]-∑i=13k-1[2(r-1+r0)-1].
Then we getr7m-8k-1=r-1 and that(2.15)r7m-8k=1-2r7m-8k-1-r7m-8k-2=1-2r-1-r7m-8k-2-[r0+3(m-2k)(r-1+r0)+∑i=13k-1[2(r-1+r0)-1]]=r0.
So, we havern=rn+7m-8k for n≥-1.
(ii)
Assume thatr-1+r0=1/2. From (2.2), we get immediately r1=r-1 and r2=r0. So, we have rn=rn+(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 {rn}n=-1∞ is periodic with prime period (8k-m).(ii)
If(8k-m)=0(mod3), then {rn}n=-1∞ is periodic with prime period ((8k-m)/3).Proof.
(i) Ifr-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 rn=rn+8k-m for n≥-1.
In the case1/2<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 that(2.16)rn≤1forn=-1,0,…,[6(m-k)-3]
and that
(2.17)r6(m-k)-2+7l>1,r6(m-k)+7l≥1,r6(m-k)-2+7(2k-m)≤1
for l=0,1,…,(2k-m-1). Thus, we get that
(2.18)r2n=r0+∑i=1n[2(r-1+r0)-1]≤1,r2n+1=r-1+∑i=1n+1[1-2(r-1+r0)]≤1
for n=1,2,…,[3(m-k)-2] and that
(2.19)r6(m-k)-2=r0+∑i=13(m-k)-1[2(r-1+r0)-1]>1,r6(m-k)-1=-(r-1+r0),r6(m-k)=1+2r-1+r0+∑i=13(m-k)-1[1-2(r-1+r0)]≥1,⋮r6(m-k)-2+7(2k-m)-1=r-1+3(2k-m)(1-r-1-r0)-∑i=13(m-k)-1[2(r-1+r0)-1]≤1,r6(m-k)-2+7(2k-m)=r0+3(2k-m)(r-1+r0-1)+∑i=13(m-k)-1[2(r-1+r0)-1]≤1,r6(m-k)+7(2k-m)-1=1-2r6(m-k)+7(2k-m)-2-r6(m-k)+7(2k-m)-3=r-1,r6(m-k)+7(2k-m)=1-2r6(m-k)+7(2k-m)-1-r6(m-k)+7(2k-m)-2=r0
So, we havern=rn+8k-m for n≥-1.
Assume that1<r-1+r0<2. From Lemmas 2.3 and 2.4, for getting the last two terms, we can assume that(2.20)rN+5l1=rN+5(l1+1)=-r,rN+5[3(k-m)]+7l2=rN+5[3(k-m)]+7(l2+1)=-r
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
(2.21)rN+4=rN-1-rN-2,rN+9=rN-1+2(-rN-2),⋮rN+5[3(k-m)-1]+4=rN-1+3(k-m)(-rN-2),
and that
(2.22)rN+5[3(k-m)]+6=rN+5[3(k-m)-1]+4-3rN-3,rN+5[3(k-m)]+13=rN+5[3(k-m)-1]+4+2(-3rN-3),⋮rN+5[3(k-m)]+7(2m-k-1)+6=rN+5[3(k-m)-1]+4+(2m-k)(-3rN-3)>1
Thus, we getrN+8k-m-1=rN-1 and rN+8k-m=rN. From rn-1=max{rn,1}-2rn-rn+1, we have rn=rn+8k-m for n≥-1. Also, it is easy to see that rN-1≠rN+5l1-1 and rN-1≠rN+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 case2<r-1+r0, the proof is similar. So, it is omitted.
(ii)
Assume thatr-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 rn=rn+(8k-m)/3 for n≥-1. The rest of proof is similar to the proof of (i) and is omitted.


## 2.2. The CaseA<1

We consider (1.1) where A<1. It is clear that the change of variables
(2.23)xn=Arnforn≥-1
reduces (1.1) to the difference equation:(2.24)rn+1=min{rn,1}-2rn-rn-1,n=0,1,2,…,
where the initial conditions r-1 and r0 are positive rational numbers.In this section we consider the behavior of the solutions of (2.24) (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{rn}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)
IfrN=r-1, then rN-1≤1 and rN+1≤1.(ii)
IfrN=r-1, then rN=rN+3 or rN=rN+5.(iii)
IfrN=r-1 and rN+2<1, then rN+3=r-1 and rN+2=rN-1+rN-1.(iv)
IfrN=r-1 and rN+2≥1, then rN+5=r-1 and rN+4=rN-1-rN-2.Lemma 2.11.
Assume that{rn}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{rn}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)
IfrN=r, then rN-1≤1 and rN+1≤1.(ii)
IfrN=r, then rN=rN+3 or rN=rN+5.(iii)
IfrN=r and rN+2<1, then rN+3=r and rN+2=rN-1+rN-1.(iv)
IfrN=r and rN+2≥1, then rN+5=r and rN+4=rN-1-rN-2.Lemma 2.13.
Assume that{rn}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 {rn}n=-1∞ is periodic with prime period (8k-7m).(ii)
If(8k-7m)=0(mod3), then {rn}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 {rn}n=-1∞ is periodic with prime period (8k+m).(ii)
If(8k+m)=0(mod3), then {rn}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].



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