J. Math. Computer Sci., 32 (2024), 318–331 Online: ISSN 2008-949X Journal Homepage: www.isr-publications.com/jmcs A detailed study of a class of recurrence equations with a generalized order Jollet Truth Kubayi, Mensah Folly-Gbetoula∗ School of Mathematics, University of the Witwatersrand, Johannesburg 2050, South Africa. Abstract In this paper, we study some family of difference equations. The study involves the use of symmetries to find exact solutions of difference equations with the aim of extending the studies that have been done in the literature. We also investigate the periodic nature and behavior of the solutions in some cases. Finally, some numerical examples illustrating our findings are presented. Keywords: Difference equation, symmetry, reduction, exact solution. 2020 MSC: 39A10, 39A99, 39A13. ©2024 All rights reserved. 1. Introduction During the nineteenth century, a prominent Norwegian mathematician, Sophus Lie (1842-1899) es- tablished remarkable work that became an important part to the theory of groups of transformations (continuous) that leave a differential equation invariant [14]. Lie aimed to create a theory of integrating ordinary differential equations that is equivalent to the Abelian theory of computing algebraic equations. He was inspired by Abel and Galois’ theory. He observed that the procedure in all exceptional cases of a universal integration on differential equations is centered on the invariance of the differential equa- tion under continuous symmetries. It is important to note that Lie’s group analysis classifies ordinary differential equations in terms of the symmetry group associated with them. Shigeru Maeda in 1987 showed that Lie’s method can be extended to also solve ordinary difference equations. He showed that a set of functional equations amounted from the linearized symmetry condi- tion of ordinary difference equations [15]. The philosophy of difference equations and their applications have cemented a central importance in applicable analysis. Later, several authors studied ordinary dif- ference equations and have obtained some interesting results, see [1–6, 8–13, 16, 17]. Maeda [15] showed how to use symmetry methods to obtain the solution of a system of first-order difference equations. It is now known that symmetries can be used to solve higher-order difference equations. ∗Corresponding author Email addresses: 710005@students.wits.ac.za (Jollet Truth Kubayi), Mensah.Folly-Gbetoula@wits.ac.za (Mensah Folly-Gbetoula) doi: 10.22436/jmcs.032.04.03 Received: 2023-02-07 Revised: 2023-07-27 Accepted: 2023-08-25 http://dx.doi.org/10.22436/jmcs.032.04.03 http://dx.doi.org/10.22436/jmcs.032.04.03 http://crossmark.crossref.org/dialog/?doi=10.22436/jmcs.032.04.03&domain=pdf J. T. Kubayi, M. Folly-Gbetoula, J. Math. Computer Sci., 32 (2024), 318–331 319 In [18], the authors investigated the solutions of the fifth-order difference equation xn+1 = xn−2xn−3xn−4 xnxn−1(±1 +±xn−2xn−3xn−4) ,n ∈N0. (1.1) In [10], the authors investigated the solutions and behavior of solutions of the difference equation xn+1 = xnxn−2xn−4 xn−1xn−3 (λ+ µxnxn−2xn−4) , (1.2) where λ and µ are real constants. In [7], the authors investigated the solutions and the properties of the difference equation xn+1 = xn−3kxn−4kxn−5k xn−kxn−2k (±1± xn−3kxn−4kxn−5k) . (1.3) We show in this paper that the left-hand side in the above equation is xn+1 and not xn as seen in [7]. Clearly, equations (1.1)-(1.3) are all special cases of xn+1 = xn−3kxn−4kxn−5k xn−kxn−2k (an + bnxn−3kxn−4kxn−5k) , (1.4) for some arbitrary real sequences an and bn. A symmetry based method will be employed to solve the generalized case (1.4) and compare the solu- tions of the corresponding special cases to those of [7, 10, 18]. To achieve this, for the sake of definitions, we will derive the solutions of the equivalent difference equation xn+5k = xnxn+kxn+2k xn+3kxn+4k (An +Bnxnxn+kxn+2k) , (1.5) where (An)n∈N and (Bn)n∈N are non-zero real random sequences, using Lie group analysis technique. Eventually, invariants of (1.5) are derived and a relationship between these invariants and the similarity variables is given. The paper is organized in the following manner. In Section 2, we revise some essential ideas that are required for computing symmetries of difference equations and order reduction. In Section 3, symmetries and solutions of (1.5) are obtained and a detailed analysis of some special cases is conducted. In Section 4, we study the periodicity and behavior of the solutions of (1.5). 2. Definitions and notation The definitions and notations in this paper are similar to those that Hydon adopted in [13]. We consider the general form of the ordinary difference equation xn+5k = ω (n, xn, xn+k, xn+2k, xn+3k, xn+4k) , (2.1) for some function ω with k ∈N. Definition 2.1. We define S to be the shift operator acting on n as S : n→ n+ 1. Consider a one-parameter Lie group of point transformations given below Ψε : (n, xn)→ (n, xn + εξ (n, xn)) , (2.2) for the continuous characteristic function ξ = ξ(n, xn). It is known that the action of the Lie group can be recovered from the corresponding infinitesimal generators. J. T. Kubayi, M. Folly-Gbetoula, J. Math. Computer Sci., 32 (2024), 318–331 320 Definition 2.2. The symmetry generator, denoted by X, is given by X = ξ(n, xn) ∂ ∂xn . (2.3) The linearized symmetry condition [13] is given by S5kξ− X̂ω = 0, (2.4) provided (2.1) holds. Note that X̂ denotes the prolongation of X to all shifts of xn appearing in the equation and is given by X̂ = ξ ∂ ∂xn + Skξ ∂ ∂xn+k + · · ·+ S4kξ ∂ ∂xn+4k . (2.5) Definition 2.3. A function v is an invariant under the group of transformation (2.2) if and only if Xv = 0. The generator in (2.3) can be used to derive the canonical coordinate which in turn can be used to obtain the invariant functions. The method for finding symmetries is explained at length in [13]. 3. Symmetry analysis and exact solutions Consider the difference equation (1.5). So, in this case, the function ω is given by ω = xnxn+kxn+2k xn+3kxn+4k (An +Bnxnxn+kxn+2k) . Assuming that the prolonged symmetry generator takes the form in (2.5), the linearized symmetry con- dition (2.4) on (1.5) gives ξ(n+ 5k, xn+5k) − ξ(n, xn) ∂ω ∂xn − ξ(n+ k, xn+k) ∂ω ∂xn+k − ξ(n+ 2k, xn+2k) ∂ω ∂xn+2k − ξ(n+ 3k, xn+3k) ∂ω ∂xn+3k − ξ(n+ 4k, xn+4k) ∂ω ∂xn+4k = 0, that is to say, ξ(n+ 5k, xn+5k) − ξ(n, xn)Anxn+kxn+2k xn+3kxn+4k(An +Bnxnxn+kxn+2k)2 − ξ(n+ k, xn+k)Anxnxn+2k xn+3kxn+4k(An +Bnxnxn+kxn+2k)2 + ξ(n+ 3k, xn+3k)xnxn+kxn+2k x2 n+3kxn+4k(An +Bnxnxn+2xn+2k) + ξ(n+ 4k, xn+4k)xnxn+kxn+2k xn+3kx 2 n+4k(An +Bnxnxn+kxn+2k) − ξ(n+ 2k, xn+2k)Anxn+k xn+3kxn+4k(An +Bnxnxn+kxn+2k)2 = 0. (3.1) We apply the operator ∂ ∂xn + Anxn+3k xn(An+Bnxnxn+kxn+2k) ∂ ∂xn+3k on (3.1). After clearing fractions and then dif- ferentiating thrice with respect to xn, we obtain the following: 2(An + 2Bnxnxn+kxn+2k)ξ (3)(n, xn) + (An +Bnxnxn+kxn+2k)xnξ (4)(n, xn) = 0. Now we separate the above, since ξ depends only on xn, to get xn+kxn+2kxn+3k : xnξ (4)(n, xn) + 4ξ(3)(n, xn) = 0, xn+3k : xnξ (4)(n, xn) + 2ξ(3)(n, xn) = 0, whose solution is given by ξ(n, xn) = βnx2 n + γnxn + λn (3.2) J. T. Kubayi, M. Folly-Gbetoula, J. Math. Computer Sci., 32 (2024), 318–331 321 for some functions βn, γn, and λn of n. Next, we substitute (3.2) into (3.1) and then separate the resulting equation by the coefficients of products of shifts of un and then setting them to zero. It turns out that λn = βn = 0 and γn must satisfy the following linear difference equation: γn + γn+k + γn+2k = 0. Solving the above equation yields γn1(m) = exp { i ( − 2nπ 3k + 2mπn k )} , γn2(m) = exp { i ( 2nπ 3k + 2mπn k )} , where m = 0, 1, . . . , k− 1. From (3.2), we have the characteristics ξ1 = γn1(m)xn and ξ2 = γn2(m)xn, m = 0, 1, . . . , k− 1. Hence, we obtain the following 2k symmetries: X1m = γn1(m)xn ∂ ∂xn and X2m = γn2(m)xn ∂ ∂xn , m = 0, 1, . . . , k− 1. The canonical coordinate that linearizes (1.5) is given by Sn = ∫ dxn ξ(n, xn) = 1 γn ln |xn| and the function given by Ṽn = γnSn + γn+kSn+k + γn+2kSn+2k is invariant under the group of transformations admitted by (1.5) since X̂(Ṽn) = 0. For the sake of convenience, we will use the invariant Vn = exp (−Ṽn) and one can easily verify that X̂(Vn) = 0. It happens that, Vn = 1 xnxn+kxn+2k (3.3) and applying the forward shift of 3k on Vn (and substituting xn+5k by its expression given in (1.5)) yields Vn+3k = AnVn +Bn. (3.4) By iterating (3.4), we obtain its solution in closed form V3kn+i = Vi  n−1∏ m1=0 A3km1+i + n−1∑ l=0 B3kl+i n−1∏ m2=l+1 A3km2+i  , (3.5) where i = 0, 1, 2, . . . , 3k− 1. It follows from (3.3) that xn+3k = Vn Vn+k xn and by iterating the above equation, we have that x3kn+i = xi ( n−1∏ s=0 V3ks+i V3ks+i+k ) , (3.6) where i = 0, 1, 2, . . . , 3k− 1. To avoid any possible confusion, we rewrite (3.6) in the following forms: x3kn+i = xi ( n−1∏ s=0 V3ks+i V3ks+i+k ) , i = 0, . . . , 2k− 1 (3.7) J. T. Kubayi, M. Folly-Gbetoula, J. Math. Computer Sci., 32 (2024), 318–331 322 and x3kn+i = xi ( n−1∏ s=0 V3ks+i V3k(s+1)+i−2k ) , i = 2k, . . . , 3k− 1. (3.8) Using (3.5) in (3.7) and (3.8), remembering that Vi = 1/(xixi+kxi+2k), we have the following solutions of (1.5): x3kn+i = xni+3k xn−1 i n−1∏ s=0 ( s−1∏ m1=0 A3km1+i ) + xixi+kxi+2k s−1∑ l=0 ( B3kl+i s−1∏ m2=l+1 A3km2+i ) ( s−1∏ m1=0 A3km1+k+i ) + xi+kxi+2kxi+3k s−1∑ l=0 ( B3kl+k+i s−1∏ m2=l+1 A3km2+k+i ) , (3.9) where i = 0, 1, 2, . . . , 2k− 1; and x3kn+i = xix n i−2kx n i−k xni+kx n i+2k × n−1∏ s=0 ( s−1∏ m1=0 A3km1+i ) + xixi+kxi+2k s−1∑ l=0 ( B3kl+i s−1∏ m2=l+1 A3km2+i ) ( s∏ m1=0 A3km1+i−2k ) + xi−2kxi−kxi s∑ l=0 ( B3kl+i−2k s∏ m2=l+1 A3km2+i−2k ) , (3.10) where i = 2k, 2k+ 1, . . . , 3k− 1. We derive the solution of (1.4) by back shifting (3.9) and (3.10) 5k− 1 times. This yields x3kn−5k+1+i = xni−2k+1 xn−1 i−5k+1 n−1∏ s=0 ( s−1∏ m1=0 a3km1+i ) + xi−5k+1xi−4k+1xi−3k+1 s−1∑ l=0 ( b3kl+i s−1∏ m2=l+1 a3km2+i ) ( s−1∏ m1=0 a3km1+k+i ) + xi−4k+1xi−3k+1xi−2k+1 s−1∑ l=0 ( b3kl+k+i s−1∏ m2=l+1 a3km2+k+i ) , (3.11) for i = 0, 1, 2, . . . , 2k− 1; and x3kn−5k+1+i = xi−5k+1x n i−7k+1x n i−6k+1 xni−4k+1x n i−3k+1 × n−1∏ s=0 ( s−1∏ m1=0 a3km1+i ) + xi−5k+1xi−4k+1xi−3k+1 s−1∑ l=0 ( b3kl+i s−1∏ m2=l+1 a3km2+i ) ( s∏ m1=0 a3km1+i−2k ) + xi−7k+1xi−6k+1xi−5k+1 s∑ l=0 ( b3kl+i−2k s∏ m2=l+1 a3km2+i−2k ) , (3.12) for i = 2k, 2k− 1, . . . , 3k− 1. 3.1. The case where an and bn are constant Here, let an = a and bn = b, where a,b ∈ R. Thus, equations (3.11) and (3.12) reduce to x3kn−5k+i+1 = xni−2k+1 xn−1 i−5k+1 n−1∏ s=0 as + xi−5k+1xi−4k+1xi−3k+1b s−1∑ l=0 al as + xi−4k+1xi−3k+1xi−2k+1b s−1∑ l=0 al , J. T. Kubayi, M. Folly-Gbetoula, J. Math. Computer Sci., 32 (2024), 318–331 323 for i = 0, 1, 2, . . . , 2k− 1; and x3kn−5k+i+1 = xi−5k+1x n i−7k+1x n i−6k+1 xni−4k+1x n i−3k+1 n−1∏ s=0 as + xi−5k+1xi−4k+1xi−3k+1b s−1∑ l=0 al as+1 + xi−7k+1xi−6k+1xi−5k+1b s∑ l=0 al , for i = 2k, 2k+ 1, . . . , 3k− 1. 3.1.1. The case where a = 1 We have x3kn−5k+i+1 = xni−2k+1 xn−1 i−5k+1 n−1∏ s=0 1 + xi−5k+1xi−4k+1xi−3k+1bs 1 + xi−4k+1xi−3k+1xi−2k+1bs , (3.13) for i = 0, 1, 2, . . . , 2k− 1; and x3kn−5k+i+1 = xi−5k+1x n i−7k+1x n i−6k+1 xni−4k+1x n i−3k+1 n−1∏ s=0 1 + xi−5k+1xi−4k+1xi−3k+1bs 1 + xi−7k+1xi−6k+1xi−5k+1b(s+ 1) , (3.14) for i = 2k, 2k+ 1, . . . , 3k− 1. Remark 3.1. The results in [7] (see Theorems 2.1.1 and 2.2.1) are special cases of ours. In fact, x3kn+i = x3k(n+1)−5k+2k+i, i = 1, 2, . . . , k = xi−3kx n+1 i−5kx n+1 i−4k xn+1 i−2kx n+1 i−k n∏ s=0 1 + xi−3kxi−2kxi−kbs 1 + xi−5kxi−4kxi−3kb(s+ 1) = xi−3k n∏ s=0 xi−5kxi−4k + xi−5kxi−4kxi−3kxi−2kxi−kbs xi−2kxi−k + xi−5kxi−4kxi−3kxi−2kxi−kb(s+ 1) and similarly, x3kn+i = x3k(n+2)−5k−k+i, i = k+ 1, 2, . . . , 3k− 1 = xn+2 i−3k xn+1 i−6k n+1∏ s=0 1 + xi−6kxi−5kxi−4kbs 1 + xi−5kxi−4kxi−3kbs = xi−3k n∏ s=0 xi−3k + xi−6kxi−5kxi−4kxi−3kb(s+ 1) xi−6k + xi−6kxi−5kxi−4kxi−3kb(s+ 1) . Consequently, Corollaries 3.1.1 and 3.2.1 are easily recovered from (3.13) and (3.14) by setting k = 2. 3.1.2. The case where a 6= 1 We have x3kn−5k+i+1 = xni−2k+1 xn−1 i−5k+1 n−1∏ s=0 as + xi−5k+1xi−4k+1xi−3k+1b (1−as 1−a ) as + xi−4k+1xi−3k+1xi−2k+1b (1−as 1−a ) , for i = 0, 1, 2, . . . , 2k− 1; and x3kn−5k+i+1 = xi−5k+1x n i−7k+1x n i−6k+1 xni−4k+1x n i−3k+1 n−1∏ s=0 as + xi−5k+1xi−4k+1xi−3k+1b (1−as 1−a ) as+1 + xi−7k+1xi−6k+1xi−5k+1b ( 1−as+1 1−a ) , for i = 2k, 2k+ 1, . . . , 3k− 1. J. T. Kubayi, M. Folly-Gbetoula, J. Math. Computer Sci., 32 (2024), 318–331 324 3.1.3. The case where k = 1 Assume an = λ and bn = µ, where λ,µ ∈ R. Thus, equations (3.11) and (3.12) simplify to x3n−4 = xn−1 xn−1 −4 n−1∏ s=0 λs + x−4x−3x−2µ s−1∑ l=0 λl λs + x−3x−2x−1µ s−1∑ l=0 λl , x3n−3 = xn0 xn−1 −3 n−1∏ s=0 λs + x−3x−2x−1µ s−1∑ l=0 λl λs + x−2x−1x0µ s−1∑ l=0 λl , x3n−2 = x−2x n −4x n −3 xn−1x n 0 n−1∏ s=0 λs + x−2x−1x0µ s−1∑ l=0 λl λs+1 + x−4x−3x−2µ s∑ l=0 λl . For λ = 1, using equations (3.11) and (3.12), we have that x3n−4 = xn−1 xn−1 −4 n−1∏ s=0 1 + x−4x−3x−2µs 1 + x−3x−2x−1µs , x3n−3 = xn0 xn−1 −3 n−1∏ s=0 1 + x−3x−2x−1µs 1 + x−2x−1x0µs , x3n−2 = x−2x n −4x n −3 xn−1x n 0 n−1∏ s=0 1 + x−2x−1x0µs 1 + x−4x−3x−2µ(s+ 1) . (3.15) The results in (3.15) were obtained by Yazlik in [18] (see Theorems 5 and 9). For λ 6= 1, equations (3.11) and (3.12) become x3n−4 = xn−1 xn−1 −4 n−1∏ s=0 λs + x−4x−3x−2µ (1−λs 1−λ ) λs + x−3x−2x−1µ (1−λs 1−λ ) , x3n−3 = xn0 xn−1 −3 n−1∏ s=0 λs + x−3x−2x−1µ (1−λs 1−λ ) l λs + x−2x−1x0µ (1−λs 1−λ ) , x3n−2 = x−2x n −4x n −3 xn−1x n 0 n−1∏ s=0 λs + x−2x−1x0µ (1−λs 1−λ ) λs+1 + x−4x−3x−2µ ( 1−λs+1 1−λ ) . In particular, when λ = −1, we have x6n−4 = x2n −1 x2n−1 −4 ( −1 + x−4x−3x−2µ −1 + x−3x−2x−1µ )n , x6n−3 = x2n 0 x2n−1 −3 ( −1 + x−3x−2x−1µ −1 + x−2x−1x0µ )n , x6n−2 = x−2x 2n −4x 2n −3 x2n −1x 2n 0 ( −1 + x−2x−1x0µ −1 + x−4x−3x−2µ )n , x6n−1 = x2n+1 −1 x2n −4 ( −1 + x−4x−3x−2µ −1 + x−3x−2x−1µ )n , x6n = x2n+1 0 x2n −3 ( −1 + x−3x−2x−1µ −1 + x−2x−1x0µ )n , x6n+1 = x−2x 2n+1 −4 x2n+1 −3 x2n+1 −1 x2n+1 0 (−1 + x−2x−1x0µ) n (−1 + x−4x−3x−2µ)n+1 . (3.16) The results in (3.16) were obtained by Yazlik in [18] (see Theorems 7 and 11). Also, by replacing n with 2n or 2n+ 1 in (3.11) and (3.12), we recover the results in equation (67) of [10]. J. T. Kubayi, M. Folly-Gbetoula, J. Math. Computer Sci., 32 (2024), 318–331 325 3.2. The case when k = 2 When k = 2, thanks to (3.11) and (3.12), the solution of xn+1 = xn−6xn−8xn−10 xn−2xn−4 (an + bnxn−6xn−8xn−10) is given by x6n−9+i = xni−3 xn−1 i−9 n−1∏ s=0 ( s−1∏ m1=0 a6m1+i ) + xi−9xi−7xi−5 s−1∑ l=0 ( b6l+i s−1∏ m2=l+1 a6m2+i ) ( s−1∏ m1=0 a6m1+2+i ) + xi−7xi−5xi−3 s−1∑ l=0 ( b6l+2+i s−1∏ m2=l+1 a6m2+2+i ) , for i = 0, 1, 2, 3; and x6n−9+i = xi−9x n i−13x n i−11 xni−7x n i−5 n−1∏ s=0 ( s−1∏ m1=0 a6m1+i ) + xi−9xi−7xi−5 s−1∑ l=0 ( b6l+i s−1∏ m2=l+1 a6m2+i ) ( s∏ m1=0 a6m1+i−4 ) + xi−13xi−11xi−9 s∑ l=0 ( b6l+i−4 s∏ m2=l+1 a6m2+i−4 ) , for i = 4, 5. 3.2.1. The case where an = a and bn = b are constant The case where a = 1 we have x6n−9+i = xni−3 xn−1 i−9 n−1∏ s=0 1 + xi−9xi−7xi−5bs 1 + xi−7xi−5xi−3bs , for i = 0, 1, 2, 3; and x6n−9+i = xi−9x n i−13x n i−11 xni−7x n i−5 n−1∏ s=0 1 + xi−9xi−7xi−5bs 1 + xi−13xi−11xi−9b(s+ 1) . for i = 4, 5. More explicitly, x6n−9 = xn−3 xn−1 −9 n−1∏ s=0 1 + x−9x−7x−5bs 1 + x−7x−5x−3bs , x6n−8 = xn−2 xn−1 −8 n−1∏ s=0 1 + x−8x−6x−4bs 1 + x−6x−4x−2bs , x6n−7 = xn−1 xn−1 −7 n−1∏ s=0 1 + x−7x−5x−3bs 1 + x−5x−3x−1bs , x6n−6 = xn0 xn−1 −6 n−1∏ s=0 1 + x−6x−4x−2bs 1 + x−4x−2x0bs , x6n−5 = x−5x n −9x n −7 xn−3x n −1 n−1∏ s=0 1 + x−5x−3x−1bs 1 + x−9x−7x−5b(s+ 1) , x6n−4 = x−4x n −8x n −6 xn−2x n 0 n−1∏ s=0 1 + x−4x−2x0bs 1 + x−8x−6x−4b(s+ 1) . Setting b = ±1 and replacing n with n + 1 or n + 2 in the above equations, we recover the results in [7] (see Corollaries 3.1.1 and 3.2.1). We note some typos in the formulas for x6n+3 (dn should be dn+2) in Corollaries 3.1.1 and 3.2.1. In fact, x6n+3 = x6(n+2)−9 and it follows from the above expressions that the power of x−3 must then be n+ 2. Another way to confirm that it should be n+ 2 is to set k = 2 in equations (2.1.3) and (2.2.3) of [7]. J. T. Kubayi, M. Folly-Gbetoula, J. Math. Computer Sci., 32 (2024), 318–331 326 For the case where a 6= 1 we have x6n−9+i = xni−3 xn−1 i−9 n−1∏ s=0 as + xi−9xi−7xi−5b (1−as 1−a ) as + xi−7xi−5xi−3b (1−as 1−a ) , for i = 0, 1, 2, 3; and x6n−9+i = xi−9x n i−13x n i−11 xni−7x n i−5 n−1∏ s=0 as + xi−9xi−7xi−5b (1−as 1−a ) as+1 + xi−13xi−11xi−9b ( 1−as+1 1−a ) , for i = 4, 5. More explicitly, x6n−9 = xn−3 xn−1 −9 n−1∏ s=0 as + x−9x−7x−5b ( 1−as 1−a ) as + x−7x−5x−3b ( 1−as 1−a ) , x6n−8 = xn−2 xn−1 −8 n−1∏ s=0 as + x−8x−6x−4b ( 1−as 1−a ) as + x−6x−4x−2b ( 1−as 1−a ) , x6n−7 = xn−1 xn−1 −7 n−1∏ s=0 as + x−7x−5x−3b ( 1−as 1−a ) as + x−5x−3x−1b ( 1−as 1−a ) , x6n−6 = xn0 xn−1 −6 n−1∏ s=0 as + x−6x−4x−2b ( 1−as 1−a ) as + x−4x−2x0b ( 1−as 1−a ) , x6n−5 = x−5x n −9x n −7 xn−3x n −1 n−1∏ s=0 as + x−5x−3x−1b ( 1−as 1−a ) as+1 + x−9x−7x−5b ( 1−as+1 1−a ) , x6n−4 = x−4x n −8x n −6 xn−2x n 0 n−1∏ s=0 as + x−4x−2x0b ( 1−as 1−a ) as+1 + x−8x−6x−4b ( 1−as+1 1−a ) . For a = −1, the formulas reduce to x12n−9 = x2n −3 x2n−1 −9 ( −1 + x−9x−7x−5b −1 + x−7x−5x−3b )n , x12n−8 = x2n −2 x2n−1 −8 ( −1 + x−8x−6x−4b −1 + x−6x−4x−2b )n , x12n−7 = x2n −1 x2n−1 −7 ( −1 + x−7x−5x−3b −1 + x−5x−3x−1b )n , x12n−6 = x2n 0 x2n−1 −6 ( −1 + x−6x−4x−2b −1 + x−4x−2x0b )n , x12n−5 = x−5x 2n −9x 2n −7 x2n −3x 2n −1 ( −1 + x−5x−3x−1b −1 + x−9x−7x−5b )n , x12n−4 = x−4x 2n −8x 2n −6 x2n −2x 2n 0 ( −1 + x−4x−2x0b −1 + x−8x−6x−4b )n , x12n−3 = x2n+1 −3 x2n −9 ( −1 + x−9x−7x−5b −1 + x−7x−5x−3b )n , x12n−2 = x2n+1 −2 x2n −8 ( −1 + x−8x−6x−4b −1 + x−6x−4x−2b )n , x12n−1 = x2n+1 −1 x2n −7 ( −1 + x−7x−5x−3b −1 + x−5x−3x−1b )n , x12n = x2n+1 0 x2n −6 ( −1 + x−6x−4x−2b −1 + x−4x−2x0b )n , x12n+1 = x−5x 2n+1 −9 x2n+1 −7 x2n+1 −3 x2n+1 −1 (−1 + x−5x−3x−1b) n (−1 + x−9x−7x−5b)n+1 , x12n+2 = x−4x 2n+1 −8 x2n+1 −6 x2n+1 −2 x2n+1 0 (−1 + x−4x−2x0b) n (−1 + x−8x−6x−4b)n+1 . If we set b = ±1 and we replace n with n + 1 in the above equations, we recover the results [7] (see Corollaries 3.3.1 and 3.4.1). We note that the x12n+5 in the last equation in Corollaries 3.3.1 and 3.4.1 should be x12n+11. 4. Periodic nature and behavior of the solutions Theorem 4.1. Let xn be a solution of xn+5k = xnxn+kxn+2k xn+3kxn+4k(A+Bxnxn+kxn+2k) (4.1) for some constants A 6= 1 and B. If the initial conditions xi, i = 0, . . . , 5k − 1, are such that x3 i = x3 i+k = (1 −A)/B, then xn = x = [(1 −A)/B]1/3 for all n. J. T. Kubayi, M. Folly-Gbetoula, J. Math. Computer Sci., 32 (2024), 318–331 327 Proof. Suppose the initial conditions are such that xi = xi+k and x3 i = (1−A)/B for i = 0, . . . ,k− 1. From (3.9) and (3.10), we have that x3kn+i = xi n−1∏ s=0 As + x3 iB s−1∑ l=0 Al As + x3 iB s−1∑ l=0 Al = xi, where i = 0, 1, 2, . . . , 2k− 1; and x3kn+i = xi n−1∏ s=0 As + x3 iB s−1∑ l=0 Al As+1 + x3 iB s∑ l=0 Al = xi An + x3 iB n−1∑ l=0 Al = xi, where i = 2k, 2k+ 1, . . . , 3k− 1. That is, x3kn+i = xi, i = 0, . . . , 3k− 1, and x3kn+i+k = xi, for all k. Figure 1 illustrates Theorem 4.1. Note that x in Theorem 4.1 is a fixed point of (4.1). This theorem is interesting in the sense that if any of the initial condition does not satisfy x3 i = (1 −A)/B, xn can neither be a constant nor periodic even if the initial conditions are all the same (see Figure 2). Figure 1: Graph of xn+10 = xnxn+4xn+2 xn+6xn+8(3 + 0.25xnxn+2xn+4) , where x0 = x1 = · · · = x9 = −2 = ((1 −A)/B)1/3. Figure 2: Graph of xn+10 = xnxn+4xn+2 xn+6xn+8(3 + 0.25xnxn+2xn+4) , where x0 = x1 = · · · = x9 = −3 6= ((1 −A)/B)1/3. Theorem 4.2. Let xn be non-zero solutions of xn+5k = xnxn+kxn+2k xn+3kxn+4k(1 +Bxnxn+kxn+2k) J. T. Kubayi, M. Folly-Gbetoula, J. Math. Computer Sci., 32 (2024), 318–331 328 for some constant B. If the initial conditions xi, i = 0, . . . , 5k− 1, are such that xi = xi+k, then the solution can not be periodic. Furthermore, the limit of xn, as n→∞, does not exist. Proof. Suppose the non-zero initial conditions are such that xi = xi+k. From (3.9) and (3.10), we have that x3kn+i = xi n−1∏ s=0 1 + x3 iBs 1 + x3 iBs = xi, where i = 0, 1, 2, . . . , 2k− 1; and x3kn+i = xi n−1∏ s=0 1 + x3 iBs 1 + x3 iB(s+ 1) = xi 1 + x3 iBn 6= xi, where i = 2k, 2k+ 1, . . . , 3k− 1. It follows that lim n→∞ x3kn+i = xi for i = 0, . . . , 2k− 1; and lim n→∞ x3kn+i = 0 for i = 2k, . . . , 3k− 1. Thus, the limit does not exist. Figure 3 illustrates Theorem 4.2. Figure 3: Graph of xn+10 = xnxn+4xn+2 xn+6xn+8(1 + xnxn+2xn+4) , where x0 = x1 = · · · = x9 = −2. Theorem 4.3. Let xn be a solution of xn+5k = xnxn+kxn+2k xn+3kxn+4k(A+Bxnxn+kxn+2k) for some constants A 6= 1 and B. If the initial conditions xi, i = 0, . . . , 5k− 1 are such that xi = xi+3k, then the solution is periodic with period 3k if and only if xixi+kxi+2k = (1 −A)/B. Proof. Suppose the initial conditions are such that xi = xi+3k and xixi+kxi+2k = (1 −A)/B. From (3.9) and (3.10), we have that x3kn+i = xi n−1∏ s=0 As + xixi+kxi+2kB s−1∑ l=0 Al As + xixi+kxi+2kB s−1∑ l=0 Al = xi, where i = 0, 1, 2, . . . , 2k− 1; and x3kn+i = xi n−1∏ s=0 As + xixi+kxi+2kB s−1∑ l=0 Al As+1 + xixi+kxi+2kB s∑ l=0 Al = xi An + xixi+kxi+2kB n−1∑ l=0 Al = xi, where i = 2k, 2k+ 1, . . . , 3k− 1. That is, x3kn+i = xi, i = 0, . . . , 3k− 1, and x3kn+i+3k = xi, for all k. J. T. Kubayi, M. Folly-Gbetoula, J. Math. Computer Sci., 32 (2024), 318–331 329 Figures 4 and 5 illustrate Theorem 4.3. Figure 4: Graph of xn+10 = xnxn+4xn+2 xn+6xn+8(3 + 0.25xnxn+2xn+4) , where x0 = x6 = −2, x1 = x7 = −2/3, x2 = x8 = 1, x3 = x9 = −3, x4 = 4, x5 = −4 and are such that x0x2x4 = x1x3x5 = (1 −A)/B. Figure 5: Graph of xn+10 = xnxn+4xn+2 xn+6xn+8(3 + 0.25xnxn+2xn+4) , where x0 = −x6 = −7, x1 = −x7 = −7/3, x2 = −x8 = 1, x3 = −x9 = −3, x4 = 4, x5 = −4 and are such that x0x2x4 6= x1x3x5 6= (1 −A)/B. Theorem 4.4. Let xn be a non-zero solution of xn+5k = xnxn+kxn+2k xn+3kxn+4k(1 +Bxnxn+kxn+2k) for some constant B. If the initial conditions xi, i = 0, . . . , 5k− 1 are such that xi = xi+3k, then the solution can not be periodic. Furthermore, the limit of xn, as n→∞, does not exist. Proof. Suppose the non-zero initial conditions are such that xi = xi+3k. From (3.9) and (3.10), we have that x3kn+i = xi n−1∏ s=0 1 + xixi+kxi+2kBs 1 + xixi+kxi+2kBs = xi, where i = 0, 1, 2, . . . , 2k− 1; and x3kn+i = xi n−1∏ s=0 1 + xixi+kxi+2kBs 1 + xixi+kxi+2kB(s+ 1) = xi 1 + xixi+kxi+2kBn 6= xi, where i = 2k, 2k+ 1, . . . , 3k− 1. It follows that lim n→∞ x3kn+i = xi for i = 0, . . . , 2k− 1; and lim n→∞ x3kn+i = 0 for i = 2k, . . . , 3k− 1. Hence the limit does not exist. Figure 6 illustrates Theorem 4.4. J. T. Kubayi, M. Folly-Gbetoula, J. Math. Computer Sci., 32 (2024), 318–331 330 Figure 6: Graph of xn+10 = xnxn+4xn+2 xn+6xn+8(1 + 0.01xnxn+2xn+4) , where x0 = x6 = −2, x1 = x7 = −2/3, x2 = x8 = 1, x3 = x9 = −3, x4 = 4, x5 = −4. 5. Conclusion We investigated the difference equation (1.4) by finding the symmetry generators and we used the canonical coordinates to find its invariants which led to the solutions in closed form. We showed that the findings in [7, 10, 18] are special cases of our results and we pointed out some errors in [7]. As a matter of fact, all the formulas solutions found in [7] are solutions of (1.4), when an = bn = 1, and not (1.3). Finally, we studied the periodic nature and behavior of the solutions in some cases. Acknowledgment This work is based on the research supported by the National Research Foundation of South Africa (Grant Number: 132108). References [1] L. S. Aljoufi, M. B. Almatrafi, A. R. Seadawy, Dynamical analysis of discrete time equations with a generalized order, Alex. Eng. J., 64 (2023), 937–945. 1 [2] M. B. Almatrafi, Analysis of solutions of some discrete systems of rational difference equations, J. Comput. Anal. Appl., 29 (2021), 355–368. [3] M. B. Almatrafi, M. M. Alzubaidi, Stability analysis for a rational difference equation, Arab J. Basic Appl. Sci., 27 (2020), 114–120. [4] M. B. Almatrafi, M. M. Alzubaidi, The solution and dynamic behaviour of some difference equations of seventh order, J. Appl. Nonlinear Dyn., 10 (2021), 709–719. [5] M. B. Almatrafi, E. M. Elsayed, Solutions and formulae for some systems of difference equations, MathLAB J., 1 (2018), 356–369. [6] M. B. Almatrafi, E. M. Elsayed, F. Alzahrani, The Solution and Dynamic Behavior of Some Difference Equations of Fourth Order, Dyn. Contin. Discrete Impuls. Syst. Ser. A Math. Anal., 29 (2022), 33–50. 1 [7] G. Çinar, A. Gelişken, O. Özkan, Well-defined solutions of the difference equation xn = xn−3kxn−4kxn−5k xn−kxn−2k(±1±xn−3kxn−4kxn−5k) , Asian-Eur. J. Math., 12 (2019), 13 pages. 1, 1, 1, 3.1, 3.2.1, 5 [8] M. Folly-Gbetoula, A. H. Kara, Invariance analysis and reduction of discrete Painlevé equations, J. Difference Equ. Appl., 22 (2016), 1378–1388. 1 [9] M. Folly-Gbetoula, K. Mkhwanazi, D. Nyirenda, On a study of a family of higher order recurrence relations, Math. Probl. Eng., 2022 (2022), 11pages. [10] M. Folly-Gbetoula, N. Mnguni, A. H. Kara, A group theory approach towards some rational difference equations, J. Math., 2019 (2019), 9 pages. 1, 1, 3.1.3, 5 [11] M. Folly-Gbetoula, D. Nyirenda, Lie symmetry analysis and explicit formulas for solutions of some third-order difference equations, Quaest. Math., 42 (2019), 907–917. [12] M. Folly-Gbetoula, D. Nyirenda, Explicit formulas for solutions of some (k+ 3)th-order difference equations, Differ. Equ. Dyn. Syst., 31 (2023), 345–356. [13] P. E. Hydon, Difference equations by differential equation methods, Cambridge University Press, Cambridge, (2014). 1, 2, 2, 2 https://doi.org/10.1016/j.aej.2022.09.025 https://doi.org/10.1016/j.aej.2022.09.025 http://www.eudoxuspress.com/images/JOCAAA-VOL-29-2021-ISSUE-2.pdf#page=151 http://www.eudoxuspress.com/images/JOCAAA-VOL-29-2021-ISSUE-2.pdf#page=151 https://www.tandfonline.com/doi/abs/10.1080/25765299.2020.1726557 https://www.tandfonline.com/doi/abs/10.1080/25765299.2020.1726557 https://doi.org/10.5890/JAND.2021.12.010 https://doi.org/10.5890/JAND.2021.12.010 https://www.researchgate.net/profile/M-Almatrafi/publication/331413727_Solutions_And_Formulae_For_Some_Systems_Of_Difference_Equations/links/5c784874458515831f781a8a/Solutions-And-Formulae-For-Some-Systems-Of-Difference-Equations.pdf https://www.researchgate.net/profile/M-Almatrafi/publication/331413727_Solutions_And_Formulae_For_Some_Systems_Of_Difference_Equations/links/5c784874458515831f781a8a/Solutions-And-Formulae-For-Some-Systems-Of-Difference-Equations.pdf https://www.researchgate.net/profile/Elsayed-Elsayed-7/amp https://www.researchgate.net/profile/Elsayed-Elsayed-7/amp https://doi.org/10.1142/S1793557120400161 https://doi.org/10.1142/S1793557120400161 https://doi.org/10.1080/10236198.2016.1198342 https://doi.org/10.1080/10236198.2016.1198342 https://search.proquest.com/openview/aae0869f9f4d194a012d095724d167ea/1?pq-origsite=gscholar&cbl=237775 https://search.proquest.com/openview/aae0869f9f4d194a012d095724d167ea/1?pq-origsite=gscholar&cbl=237775 https://doi.org/10.1155/2019/1505619 https://doi.org/10.1155/2019/1505619 https://doi.org/10.2989/16073606.2018.1499563 https://doi.org/10.2989/16073606.2018.1499563 https://doi.org/10.1007/s12591-019-00494-8 https://doi.org/10.1007/s12591-019-00494-8 https://doi.org/10.1017/CBO9781139016988 J. T. Kubayi, M. Folly-Gbetoula, J. Math. Computer Sci., 32 (2024), 318–331 331 [14] S. Lie, Vorlesungen über Differentialgleichungen: Mit Bekannten Infinitesimalen Transformationen, BG Teubner: Leipzig, Germany, (1891). 1 [15] S. Maeda, The similarity method for difference equations, IMA J. Appl. Math., 38 (1987), 129–134. 1 [16] N. Mnguni, M. Folly-Gbetoula, Invariance analysis of a third-order difference equation with variable coefficients, Dyn. Contin. Discrete Impuls. Syst. Ser. B Appl. Algorithms, 25 (2018), 63–73. 1 [17] N. Mnguni, D. Nyirenda, M. Folly-Gbetoula, Symmetry Lie algebra and exact solutions of some fourth-order difference equations, J. Nonlinear Sci. Appl., 11 (2018), 1262–1270. 1 [18] Y. Yazlik, On the solutions and behavior of rational difference equations, J. Comput. Anal. Appl., 17 (2014), 584–594. 1, 1, 3.1.3, 3.1.3, 5 https://books.google.com/books?hl=en&lr=&id=FDcNAAAAYAAJ&oi=fnd&pg=PA1&dq=Vorlesungen+%C3%BCber+differentialgleichungen:+mit+bekannten+infinitesimalen+transformationen&ots=HsLh9GeF0o&sig=6xFMlZ0QTzbSOkj1ZHuKyNZ6aSQ https://books.google.com/books?hl=en&lr=&id=FDcNAAAAYAAJ&oi=fnd&pg=PA1&dq=Vorlesungen+%C3%BCber+differentialgleichungen:+mit+bekannten+infinitesimalen+transformationen&ots=HsLh9GeF0o&sig=6xFMlZ0QTzbSOkj1ZHuKyNZ6aSQ https://doi.org/10.1093/imamat/38.2.129 http://online.watsci.org/abstract_pdf/2018v25/v25n1b-pdf/4.pdf http://online.watsci.org/abstract_pdf/2018v25/v25n1b-pdf/4.pdf https://doi.org/10.22436/jnsa.011.11.06 https://doi.org/10.22436/jnsa.011.11.06 https://search.ebscohost.com/login.aspx?direct=true&profile=ehost&scope=site&authtype=crawler&jrnl=15211398&AN=92883506&h=XXFME%2FLpoBaIUWM2gkfEZ4%2BFfHt6u40waHIWTnVwGlxT3%2BOkejA2q5vT7tGOHzmgxePU3wBM2YqcaWAaIUh4Ig%3D%3D&crl=c Introduction Definitions and notation Symmetry analysis and exact solutions The case where are constant The case where The case where The case where The case when The case where are constant Periodic nature and behavior of the solutions Conclusion