# On a Study of a Family of Higher-Order Recurrence Relations

Authors: Mensah Folly-Gbetoula; Kgatliso Mkhwanazi; Darlison Nyirenda
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

In this work, group theory is applied to obtain generators of the Lie algebra for a class of rational difference equations of orderk+l where k and l are positive integers. This is followed by deriving invariants and solutions via some linear recurrences and with the introduction of some number-theoretic arithmetical functions. Furthermore, sufficient conditions for the existence of solutions for periods two and four are provided in special cases. The results in this paper generalize certain known results.


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## Body

## 1. Introduction

There has been great progress in the study of difference equations. Several techniques for solving difference equations have been developed. One of them is Lie analysis [1–3]. Lie analysis was introduced by Sophus Lie, a Norwegian mathematician who realized that many of the methodologies developed for solving differential equations are special cases of an integration procedure that is intrinsic to a continuous group of point transformations admitted by these differential equations (see [4]). Despite Hydon providing an algorithm for determining Lie symmetries of difference equations (see [1]), a huge amount of work involving complex calculations is involved. It is important to point out that using symmetries in reducing the order of difference equations is somewhat recent, particularly for difference equations of higher order.Difference equations have real-life applications that include compound interest, loan payments, and biological population dynamics. With regard to biological semelparous populations such as insects, it turns out that(1)rn+1=p0rn1+trn,where rn denotes the number of adults in the nth breeding season and p0 is the average number of eggs laid by an adult and t is some special constant [5].In [6], the authors discussed the Beverton-Holt equation of order l∈ℕ given by(2)zn+l=μnKnznKn+μn−1zn,n∈ℕ0,where μn>1 is the growth rate, Kn>0 is the carrying capacity, and the initial conditions z0=z0,z1,…,zl−1T, zi>0.Settingzn=xnxn+k, an=1/μn and bn=μn−1/μnKn, (1) takes the form(3)xn+k+l=xnxn+kxn+lan+bnxnxn+k,or equivalently(4)xn+1=xn−k−l+1xn−l+1xn−k+1An+Bnxn−k−l+1xn−l+1.Special cases of (4) attracted the interest of many researchers and interesting results can be found in the literature, see [7–17] among others. In this paper, we generalize some results in these articles by studying (2) with initial values x0,x1,…,xk+l−1 and real sequences an, bn. We derive symmetry generators and explicit formulas for solutions of (2). We also look at periodicity in some cases of well-defined solutions to the equation.It is now known that the Lie analysis method can be successfully used to solve ordinary difference equations, even in the case of systems of difference equations (see [1]). For more work, see [18–22].The main advantage of our results is that they generalize several results in the literature. This study is important in the sense that it involves an application of the symmetry method to solving difference equations of higher order as well as the involvement of number theoretic techniques. The presentation of the solution in a unified manner for some cases strikes an important departure from how the solution is presented in most of the existing literature.


### 1.1. Definitions and Notation

There is a lot of literature on using geometric methods such as symmetries and first integrals to derive solutions of difference equations (see [1, 3, 23, 24]).Hydon presents a systematic method for determining point symmetries of difference equations in [1]. Our notation is similar to that of Hydon. Hydon’s algorithm is valid for any order although he applied it to lower order difference equations.Definition 1.
[25]. Let G be a connected group of transformations acting on a manifold M. A smooth real-valued function ℐ:M⟶ℝ is an invariant function for G if and only if(5)Vℐ=0for allx∈M,and every (prolonged) infinitesimal generator V of G.Definition 2.
[1]. A parameterized set of point transformations,(6)Γε:x↦x^x;ε,where x=xi,i=1,…,p are continuous variables, is a one-parameter local Lie group of transformations if the following conditions are satisfied:(1)
Γ0 is the identity map if x^=x when ε=0,(2)
ΓaΓb=Γa+b for every a and b sufficiently close to 0,(3)
Eachxi^ can be represented as a Taylor series (in a neighborhood of ε=0 that is determined by x), and therefore,(7)xi^x:ε=xi+εξix+Oε2,i=1,…,p.
For the discrete analog, we introduce the shift forward operator(8)Si:n⟶n+i.
Consider the following family of higher-order difference equations:(9)xn+k+l=Fn,xn,xn+k,xn+l,for some differentiable function F. Symmetry groups are related to infinitesimal transformations. Suppose that the group of infinitesimal transformations of (9) assumes the shape(10)x^n=xn+ϵQn,xn+Oϵ2,where ϵ is the one-parameter family of the group of transformations, with the corresponding vector field(11)X=Qn,xn∂∂xn.
The functionQ is called the characteristic of the group of point transformations. In order to know Q, we must know the k+l−1-th extension, Xk+l−1, of the vector field X.(12)Xk+l−1=Qn,xn∂∂xn+SQn,xn∂∂xn+1+⋯+Sk+l−1Qn,xn∂∂xn+k+l−1.
Note that the necessary and sufficient condition for (5) to be a symmetry group is that the linearized symmetry condition(13)Qn+k+l,F−Xk+l−1F=0|xn+k+l=Fn,xn,xn+l,xn+k,be satisfied. Assume the characteristic is derived by solving the functional (13). One can utilize the canonical coordinate [26](14)wn=∫dxnQn,xn,to deduce the similarity variables. The similarity variables can then be used in reducing the order of the difference equation.
Indeed with this particular choice of the canonical coordinate, as pointed out by Hydon (see [1]), the difference equation can be written in the form(15)wn+1−wn=bn.
Thus, the solution of (15) is(16)wn=∑k=n0nbk+c1,for some constant c1. At this stage, it is not difficult to get the solution expressed in terms of the original variables.
This paper is organized as follows: In Section2, symmetries and explicit expressions of solutions are deduced. In Sections 3 and 4, we explore special cases for values of l and k when sequences an and bn are constant. In such cases, apart from the form of the solution, we also look at periodicity. In Section 5, we discuss the results in relation to what has been done in the literature. In this work, the process of finding the solution using the canonical coordinate involves a different method.


## 1.1. Definitions and Notation

There is a lot of literature on using geometric methods such as symmetries and first integrals to derive solutions of difference equations (see [1, 3, 23, 24]).Hydon presents a systematic method for determining point symmetries of difference equations in [1]. Our notation is similar to that of Hydon. Hydon’s algorithm is valid for any order although he applied it to lower order difference equations.Definition 1.
[25]. Let G be a connected group of transformations acting on a manifold M. A smooth real-valued function ℐ:M⟶ℝ is an invariant function for G if and only if(5)Vℐ=0for allx∈M,and every (prolonged) infinitesimal generator V of G.Definition 2.
[1]. A parameterized set of point transformations,(6)Γε:x↦x^x;ε,where x=xi,i=1,…,p are continuous variables, is a one-parameter local Lie group of transformations if the following conditions are satisfied:(1)
Γ0 is the identity map if x^=x when ε=0,(2)
ΓaΓb=Γa+b for every a and b sufficiently close to 0,(3)
Eachxi^ can be represented as a Taylor series (in a neighborhood of ε=0 that is determined by x), and therefore,(7)xi^x:ε=xi+εξix+Oε2,i=1,…,p.
For the discrete analog, we introduce the shift forward operator(8)Si:n⟶n+i.
Consider the following family of higher-order difference equations:(9)xn+k+l=Fn,xn,xn+k,xn+l,for some differentiable function F. Symmetry groups are related to infinitesimal transformations. Suppose that the group of infinitesimal transformations of (9) assumes the shape(10)x^n=xn+ϵQn,xn+Oϵ2,where ϵ is the one-parameter family of the group of transformations, with the corresponding vector field(11)X=Qn,xn∂∂xn.
The functionQ is called the characteristic of the group of point transformations. In order to know Q, we must know the k+l−1-th extension, Xk+l−1, of the vector field X.(12)Xk+l−1=Qn,xn∂∂xn+SQn,xn∂∂xn+1+⋯+Sk+l−1Qn,xn∂∂xn+k+l−1.
Note that the necessary and sufficient condition for (5) to be a symmetry group is that the linearized symmetry condition(13)Qn+k+l,F−Xk+l−1F=0|xn+k+l=Fn,xn,xn+l,xn+k,be satisfied. Assume the characteristic is derived by solving the functional (13). One can utilize the canonical coordinate [26](14)wn=∫dxnQn,xn,to deduce the similarity variables. The similarity variables can then be used in reducing the order of the difference equation.
Indeed with this particular choice of the canonical coordinate, as pointed out by Hydon (see [1]), the difference equation can be written in the form(15)wn+1−wn=bn.
Thus, the solution of (15) is(16)wn=∑k=n0nbk+c1,for some constant c1. At this stage, it is not difficult to get the solution expressed in terms of the original variables.
This paper is organized as follows: In Section2, symmetries and explicit expressions of solutions are deduced. In Sections 3 and 4, we explore special cases for values of l and k when sequences an and bn are constant. In such cases, apart from the form of the solution, we also look at periodicity. In Section 5, we discuss the results in relation to what has been done in the literature. In this work, the process of finding the solution using the canonical coordinate involves a different method.


## 2. Symmetries and Explicit Solutions

We start by deducing the characteristicQ (which will lead to the knowledge of the vector field) admitted by (3). Recall equation (3):(17)xn+k+l=xnxn+kxn+lan+bnxnxn+k.When the constraint (13) is imposed on (2), equation (6) is reduced to(18)Qn+k+l,xn+k+l+xnxn+kQn+l,xn+lxn+l2an+bnxnxn+k−anxnQn+k,xn+k+xn+kQn,xnxn+lan+bnxnxn+k2=0.The steps for computingQ are given as follows:(i)
Apply the differential operator∂/∂xn−xn+k/xn∂/∂xn+l on (18).(ii)
Rearrange the equation resulting from the previous step via multiplication byxn+lan+bnxnxn+k2 and then differentiate the result with respect to xn. You will obtain a differential equation in Q.(iii)
Solve the differential equation obtained in the previous step forQ.(iv)
Remove the dependency among the arbitrary constants by back-substituting the general expression ofQ in (18).Due to the amount of computation involved in executing the above steps, we leave out the details and state what is obtained. After a long calculation, we get(19)Q=βnxn,where βn satisfies(20)βn+βn+k=0.The solutions of (20) are given by βn=expi2r+1nπ/k, r=0,1,…,k−1.Thus we end up with a Lie algebra spanned by(21)Xr=expi2r+1nπkxn∂∂xn,r=0,1,…,k−1.In order to come up with the similarity variable, we make use of the canonical coordinate(22)w˜n=∫dxnβnxn=1βnlnxnand welet,wn=1expw˜nβn+w˜n+kβn+k.Clearly,wn is a group invariant of (3) because(23)Xrk+l−1wn=0,r=0,1,…,k−1.Observe thatwn yields(24)wn=1xnxn+k,in simplification, so that(25)wn+l=anwn+bn.Iterating (11), we obtain(26)wln+j=wj∏k1=0n−1alk1+j+∑m=0n−1blm+j∏k2=m+1n−1alk2+j,j=0,1,…,l−1.From (24), that is, wn=1/xnxn+k, one can check that(27)x2kn+j=xj∏s=0n−1w2ks+jw2s+1k+j,0≤j≤2k−1.Lett=lcm2k,l (the least common multiple). For a nonnegative integer a, let τa=la/l where  is the greatest integer function. Then a=τa+v where v=a−τa∈0,1,2,…l−1. We have(28)xtn+j=xj∏s=0tn/2k−1w2ks+jw2s+1k+j,0≤j≤t−1=xj∏s=0n−1wts+jwts+j+kwts+j+2kwts+j+3k…wts+j+t/2k−12kwts+j+t/2k−12k+k=xj∏s=0n−1∏r=0t/2k−1wts+j+2rkwts+j+2rk+k=xj∏s=0n−1∏r=0t/2k−1wts+j+2rkwts+j+2r+1k.Using (12) and the fact that wn=1/xnxn+k, the solution of (3) is given by(29)xtn+j=xj∏s=0n−1∏r=0t/2k−1wlt/ls+j+2rk/l+j+2rk−τj+2rkwlt/ls+j+2r+1k/l+j+2r+1k−τj+2r+1k,0≤j≤t−1,where the numerator(30)wltls+j+2rkl+j+2rk−τj+2rk=1xj+2rk−τj+2rkxj+2rk−τj+2rk+k∏k1=0ts/l+j+2rk/l−1alk1+j+2rk−τj+2rk+∑m=0ts/l+j+2rk/l−1blm+j+2rk−τj+2rk∏k2=m+1ts/l+j+2rk/l−1alk2+j+2rk−τj+2rk,and the denominator(31)wltls+j+2r+1kl+j+2r+1k−τj+2r+1k=1xj+2r+1k−τj+2r+1kxj+2r+1k−τj+2r+1k+k∏k1=0ts/l+j+2r+1k/l−1alk1+j+2r+1k−τj+2r+1k+∑m=0ts/l+j+2r+1k/l−1blm+j+2r+1k−τj+2r+1k∏k2=m+1ts/l+j+2r+1k/l−1alk2+j+2r+1k−τj+2r+1k.Assume thatan=a and bn=b. Thanks to (29), the solution to (3) is given by(32)xtn+j=xj∏s=0n−1∏r=0t/2k−11/xj+2rk−τj+2rkxj+2rk−τj+2rk+kats/l+⌊j+2rk/l⌋+b∑m=0ts/l+⌊j+2rk/l⌋−1am1/xj+2r+1k−τj+2r+1kxj+2r+1k−τj+2r+1k+kats/l+⌊j+2r+1k/l⌋+b∑m=0ts/l+⌊j+2r+1k/l⌋−1am,0≤j≤t−1.Equivalently, the solutions of (4) are obtained by back shifting (32) k+l−1 times as follows:(33)xtn+j−k−l+1=xj−k−l+1∏s=0n−1∏r=0t/2k−11/xj+2rk−τj+2rk−k−l+1xj+2rk−τj+2rk−l+1ats/l+⌊j+2rk/l⌋+b∑m=0ts/l+⌊j+2rk/l⌋−1am1/xj+2r+1k−τj+2r+1k−k−l+1xj+2r+1k−τj+2r+1k−l+1ats/l+⌊j+2r+1k/l⌋+b∑m=0ts/l+⌊j+2r+1k/l⌋−1am,0≤j≤t−1 when an and bn are constant.In the following sections, we look at some particular cases.


## 3. The Casesl=8, k=2 and an and bn Are Constant Sequences

### 3.1. Simplified Solutions

We setan=a and bn=b, and thus, substitute l=8, k=2 and t=lcm4,8=8 in (32) to obtain the solution(34)x8n+j=xjx2+j−τ2+jnx2+j−τ2+j+2nxj−τjnxj−τj+2nx6+j−τ6+jnx6+j−τ6+j+2nx4+j−τ4+jnx4+j−τ4+j+2n×∏s=0n−1as+j/8+bxj−τjxj−τj+2∑m=0s+j/8−1amas+2+j/8+bx2+j−τ2+jx2+j−τ2+j+2∑m=0s+2+j/8−1am×∏s=0n−1as+4+j/8+bx4+j−τ4+jx4+j−τ4+j+2∑m=0s+4+j/8−1amas+6+j/8+bx6+j−τ6+jx6+j−τ6+j+2∑m=0s+6+j/8−1am,where j=0,1,2,…,7.More explicitly, we have(35)x8n=x01−nx8n∏s=0n−1as+bx0x2∑m=0s−1amas+bx2x4∑m=0s−1amas+bx4x6∑m=0s−1amas+bx6x8∑m=0s−1am,(36)x8n+1=x11−nx9n∏s=0n−1as+bx1x3∑m=0s−1amas+bx3x5∑m=0s−1amas+bx5x7∑m=0s−1amas+bx7x9∑m=0s−1am,(37)x8n+2=x2x0nx8n∏s=0n−1as+bx2x4∑m=0s−1amas+bx4x6∑m=0s−1amas+bx6x8∑m=0s−1amas+1+bx0x2∑m=0sam,(38)x8n+3=x3x1nx9n∏s=0n−1as+bx3x5∑m=0s−1amas+bx5x7∑m=0s−1amas+bx7x9∑m=0s−1amas+1+bx1x3∑m=0sam,(39)x8n+4=x4x8nx0n∏s=0n−1as+bx4x6∑m=0s−1amas+bx6x8∑m=0s−1amas+1+bx0x2∑m=0samas+1+bx2x4∑m=0sam,(40)x8n+5=x5x9nx1n∏s=0n−1as+bx5x7∑m=0s−1amas+bx7x9∑m=0s−1amas+1+bx1x3∑m=0samas+1+bx3x5∑m=0sam,(41)x8n+6=x6x0nx8n∏s=0n−1as+bx6x8∑m=0s−1amas+1+bx0x2∑m=0samas+1+bx2x4∑m=0samas+1+bx4x6∑m=0sam,(42)x8n+7=x7x1nx9n∏s=0n−1as+bx7x9∑m=0s−1amas+1+bx1x3∑m=0samas+1+bx3x5∑m=0samas+1+bx5x7∑m=0sam.


### 3.2. Solutions Written in Unified Manner

One can write the solutions given in equations (17)–(24) in the following unified manner:(43)xn=Hnexp14∑s=0n−11+−1n−s+2cosn−sπ2lnwsws+2,where the Hn’s are such that H0=x0,H1=x1,H2=x2,H3=x3 and H4n+j=xjforj=0,1,2,3. Note that, here, the ws’s satisfy wn+8=awn+b, that is, they are given by w8n+j=wjan+b∑l=0n−1al.In fact, we have(44)x4n+j=Hjexp14∑s=04n+j−11+−1j−s+2cosj−sπ2lnwsws+2x4n+j=xj∏s=0n−1w4s+jw4s+2+j,0≤j≤3.Then from (44), we have(45)x8n+j=xj∏s=02n−1w4s+jw4s+2+j,0≤j≤7,=xj∏s=0n−1∏r=01w8s+4r+jw8s+4r+2+j.We have4r+j=24r+j/2+4r+j−τ4r+j so that, by (26),(46)w8s+4r+j=w24s+4r+j2+4r+j−τ4r+j=w4r+j−τ4r+ja4s+4r+j2+b∑m=04s+4r+j/2−1a4s+4r+j2−m−1=1x4r+j−τ4r+jx4r+j−τ4r+j+2a4s+4r+j2+b∑m=04s+4r+j/2−1a4s+4r+j2−m−1=a4s+4r+j/2x4r+j−τ4r+jx4r+j−τ4r+j+2+b∑m=04s+4r+j/2−1am=1x4r+j−τ4r+jx4r+j−τ4r+j+2a4s+4r+j2+bx4r+j−τ4r+jx4r+j−τ4r+j+2∑m=04s+4r+j/2−1am.Similarly, we get(47)w8s+4r+2+j=1x4r+2+j−τ4r+2+jx4r+2+j−τ4r+2+j+2a4s+4r+2+j2+bx4r+2+j−τ4r+2+jx4r+2+j−τ4r+2+j+2∑m=04s+4r+2+j/2−1am.So that the solution of (3), when l=8,k=2 and for constant coefficients, is given by(48)x8n+j=xj∏r=01x4r+2+j−τ4r+2+jnx4r+2+j−τ4r+2+j+2nx4r+j−τ4r+jnx4r+j−τ4r+j+2n×∏s=0n−1a4s+⌊4r+j/2⌋+bx4r+j−τ4r+jx4r+j−τ4r+j+2∑m=04s+⌊4r+j/2⌋−1ama4s+⌊4r+2+j/2⌋+bx4r+2+j−τ4r+2+jx4r+2+j−τ4r+2+j+2∑m=04s+⌊4r+2+j/2⌋−1am,which is the same as (34).


### 3.3. Existence of Periodic Solutions

Theorem 1.
Ifa,b≠0, xi=x4+i for 0≤i≤5, x0x2=x1x3=1−a/b, then the difference equation(49)xn+10=xnxn+2xn+8a+bxnxn+2,with initial conditions x0,x1,…,x8,x9 is periodic with period four.Proof.
Note that considering (35) and (39) subject to the condition x0=x4=x8, we have(50)x8n=x8n+4=x4for alln∈ℤ≥0.
Equations (36) and (40) together with x1=x5=x9 result in(51)x8n+1=x8n+5=x1for alln∈ℤ≥0.
Supposex0=x8 and x2=x6. Note that x0x2=1−a/b implies asbx0x2=as−as+1, i.e., as+1+bx0x2as=as so that(52)as+1+bx0x2as+bx0x2∑m=0s−1am=as+bx0x2∑m=0s−1am=as+bx6x8∑m=0s−1am,i.e., as+1+bx0x2∑m=0sam=as+bx6x8∑m=0s−1am. Thus x8n+2=x8n+6=x2 for all n∈ℤ≥0. Similarly, it can be shown that the conditions x3=x7,x1=x9 and x1x3=1−a/b imply(53)x8n+3=x8n+7=x3.
Hence the sequencexnn=0∞ satisfies(54)x8n+j=xjfor allj≥0,n≥0.
Sincex4+i=xi for all i=0,1,2,3,4,5, it follows that the sequence xnn=0∞ is periodic of period four. Therefore, the proof is complete.
We illustrate Theorem1 with Figure 1 where a=3, b=5, x0=x4=x8=1/13,x1=x5=x9=1/3,x2=x6=−26/5,x3=x7=−6/5.Figure 1
The plot ofxn+10=xnxn+2/xn+83+5xnxn+2.


## 3.1. Simplified Solutions

We setan=a and bn=b, and thus, substitute l=8, k=2 and t=lcm4,8=8 in (32) to obtain the solution(34)x8n+j=xjx2+j−τ2+jnx2+j−τ2+j+2nxj−τjnxj−τj+2nx6+j−τ6+jnx6+j−τ6+j+2nx4+j−τ4+jnx4+j−τ4+j+2n×∏s=0n−1as+j/8+bxj−τjxj−τj+2∑m=0s+j/8−1amas+2+j/8+bx2+j−τ2+jx2+j−τ2+j+2∑m=0s+2+j/8−1am×∏s=0n−1as+4+j/8+bx4+j−τ4+jx4+j−τ4+j+2∑m=0s+4+j/8−1amas+6+j/8+bx6+j−τ6+jx6+j−τ6+j+2∑m=0s+6+j/8−1am,where j=0,1,2,…,7.More explicitly, we have(35)x8n=x01−nx8n∏s=0n−1as+bx0x2∑m=0s−1amas+bx2x4∑m=0s−1amas+bx4x6∑m=0s−1amas+bx6x8∑m=0s−1am,(36)x8n+1=x11−nx9n∏s=0n−1as+bx1x3∑m=0s−1amas+bx3x5∑m=0s−1amas+bx5x7∑m=0s−1amas+bx7x9∑m=0s−1am,(37)x8n+2=x2x0nx8n∏s=0n−1as+bx2x4∑m=0s−1amas+bx4x6∑m=0s−1amas+bx6x8∑m=0s−1amas+1+bx0x2∑m=0sam,(38)x8n+3=x3x1nx9n∏s=0n−1as+bx3x5∑m=0s−1amas+bx5x7∑m=0s−1amas+bx7x9∑m=0s−1amas+1+bx1x3∑m=0sam,(39)x8n+4=x4x8nx0n∏s=0n−1as+bx4x6∑m=0s−1amas+bx6x8∑m=0s−1amas+1+bx0x2∑m=0samas+1+bx2x4∑m=0sam,(40)x8n+5=x5x9nx1n∏s=0n−1as+bx5x7∑m=0s−1amas+bx7x9∑m=0s−1amas+1+bx1x3∑m=0samas+1+bx3x5∑m=0sam,(41)x8n+6=x6x0nx8n∏s=0n−1as+bx6x8∑m=0s−1amas+1+bx0x2∑m=0samas+1+bx2x4∑m=0samas+1+bx4x6∑m=0sam,(42)x8n+7=x7x1nx9n∏s=0n−1as+bx7x9∑m=0s−1amas+1+bx1x3∑m=0samas+1+bx3x5∑m=0samas+1+bx5x7∑m=0sam.


## 3.2. Solutions Written in Unified Manner

One can write the solutions given in equations (17)–(24) in the following unified manner:(43)xn=Hnexp14∑s=0n−11+−1n−s+2cosn−sπ2lnwsws+2,where the Hn’s are such that H0=x0,H1=x1,H2=x2,H3=x3 and H4n+j=xjforj=0,1,2,3. Note that, here, the ws’s satisfy wn+8=awn+b, that is, they are given by w8n+j=wjan+b∑l=0n−1al.In fact, we have(44)x4n+j=Hjexp14∑s=04n+j−11+−1j−s+2cosj−sπ2lnwsws+2x4n+j=xj∏s=0n−1w4s+jw4s+2+j,0≤j≤3.Then from (44), we have(45)x8n+j=xj∏s=02n−1w4s+jw4s+2+j,0≤j≤7,=xj∏s=0n−1∏r=01w8s+4r+jw8s+4r+2+j.We have4r+j=24r+j/2+4r+j−τ4r+j so that, by (26),(46)w8s+4r+j=w24s+4r+j2+4r+j−τ4r+j=w4r+j−τ4r+ja4s+4r+j2+b∑m=04s+4r+j/2−1a4s+4r+j2−m−1=1x4r+j−τ4r+jx4r+j−τ4r+j+2a4s+4r+j2+b∑m=04s+4r+j/2−1a4s+4r+j2−m−1=a4s+4r+j/2x4r+j−τ4r+jx4r+j−τ4r+j+2+b∑m=04s+4r+j/2−1am=1x4r+j−τ4r+jx4r+j−τ4r+j+2a4s+4r+j2+bx4r+j−τ4r+jx4r+j−τ4r+j+2∑m=04s+4r+j/2−1am.Similarly, we get(47)w8s+4r+2+j=1x4r+2+j−τ4r+2+jx4r+2+j−τ4r+2+j+2a4s+4r+2+j2+bx4r+2+j−τ4r+2+jx4r+2+j−τ4r+2+j+2∑m=04s+4r+2+j/2−1am.So that the solution of (3), when l=8,k=2 and for constant coefficients, is given by(48)x8n+j=xj∏r=01x4r+2+j−τ4r+2+jnx4r+2+j−τ4r+2+j+2nx4r+j−τ4r+jnx4r+j−τ4r+j+2n×∏s=0n−1a4s+⌊4r+j/2⌋+bx4r+j−τ4r+jx4r+j−τ4r+j+2∑m=04s+⌊4r+j/2⌋−1ama4s+⌊4r+2+j/2⌋+bx4r+2+j−τ4r+2+jx4r+2+j−τ4r+2+j+2∑m=04s+⌊4r+2+j/2⌋−1am,which is the same as (34).


## 3.3. Existence of Periodic Solutions

Theorem 1.
Ifa,b≠0, xi=x4+i for 0≤i≤5, x0x2=x1x3=1−a/b, then the difference equation(49)xn+10=xnxn+2xn+8a+bxnxn+2,with initial conditions x0,x1,…,x8,x9 is periodic with period four.Proof.
Note that considering (35) and (39) subject to the condition x0=x4=x8, we have(50)x8n=x8n+4=x4for alln∈ℤ≥0.
Equations (36) and (40) together with x1=x5=x9 result in(51)x8n+1=x8n+5=x1for alln∈ℤ≥0.
Supposex0=x8 and x2=x6. Note that x0x2=1−a/b implies asbx0x2=as−as+1, i.e., as+1+bx0x2as=as so that(52)as+1+bx0x2as+bx0x2∑m=0s−1am=as+bx0x2∑m=0s−1am=as+bx6x8∑m=0s−1am,i.e., as+1+bx0x2∑m=0sam=as+bx6x8∑m=0s−1am. Thus x8n+2=x8n+6=x2 for all n∈ℤ≥0. Similarly, it can be shown that the conditions x3=x7,x1=x9 and x1x3=1−a/b imply(53)x8n+3=x8n+7=x3.
Hence the sequencexnn=0∞ satisfies(54)x8n+j=xjfor allj≥0,n≥0.
Sincex4+i=xi for all i=0,1,2,3,4,5, it follows that the sequence xnn=0∞ is periodic of period four. Therefore, the proof is complete.
We illustrate Theorem1 with Figure 1 where a=3, b=5, x0=x4=x8=1/13,x1=x5=x9=1/3,x2=x6=−26/5,x3=x7=−6/5.Figure 1
The plot ofxn+10=xnxn+2/xn+83+5xnxn+2.


## 4. The Casesl=5, k=1, an and bn Are Constants

### 4.1. Simplified Solutions

In this case, we havet=lcm2,5=10 so that t/2k=5. Hence, for an=a and bn=b constants, the solution, in terms of initial values x0,x1,x2,…,x5, is given (using (32)) by(55)x10n+j=xj∏s=0n−1∏r=04xj+2r+1−τj+2r+1xj+2r+1−τj+2r+1+1xj+2r−τj+2rxj+2r−τj+2r+1×a2s+j+2r/5+bxj+2r−τj+2rxj+2r−τj+2r+1∑m=02s+j+2r/5−1ama2s+j+2r+1/5+bxj+2r+1−τj+2r+1xj+2r+1−τj+2r+1+1∑m=02s+j+2r+1/5−1am=xjxj+1−τj+1xj+1−τj+1+1xj−τjxj−τj+1xj+3−τj+3xj+3−τj+3+1xj+2−τj+2xj+2−τj+2+1xj+5−τj+5xj+5−τj+5+1xj+4−τj+4xj+4−τj+4+1xj+7−τj+7xj+7−τj+7+1xj+6−τj+6xj+6−τj+6+1xj+9−τj+9xj+9−τj+9+1xj+8−τj+8xj+8−τj+8+1n×∏s=0n−1a2s+j/5+bxj−τjxj−τj+1∑m=02s+j/5−1ama2s+j+1/5+bxj+1−τj+1xj+1−τj+1+1∑m=02s+j+1/5−1ama2s+j+2/5+bxj+2−τj+2xj+2−τj+2+1∑m=02s+j+2/5−1ama2s+j+3/5+bxj+3−τj+3xj+3−τj+3+1∑m=02s+j+3/5−1ama2s+j+4/5+bxj+4−τj+4xj+4−τj+4+1∑m=02s+j+4/5−1ama2s+j+5/5+bxj+5−τj+5xj+5−τj+5+1∑m=02s+j+5/5−1ama2s+j+6/5+bxj+6−τj+6xj+6−τj+6+1∑m=02s+j+6/5−1ama2s+j+7/5+bxj+7−τj+7xj+7−τj+7+1∑m=02s+j+7/5−1am×a2s+j+8/5+bxj+8−τj+8xj+8−τj+8+1∑m=02s+j+8/5−1ama2s+j+9/5+bxj+9−τj+9xj+9−τj+9+1∑m=02s+j+9/5−1am,where j=0,1,2,…,9.By substituting the valuesj=0,1,2,3,…,9, we obtain the following more explicit description of the solution:(56)x10n=x0x1x2x0x1x3x4x2x3x0x1x4x5x2x3x1x2x4x5x3x4n×∏s=0n−1a2s+bx0x1∑m=02s−1ama2s+bx1x2∑m=02s−1ama2s+bx2x3∑m=02s−1ama2s+bx3x4∑m=02s−1ama2s+bx4x5∑m=02s−1ama2s+1+bx0x1∑m=02sama2s+1+bx1x2∑m=02sama2s+1+bx2x3∑m=02sam×a2s+1+bx3x4∑m=02sama2s+1+bx4x5∑m=02sam,i.e.(57)x10n=x0∏s=0n−1a2s+bx0x1∑l=02s−1ala2s+1+bx0x1∑m=02sama2s+bx2x3∑m=02s−1ama2s+1+bx2x3∑m=02sama2s+bx4x5∑m=02s−1ama2s+1+bx4x5∑m=02sama2s+1+bx1x2∑m=02sama2s+bx1x2∑m=02s−1am×a2s+1+bx3x4∑m=02sama2s+bx3x4∑m=02s−1am.We also have(58)x10n+1=x1x2x3x1x2x4x5x3x4x1x2x0x1x3x4x2x3x0x1x4x5n×∏s=0n−1a2s+bx1x2∑m=02s−1ama2s+bx2x3∑m=02s−1ama2s+bx3x4∑m=02s−1ama2s+bx4x5∑m=02s−1ama2s+1+bx0x1∑m=02sama2s+1+bx1x2∑m=02sam×a2s+1+bx2x3∑m=02sama2s+1+bx3x4∑m=02sama2s+1+bx4x5∑m=02sama2s+2+bx0x1∑m=02s+1am,i.e.,(59)x10n+1=x1∏s=0n−1a2s+bx1x2∑m=02s−1ama2s+1+bx1x2∑m=02sama2s+bx3x4∑m=02s−1ama2s+1+bx3x4∑m=02sama2s+1+bx0x1∑m=02sama2s+2+bx0x1∑m=02s+1am×a2s+1+bx2x3∑m=02sama2s+bx2x3∑m=02s−1ama2s+1+bx4x5∑m=02sama2s+bx4x5∑m=02s−1am.Similarly, we have(60)x10n+2=x2∏s=0n−1a2s+bx2x3∑m=02s−1ama2s+1+bx2x3∑m=02sama2s+bx4x5∑m=02s−1ama2s+1+bx4x5∑m=02sama2s+1+bx1x2∑m=02sama2s+2+bx1x2∑m=02s+1ama2s+1+bx3x4∑m=02sama2s+bx3x4∑m=02s+1am×a2s+2+bx0x1∑m=02s+1ama2s+1+bx0x1∑m=02sam,(61)x10n+3=x3∏s=0n−1a2s+bx3x4∑m=02s−1ama2s+1+bx3x4∑m=02sama2s+1+bx0x1∑m=02sama2s+2+bx0x1∑m=02s+1ama2s+1+bx2x3∑m=02sama2s+2+bx2x3∑m=02s+1ama2s+1+bx4x5∑m=02sama2s+bx4x5∑m=02s−1am×a2s+2+bx1x2∑m=02s+1ama2s+1+bx1x2∑m=02sam,(62)x10n+4=x4∏s=0n−1a2s+bx4x5∑m=02s−1ama2s+1+bx4x5∑m=02sama2s+1+bx1x2∑m=02sama2s+2+bx1x2∑m=02s+1ama2s+1+bx3x4∑m=02sama2s+2+bx3x4∑m=02s+1ama2s+2+bx0x1∑m=02s+1ama2s+1+bx0x1∑m=02sam×a2s+2+bx2x3∑m=02s+1ama2s+1+bx2x3∑m=02sam,(63)x10n+5=x5∏s=0n−1a2s+1+bx0x1∑m=02sama2s+2+bx0x1∑m=02s+1ama2s+1+bx2x3∑m=02sama2s+2+bx2x3∑m=02s+1ama2s+1+bx4x5∑m=02sama2s+2+bx4x5∑m=02s+1ama2s+2+bx1x2∑m=02s+1ama2s+1+bx1x2∑m=02sam×a2s+2+bx3x4∑m=02s+1ama2s+1+bx3x4∑m=02sam,(64)x10n+6=x0x1x5a+bx0x1∏s=0n−1a2s+1+bx1x2∑m=02sama2s+2+bx1x2∑m=02s+1ama2s+1+bx3x4∑m=02sama2s+2+bx3x4∑m=02s+1ama2s+2+bx0x1∑m=02s+1ama2s+3+bx0x1∑m=02s+2am×a2s+2+bx2x3∑m=02s+1ama2s+1+bx2x3∑m=02sama2s+2+bx4x5∑m=02s+1ama2s+1+bx4x5∑m=02sam,(65)x10n+7=x2x5a+bx0x1x0a+bx1x2∏s=0n−1a2s+1+bx2x3∑m=02sama2s+2+bx2x3∑m=02s+1ama2s+1+bx4x5∑m=02sama2s+2+bx4x5∑m=02s+1ama2s+2+bx1x2∑m=02s+1ama2s+3+bx1x2∑m=02s+2am×a2s+2+bx3x4∑m=02s+1ama2s+1+bx3x4∑m=02sama2s+3+bx0x1∑m=02s+2ama2s+2+bx0x1∑m=02s+1am,(66)x10n+8=x0x3a+bx1x2x5a+bx0x1a+bx2x3∏s=0n−1a2s+1+bx3x4∑m=02sama2s+2+bx3x4∑m=02s+1ama2s+2+bx0x1∑m=02s+1ama2s+3+bx0x1∑m=02s+2am×a2s+2+bx2x3∑m=02s+1ama2s+3+bx2x3∑m=02s+2ama2s+2+bx4x5∑m=02s+1ama2s+1+bx4x5∑m=02sama2s+3+bx1x2∑m=02s+2ama2s+2+bx1x2∑m=02s+1am,(67)x10n+9=x4x5a+bx0x1a+bx2x3x0a+bx1x2a+bx3x4∏s=0n−1a2s+1+bx4x5∑m=02sama2s+2+bx4x5∑m=02s+1ama2s+2+bx1x2∑m=02s+1ama2s+3+bx1x2∑m=02s+2am×a2s+2+bx3x4∑m=02s+1ama2s+3+bx3x4∑m=02s+2ama2s+3+bx0x1∑m=02s+2ama2s+2+bx0x1∑m=02s+1ama2s+3+bx2x3∑m=02s+2ama2s+2+bx2x3∑m=02s+1am.


#### 4.1.1. The Case Whena=−1 and b Constant

We investigate the case whenk=1, a=−1 and b constant. In this case, the solution is given by(68)x10n=x0−1+bx1x2n−1+bx3x4n−1+bx0x1n−1+bx2x3n−1+bx4x5n,(69)x10n+1=x1−1+bx2x3n−1+bx4x5n−1+bx0x1n−1+bx3x4n−1+bx1x2n,(70)x10n+2=x2−1+bx1x2n−1+bx3x4n−1+bx0x1n−1+bx2x3n−1+bx4x5n,(71)x10n+3=x3−1+bx2x3n−1+bx4x5n−1+bx0x1n−1+bx3x4n−1+bx1x2n,(72)x10n+4=x4−1+bx1x2n−1+bx3x4n−1+bx0x1n−1+bx2x3n−1+bx4x5n,(73)x10n+5=x5−1+bx2x3n−1+bx4x5n−1+bx0x1n−1+bx3x4n−1+bx1x2n,(74)x10n+6=x0x1x5−1+bx1x2n−1+bx3x4n−1+bx0x1n+1−1+bx2x3n−1+bx4x5n,(75)x10n+7=x2x5x0−1+bx2x3n−1+bx4x5n−1+bx0x1n+1−1+bx3x4n−1+bx1x2n+1,(76)x10n+8=x0x3x5−1+bx1x2n+1−1+bx3x4n−1+bx0x1n+1−1+bx2x3n+1−1+bx4x5n,(77)x10n+9=x4x5x0−1+bx2x3n+1−1+bx4x5n−1+bx0x1n+1−1+bx3x4n+1−1+bx1x2n+1.Corollary 1.
The solutionxn satisfies(78)x10n+jxj=x10n+j+2xj+2for allj≥0.The proof is clear by considering the equations (36)–(45).


### 4.2. Periodicity

Theorem 2.
Leta,b≠0. If xixi+1=1−a/b for all i=0,1,2,3,4, then the solution xnn=0∞ to the difference equation(79)xn+6=xnxn+1xn+5a+bxnxn+1,is such that x10n+j=xj for all n,j≥0. Furthermore, the solution is periodic with period two.Proof.
Note that fori=0,1,2,3,4, xixi+1=1−a/b implies 1−a=bxixi+1 so that a2s+r1−a=a2s+rbxixi+1, i.e., a2s+r=a2s+r+1+a2s+rbxixi+1. This implies(80)a2s+r+bxixi+1∑m=02s+r−1am=a2s+r+1+a2s+rbxixi+1+bxixi+1∑m=02s+r−1am=a2s+r+1+bxixi+1∑m=02s+ram.
Now settingr=0,1,2,3 and making an appropriate substitution in equations (26)–(35) shows that(81)x10n+j=xj∀j≥0,n≥0.
Furthermore,xixi+1=1−b/a yields x0=x2=x4,x1=x3=x5. From xn+6=xnxn+1/1+bxnxn+1, we have xn+6xn+5=xnxn+1/a+bxnxn+1 which implies that xjxj+1=1−b/a for j=5,6,7,8,9. Since every n can be written as 10m+j for some nonnegative integer m and j∈0,1,2,3,4…,9 and using the fact that x10m+jx10m+j+1=xjxj+1, it follows that(82)xnxn+1=1−bafor alln≥0.
Hencex2j=x0 and x2j+1=x1 for all j≥0. In this case, the sequence xnn=0∞ is periodic with period two.
Figure2 demonstrates the phenomenon. In the figure, we use a=9, b=13.
Figures3 and 4 demonstrate two examples of stable and unstable solutions.Figure 2
The graph ofxn+6=xnxn+1/xn+59+13xnxn+1 with x0=x2=x4=−4/7, x1=x3=x5=14/13.Figure 3
The graph ofxn+5=xnxn+3/xn+22+4xnxn+3 with x0=0.71,x1=0.88,x2=0.95,x3=0.3, x4=0.81.Figure 4
The graph ofxn+5=xnxn+1/xn+40.96+0.56xnxn+1 with x0=0.44,x1=0.19,x2=0.1,,x3=0.92,x4=0.25.


## 4.1. Simplified Solutions

In this case, we havet=lcm2,5=10 so that t/2k=5. Hence, for an=a and bn=b constants, the solution, in terms of initial values x0,x1,x2,…,x5, is given (using (32)) by(55)x10n+j=xj∏s=0n−1∏r=04xj+2r+1−τj+2r+1xj+2r+1−τj+2r+1+1xj+2r−τj+2rxj+2r−τj+2r+1×a2s+j+2r/5+bxj+2r−τj+2rxj+2r−τj+2r+1∑m=02s+j+2r/5−1ama2s+j+2r+1/5+bxj+2r+1−τj+2r+1xj+2r+1−τj+2r+1+1∑m=02s+j+2r+1/5−1am=xjxj+1−τj+1xj+1−τj+1+1xj−τjxj−τj+1xj+3−τj+3xj+3−τj+3+1xj+2−τj+2xj+2−τj+2+1xj+5−τj+5xj+5−τj+5+1xj+4−τj+4xj+4−τj+4+1xj+7−τj+7xj+7−τj+7+1xj+6−τj+6xj+6−τj+6+1xj+9−τj+9xj+9−τj+9+1xj+8−τj+8xj+8−τj+8+1n×∏s=0n−1a2s+j/5+bxj−τjxj−τj+1∑m=02s+j/5−1ama2s+j+1/5+bxj+1−τj+1xj+1−τj+1+1∑m=02s+j+1/5−1ama2s+j+2/5+bxj+2−τj+2xj+2−τj+2+1∑m=02s+j+2/5−1ama2s+j+3/5+bxj+3−τj+3xj+3−τj+3+1∑m=02s+j+3/5−1ama2s+j+4/5+bxj+4−τj+4xj+4−τj+4+1∑m=02s+j+4/5−1ama2s+j+5/5+bxj+5−τj+5xj+5−τj+5+1∑m=02s+j+5/5−1ama2s+j+6/5+bxj+6−τj+6xj+6−τj+6+1∑m=02s+j+6/5−1ama2s+j+7/5+bxj+7−τj+7xj+7−τj+7+1∑m=02s+j+7/5−1am×a2s+j+8/5+bxj+8−τj+8xj+8−τj+8+1∑m=02s+j+8/5−1ama2s+j+9/5+bxj+9−τj+9xj+9−τj+9+1∑m=02s+j+9/5−1am,where j=0,1,2,…,9.By substituting the valuesj=0,1,2,3,…,9, we obtain the following more explicit description of the solution:(56)x10n=x0x1x2x0x1x3x4x2x3x0x1x4x5x2x3x1x2x4x5x3x4n×∏s=0n−1a2s+bx0x1∑m=02s−1ama2s+bx1x2∑m=02s−1ama2s+bx2x3∑m=02s−1ama2s+bx3x4∑m=02s−1ama2s+bx4x5∑m=02s−1ama2s+1+bx0x1∑m=02sama2s+1+bx1x2∑m=02sama2s+1+bx2x3∑m=02sam×a2s+1+bx3x4∑m=02sama2s+1+bx4x5∑m=02sam,i.e.(57)x10n=x0∏s=0n−1a2s+bx0x1∑l=02s−1ala2s+1+bx0x1∑m=02sama2s+bx2x3∑m=02s−1ama2s+1+bx2x3∑m=02sama2s+bx4x5∑m=02s−1ama2s+1+bx4x5∑m=02sama2s+1+bx1x2∑m=02sama2s+bx1x2∑m=02s−1am×a2s+1+bx3x4∑m=02sama2s+bx3x4∑m=02s−1am.We also have(58)x10n+1=x1x2x3x1x2x4x5x3x4x1x2x0x1x3x4x2x3x0x1x4x5n×∏s=0n−1a2s+bx1x2∑m=02s−1ama2s+bx2x3∑m=02s−1ama2s+bx3x4∑m=02s−1ama2s+bx4x5∑m=02s−1ama2s+1+bx0x1∑m=02sama2s+1+bx1x2∑m=02sam×a2s+1+bx2x3∑m=02sama2s+1+bx3x4∑m=02sama2s+1+bx4x5∑m=02sama2s+2+bx0x1∑m=02s+1am,i.e.,(59)x10n+1=x1∏s=0n−1a2s+bx1x2∑m=02s−1ama2s+1+bx1x2∑m=02sama2s+bx3x4∑m=02s−1ama2s+1+bx3x4∑m=02sama2s+1+bx0x1∑m=02sama2s+2+bx0x1∑m=02s+1am×a2s+1+bx2x3∑m=02sama2s+bx2x3∑m=02s−1ama2s+1+bx4x5∑m=02sama2s+bx4x5∑m=02s−1am.Similarly, we have(60)x10n+2=x2∏s=0n−1a2s+bx2x3∑m=02s−1ama2s+1+bx2x3∑m=02sama2s+bx4x5∑m=02s−1ama2s+1+bx4x5∑m=02sama2s+1+bx1x2∑m=02sama2s+2+bx1x2∑m=02s+1ama2s+1+bx3x4∑m=02sama2s+bx3x4∑m=02s+1am×a2s+2+bx0x1∑m=02s+1ama2s+1+bx0x1∑m=02sam,(61)x10n+3=x3∏s=0n−1a2s+bx3x4∑m=02s−1ama2s+1+bx3x4∑m=02sama2s+1+bx0x1∑m=02sama2s+2+bx0x1∑m=02s+1ama2s+1+bx2x3∑m=02sama2s+2+bx2x3∑m=02s+1ama2s+1+bx4x5∑m=02sama2s+bx4x5∑m=02s−1am×a2s+2+bx1x2∑m=02s+1ama2s+1+bx1x2∑m=02sam,(62)x10n+4=x4∏s=0n−1a2s+bx4x5∑m=02s−1ama2s+1+bx4x5∑m=02sama2s+1+bx1x2∑m=02sama2s+2+bx1x2∑m=02s+1ama2s+1+bx3x4∑m=02sama2s+2+bx3x4∑m=02s+1ama2s+2+bx0x1∑m=02s+1ama2s+1+bx0x1∑m=02sam×a2s+2+bx2x3∑m=02s+1ama2s+1+bx2x3∑m=02sam,(63)x10n+5=x5∏s=0n−1a2s+1+bx0x1∑m=02sama2s+2+bx0x1∑m=02s+1ama2s+1+bx2x3∑m=02sama2s+2+bx2x3∑m=02s+1ama2s+1+bx4x5∑m=02sama2s+2+bx4x5∑m=02s+1ama2s+2+bx1x2∑m=02s+1ama2s+1+bx1x2∑m=02sam×a2s+2+bx3x4∑m=02s+1ama2s+1+bx3x4∑m=02sam,(64)x10n+6=x0x1x5a+bx0x1∏s=0n−1a2s+1+bx1x2∑m=02sama2s+2+bx1x2∑m=02s+1ama2s+1+bx3x4∑m=02sama2s+2+bx3x4∑m=02s+1ama2s+2+bx0x1∑m=02s+1ama2s+3+bx0x1∑m=02s+2am×a2s+2+bx2x3∑m=02s+1ama2s+1+bx2x3∑m=02sama2s+2+bx4x5∑m=02s+1ama2s+1+bx4x5∑m=02sam,(65)x10n+7=x2x5a+bx0x1x0a+bx1x2∏s=0n−1a2s+1+bx2x3∑m=02sama2s+2+bx2x3∑m=02s+1ama2s+1+bx4x5∑m=02sama2s+2+bx4x5∑m=02s+1ama2s+2+bx1x2∑m=02s+1ama2s+3+bx1x2∑m=02s+2am×a2s+2+bx3x4∑m=02s+1ama2s+1+bx3x4∑m=02sama2s+3+bx0x1∑m=02s+2ama2s+2+bx0x1∑m=02s+1am,(66)x10n+8=x0x3a+bx1x2x5a+bx0x1a+bx2x3∏s=0n−1a2s+1+bx3x4∑m=02sama2s+2+bx3x4∑m=02s+1ama2s+2+bx0x1∑m=02s+1ama2s+3+bx0x1∑m=02s+2am×a2s+2+bx2x3∑m=02s+1ama2s+3+bx2x3∑m=02s+2ama2s+2+bx4x5∑m=02s+1ama2s+1+bx4x5∑m=02sama2s+3+bx1x2∑m=02s+2ama2s+2+bx1x2∑m=02s+1am,(67)x10n+9=x4x5a+bx0x1a+bx2x3x0a+bx1x2a+bx3x4∏s=0n−1a2s+1+bx4x5∑m=02sama2s+2+bx4x5∑m=02s+1ama2s+2+bx1x2∑m=02s+1ama2s+3+bx1x2∑m=02s+2am×a2s+2+bx3x4∑m=02s+1ama2s+3+bx3x4∑m=02s+2ama2s+3+bx0x1∑m=02s+2ama2s+2+bx0x1∑m=02s+1ama2s+3+bx2x3∑m=02s+2ama2s+2+bx2x3∑m=02s+1am.


### 4.1.1. The Case Whena=−1 and b Constant

We investigate the case whenk=1, a=−1 and b constant. In this case, the solution is given by(68)x10n=x0−1+bx1x2n−1+bx3x4n−1+bx0x1n−1+bx2x3n−1+bx4x5n,(69)x10n+1=x1−1+bx2x3n−1+bx4x5n−1+bx0x1n−1+bx3x4n−1+bx1x2n,(70)x10n+2=x2−1+bx1x2n−1+bx3x4n−1+bx0x1n−1+bx2x3n−1+bx4x5n,(71)x10n+3=x3−1+bx2x3n−1+bx4x5n−1+bx0x1n−1+bx3x4n−1+bx1x2n,(72)x10n+4=x4−1+bx1x2n−1+bx3x4n−1+bx0x1n−1+bx2x3n−1+bx4x5n,(73)x10n+5=x5−1+bx2x3n−1+bx4x5n−1+bx0x1n−1+bx3x4n−1+bx1x2n,(74)x10n+6=x0x1x5−1+bx1x2n−1+bx3x4n−1+bx0x1n+1−1+bx2x3n−1+bx4x5n,(75)x10n+7=x2x5x0−1+bx2x3n−1+bx4x5n−1+bx0x1n+1−1+bx3x4n−1+bx1x2n+1,(76)x10n+8=x0x3x5−1+bx1x2n+1−1+bx3x4n−1+bx0x1n+1−1+bx2x3n+1−1+bx4x5n,(77)x10n+9=x4x5x0−1+bx2x3n+1−1+bx4x5n−1+bx0x1n+1−1+bx3x4n+1−1+bx1x2n+1.Corollary 1.
The solutionxn satisfies(78)x10n+jxj=x10n+j+2xj+2for allj≥0.The proof is clear by considering the equations (36)–(45).


## 4.1.1. The Case Whena=−1 and b Constant

We investigate the case whenk=1, a=−1 and b constant. In this case, the solution is given by(68)x10n=x0−1+bx1x2n−1+bx3x4n−1+bx0x1n−1+bx2x3n−1+bx4x5n,(69)x10n+1=x1−1+bx2x3n−1+bx4x5n−1+bx0x1n−1+bx3x4n−1+bx1x2n,(70)x10n+2=x2−1+bx1x2n−1+bx3x4n−1+bx0x1n−1+bx2x3n−1+bx4x5n,(71)x10n+3=x3−1+bx2x3n−1+bx4x5n−1+bx0x1n−1+bx3x4n−1+bx1x2n,(72)x10n+4=x4−1+bx1x2n−1+bx3x4n−1+bx0x1n−1+bx2x3n−1+bx4x5n,(73)x10n+5=x5−1+bx2x3n−1+bx4x5n−1+bx0x1n−1+bx3x4n−1+bx1x2n,(74)x10n+6=x0x1x5−1+bx1x2n−1+bx3x4n−1+bx0x1n+1−1+bx2x3n−1+bx4x5n,(75)x10n+7=x2x5x0−1+bx2x3n−1+bx4x5n−1+bx0x1n+1−1+bx3x4n−1+bx1x2n+1,(76)x10n+8=x0x3x5−1+bx1x2n+1−1+bx3x4n−1+bx0x1n+1−1+bx2x3n+1−1+bx4x5n,(77)x10n+9=x4x5x0−1+bx2x3n+1−1+bx4x5n−1+bx0x1n+1−1+bx3x4n+1−1+bx1x2n+1.Corollary 1.
The solutionxn satisfies(78)x10n+jxj=x10n+j+2xj+2for allj≥0.The proof is clear by considering the equations (36)–(45).


## 4.2. Periodicity

Theorem 2.
Leta,b≠0. If xixi+1=1−a/b for all i=0,1,2,3,4, then the solution xnn=0∞ to the difference equation(79)xn+6=xnxn+1xn+5a+bxnxn+1,is such that x10n+j=xj for all n,j≥0. Furthermore, the solution is periodic with period two.Proof.
Note that fori=0,1,2,3,4, xixi+1=1−a/b implies 1−a=bxixi+1 so that a2s+r1−a=a2s+rbxixi+1, i.e., a2s+r=a2s+r+1+a2s+rbxixi+1. This implies(80)a2s+r+bxixi+1∑m=02s+r−1am=a2s+r+1+a2s+rbxixi+1+bxixi+1∑m=02s+r−1am=a2s+r+1+bxixi+1∑m=02s+ram.
Now settingr=0,1,2,3 and making an appropriate substitution in equations (26)–(35) shows that(81)x10n+j=xj∀j≥0,n≥0.
Furthermore,xixi+1=1−b/a yields x0=x2=x4,x1=x3=x5. From xn+6=xnxn+1/1+bxnxn+1, we have xn+6xn+5=xnxn+1/a+bxnxn+1 which implies that xjxj+1=1−b/a for j=5,6,7,8,9. Since every n can be written as 10m+j for some nonnegative integer m and j∈0,1,2,3,4…,9 and using the fact that x10m+jx10m+j+1=xjxj+1, it follows that(82)xnxn+1=1−bafor alln≥0.
Hencex2j=x0 and x2j+1=x1 for all j≥0. In this case, the sequence xnn=0∞ is periodic with period two.
Figure2 demonstrates the phenomenon. In the figure, we use a=9, b=13.
Figures3 and 4 demonstrate two examples of stable and unstable solutions.Figure 2
The graph ofxn+6=xnxn+1/xn+59+13xnxn+1 with x0=x2=x4=−4/7, x1=x3=x5=14/13.Figure 3
The graph ofxn+5=xnxn+3/xn+22+4xnxn+3 with x0=0.71,x1=0.88,x2=0.95,x3=0.3, x4=0.81.Figure 4
The graph ofxn+5=xnxn+1/xn+40.96+0.56xnxn+1 with x0=0.44,x1=0.19,x2=0.1,,x3=0.92,x4=0.25.


## 5. Results and Discussions

This paper is a generalization of some work in [8, 9, 12–17]. In fact, substituting l=1, k=2, a=1 and b=±1 in (33), we obtain results in Theorems 1 and 3 by Ibrahim in [15]. Substituting l=1, k=2, an=a and bn=b in (32), we obtain the results in [12]. Substituting l=2, k=2, an=±1 and bn=±1 in (33), we obtain results in Theorems 1, 4, 6, and 9 in [8]. Substituting l=3, k=2, an=±1 and bn=±1 in (15), we obtain Theorems 1, 3, 6, and 9 by Elsayed in [9]. Substituting l=2, k=6, a=±1 and b=±1 in (15), we obtain Theorems 3.4 and 4.4 in [7]. Substituting l=2, k=1, an=±1 and bn=±1 in (15), we obtain Theorems 1, 3, 6, and 8 in [10]. Substituting l=3 and k=1 in (2), we obtain the results in [11]. Substituting l=4 and k=2 in (2), we obtain the results in [17].Lastly, it is worth indicating that some of the results in [8, 9, 15] have been studied in a more general setting by Folly-Gbetoula and Nyirenda in [12–14, 16].


## 6. Conclusion

In this paper, we have studied a class of higher order difference equations via a geometric method and symmetry analysis. Ultimately, explicit solutions have been obtained. We have shown that the equation investigated has periodic solutions under some restrictions.



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