# Solution of the Rational Difference Equation $xn+1=xn−131+xn−1xn−3xn−5xn−7xn−9xn−11${x_{n + 1}} = {{{x_{n - 13}}} \over {1 + {x_{n - 1}}{x_{n - 3}}{x_{n - 5}}{x_{n - 7}}{x_{n - 9}}{x_{n - 11}}}}

Dagistan Simsek 1 , 2 , Burak Ogul 1  and Fahreddin Abdullayev 1 , 3
• 1 Kyrgyz Turkish Manas University, Bishkek, Kyrgyzstan
• 2 Konya Technical University, Konya, Turkey
• 3 Mersin University, Mersin, Turkey
Dagistan Simsek
• Kyrgyz Turkish Manas University, Bishkek, Kyrgyzstan
• Konya Technical University, Konya, Turkey
• Search for other articles:
, Burak Ogul
and Fahreddin Abdullayev
• Kyrgyz Turkish Manas University, Bishkek, Kyrgyzstan
• Mersin University, Mersin, Turkey
• Search for other articles:

## Abstract

In this paper, solution of the following difference equation is examined

$xn+1=xn−131+xn−1xn−3xn−5xn−7xn−9xn−11,$
where the initial conditions are positive real numbers.

## 1 Introduction

Difference equations appear naturally as discrete analogs and as numerical solutions of differential and delay differential equations, having applications in biology, ecology, physics.

Recently, a high attention to studying the periodic nature of nonlinear difference equations has been attracted. For some recent results concerning the periodic nature of scalar nonlinear difference equations, among other problems, see, for example, [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33].

Amleh [1], studied the global stability, the boundedness character, and the periodic nature of the positive solutions of the difference equation:

$xn+1=α+xn−1xn$
where α ∈ [0, ∞) and where the initial conditions x−1 and x0 are arbitrary positive real numbers.

De Vault [8], studied the following problems

$xn+1=Axn+1xn−2, n=0,1,…,$
and showed every positive solution of the equation where A ∈ (0, ∞).

Elsayed [15], studied the global result, boundedness, and periodicity of solutions of the difference equation

$xn+1=a+bxn−1+cxn−kdxn−1+exn−k, n=0,1,…,$
where the parameters a, b, c, d and e are positive real numbers and the initial conditions xt,xt+1,...,x0 are positive real numbers where t = max{l,k}, lk.

Ibrahim [18], studied the solutions of non-linear difference equation

$xn+1=xnxn−2xn−1(a+bxnxn−2), n=0,1,…,$
where the initial values x0,x−1 and x−2 non-negative real numbers with bx0x−2 ≠ −a and x−1 ≠ 0. He investigated some properties for this difference equation such as the local stability and the boundedness for the solutions.

Simsek et. al. [25,26,27] and [30], studied the following problems with positive initial values

$xn+1=xn−31+xn−1, n=0,1,…,xn+1=xn−51+xn−2, n=0,1,…,xn+1=xn−51+xn−1xn−3, n=0,1,…,xn+1=xn−31+xnxn−1xn−2, n=0,1,…,$
respectively.

In this work, the following non-linear difference equation is studied

$xn+1=xn−131+xn−1xn−3xn−5xn−7xn−9xn−11, n=0,1,…,$
where x−13,x−12,...,x−1,x0 ∈ (0, ∞) is investigated.

## 2 Main Result

Let $x¯$ be the unique positive equilibrium of the equation 1, then clearly,

$x¯=x¯1+x¯.xxxxx¯⇒x¯+x¯7=x¯⇒x¯7=0⇒x¯0,$
so, $x¯=0$ can be obtained. For any k ≥ 0 and m > k, notation $i=k,m¯$ means i = k,k + 1,...,m.

Theorem 1

Consider the difference equation 1. Then the following statements are true:

1. a)The sequences (x14n−13), (x14n−12),...(x14n−1), (x14n) are decreased and a1,..., a14 ≥ 0 is existed in such that:
$limn→∞x14n−13+k=a1+k, k=0,13¯.$
2. b)(a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,...) is a solution of 1 having period fourteen.
3. c)
$∏k=06limn→∞x14n−j+2k=0, j=0,1¯;$
or
$∏k=06a2k+i=0, i=0,1¯.$
4. d)If there exist n0 ∈ ℕ such that xn+1xn−11 for all nn0, then,
$limn→∞xn=0.$
5. e)The following formulas below:
$x14n+k+1=x−13+k(1−x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k1+x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k ×∑j=0n∏i=17j11+x2i−11+kx2i−9+kx2i−7+kx2i−5+kx2i−3+kx2i−1+k),$
$x14n+k+3=x−11+k(1−x−1+kx−3+kx−5+kx−7+kx−9+kx−13+k1+x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k ×∑j=0n∏i=17j+111+x2i−11+kx2i−9+kx2i−7+kx2i−5+kx2i−3+kx2i−1+k),$
$x14n+k+5=x−9+k(1−x−1+kx−3+kx−5+kx−7+kx−11+kx−13+k1+x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k ×∑j=0n∏i=17j+211+x2i−11+kx2i−9+kx2i−7+kx2i−5+kx2i−3+kx2i−1+k),$
$x14n+k+7=x−7+k(1−x−1+kx−3+kx−5+kx−9+kx−11+kx−13+k1+x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k ×∑j=0n∏i=17j+311+x2i−11+kx2i−9+kx2i−7+kx2i−5+kx2i−3+kx2i−1+k),$
$x14n+k+9=x−5+k(1−x−1+kx−3+kx−7+kx−9+kx−11+kx−13+k1+x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k ×∑j=0n∏i=17j+411+x2i−11+kx2i−9+kx2i−7+kx2i−5+kx2i−3+kx2i−1+k),$
$x14n+k+11=x−3+k(1−x−1+kx−5+kx−7+kx−9+kx−11+kx−13+k1+x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k ×∑j=0n∏i=17j+511+x2i−11+kx2i−9+kx2i−7+kx2i−5+kx2i−3+kx2i−1+k),$
$x14n+k+13=x−1+k(1−x−3+kx−5+kx−7+kx−9+kx−11+kx−13+k1+x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k ×∑j=0n∏i=17j+611+x2i−11+kx2i−9+kx2i−7+kx2i−5+kx2i−3+kx2i−1+k),$
$k=0,1¯$ , holds.
6. f)If x14n+1+kak+1 ≠ 0, x14n+3+kak+3 ≠ 0, x14n+5+kak+5 ≠ 0, x14n+7+kak+7 ≠ 0, x14n+9+kak+9 ≠ 0, x14n+11+kak+11 ≠ 0 then x14n+13+k → 0 as n → ∞, $k=0,1¯$ .

Proof

1. a)Firstly, from 1,
$xn+1(1+xn−1xn−3xn−5xn−7xn−9xn−11)=xn−13,$
is obtained. If xn−1xn−3xn−5xn−7xn−9xn−11 ∈ (0, +∞), then 1 + xn−1xn−3xn−5xn−7xn−9xn−11 ∈ (1, +∞).
Since
$xn+1
n ∈ ℕ
$limn→∞x14n−13+k=a1+k, for k=0,13¯$
existed formulas are obtained.
2. b)(a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,...) is a solution of 1 having period fourteen.
3. c)In view of the 1,
$n=14n⇒x14n+1=x14n−131+∏k=05limn→∞x14n−11+2k,$
is obtained. If the limits are put on both sides of the above equality,
$∏k=06limn→∞x14n−13+2k=0 or ∏k=06a2k+1=0,$
is obtained. Similarly for n = 14n + 1,
$n=14n+1⇒x14n+2=x14n−121+∏k=05limn→∞x14n−10+2k,$
is obtained. If the limits are put on both sides of the above equality,
$∏k=06limn→∞x14n−12+2k=0 or ∏k=06a2k+2=0,$
is obtained.
4. d)If there exist n0 ∈ ℕ such that xn+1xn−11 for all nn0, then, a1a3a5a7a9a11a13a1, a2a4a6a8a10a12a14a2. Using (c) we get
$∏k=06a2k+i=0, i=1,2¯.$
Then we see that,
$limn→∞xn=0.$
Hence the proof of (d) completed.
5. e)Subracting xn−13 from the left and right-hand sides in 1,
$xn+1−xn−13=11+xn−1xn−3xn−5xn−7xn−9xn−11(xn−1−xn−15),$
is obtained and the following formula is produced below, for n ≥ 2
$x2n−3−x2n−17=(x1−x−13)∏i=1n−211+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11,x2n−2−x2n−16=(x1−x−12)∏i=1n−211+x2ix2i−2x2i−4x2i−6x2i−8x2i−10,$
7 j inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for $k=0,1¯$ ,
$x14n+1+k−x−13+k=(x1+k−x−13+k)∑j=0n∏i=17j11+x2i−1+kx2i−3+kx2i−5+kx2i−7+kx2i−9+kx2i−11+k.$
Also, 7 j + 1 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for $k=0,1¯$ ,
$x14n+3+k−x−11+k=(x3+k−x−11+k)∑j=0n∏i=17j+111+x2i−1+kx2i−3+kx2i−5+kx2i−7+kx2i−9+kx2i−11+k.$
Also, 7 j + 2 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for $k=0,1¯$ ,
$x14n+5+k−x−9+k=(x5+k−x−9+k)∑j=0n∏i=17j+211+x2i−1+kx2i−3+kx2i−5+kx2i−7+kx2i−9+kx2i−11+k.$
Also, 7 j + 3 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for $k=0,1¯$ ,
$x14n+7+k−x−7+k=(x7+k−x−7+k)∑j=0n∏i=17j+311+x2i−1+kx2i−3+kx2i−5+kx2i−7+kx2i−9+kx2i−11+k.$
Also, 7 j + 4 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for $k=0,1¯$ ,
$x14n+9+k−x−5+k=(x9+k−x−5+k)∑j=0n∏i=17j+411+x2i−1+kx2i−3+kx2i−5+kx2i−7+kx2i−9+kx2i−11+k.$
Also, 7 j + 5 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for $k=0,1¯$ ,
$x14n+11+k−x−3+k=(x11+k−x−3+k)∑j=0n∏i=17j+511+x2i−1+kx2i−3+kx2i−5+kx2i−7+kx2i−9+kx2i−11+k.$
Also, 7 j + 6 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for $k=0,1¯$ ,
$x14n+13+k−x−1+k=(x13+k−x−1+k)∑j=0n∏i=17j+611+x2i−1+kx2i−3+kx2i−5+kx2i−7+kx2i−9+kx2i−11+k.$
Now we obtained of the above formulas:
$x14n+k+1=x−13+k(1−x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k1+x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k ×∑j=0n∏i=17j11+x2i−11+kx2i−9+kx2i−7+kx2i−5+kx2i−3+kx2i−1+k),$
$x14n+k+3=x−11+k(1−x−1+kx−3+kx−5+kx−7+kx−9+kx−13+k1+x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k ×∑j=0n∏i=17j+111+x2i−11+kx2i−9+kx2i−7+kx2i−5+kx2i−3+kx2i−1+k),$
$x14n+k+5=x−9+k(1−x−1+kx−3+kx−5+kx−7+kx−11+kx−13+k1+x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k ×∑j=0n∏i=17j+211+x2i−11+kx2i−9+kx2i−7+kx2i−5+kx2i−3+kx2i−1+k),$
$x14n+k+7=x−7+k(1−x−1+kx−3+kx−5+kx−9+kx−11+kx−13+k1+x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k ×∑j=0n∏i=17j+311+x2i−11+kx2i−9+kx2i−7+kx2i−5+kx2i−3+kx2i−1+k),$
$x14n+k+9=x−5+k(1−x−1+kx−3+kx−7+kx−9+kx−11+kx−13+k1+x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k ×∑j=0n∏i=17j+411+x2i−11+kx2i−9+kx2i−7+kx2i−5+kx2i−3+kx2i−1+k),$
$x14n+k+11=x−3+k(1−x−1+kx−5+kx−7+kx−9+kx−11+kx−13+k1+x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k ×∑j=0n∏i=17j+511+x2i−11+kx2i−9+kx2i−7+kx2i−5+kx2i−3+kx2i−1+k),$
$x14n+k+13=x−1+k(1−x−3+kx−5+kx−7+kx−9+kx−11+kx−13+k1+x−1+kx−3+kx−5+kx−7+kx−9+kx−11+k ×∑j=0n∏i=17j+611+x2i−11+kx2i−9+kx2i−7+kx2i−5+kx2i−3+kx2i−1+k).$
6. f)Suppose that a1 = a3 = a5 = a7 = a9 = a11 = a13 = 0. By (e), we have
$limn→∞x14n+1=limn→∞x−13(1−x−1x−3x−5x−7x−9x−111+x−1x−3x−5x−7x−9x−11∑j=0n∏i=17j11+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11),a1=x−13(1−x−1x−3x−5x−7x−9x−111+x−1x−3x−5x−7x−9x−11∑j=0n∏i=17j11+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11)a1=0⇒1+x−1x−3x−5x−7x−9x−11x−1x−3x−5x−7x−9x−11=∑j=0n∏i=17j11+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11.$
Similarly,
$a3=0⇒1+x−1x−3x−5x−7x−9x−11x−1x−3x−5x−7x−9x−13=∑j=0n∏i=17j+111+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11.$
Similarly,
$a5=0⇒1+x−1x−3x−5x−7x−9x−11x−1x−3x−5x−7x−11x−13=∑j=0n∏i=17j+211+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11.$
Similarly,
$a7=0⇒1+x−1x−3x−5x−7x−9x−11x−1x−3x−5x−9x−11x−13=∑j=0n∏i=17j+311+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11.$
Similarly,
$a9=0⇒1+x−1x−3x−5x−7x−9x−11x−1x−3x−7x−9x−11x−13=∑j=0n∏i=17j+411+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11.$
Similarly,
$a11=0⇒1+x−1x−3x−5x−7x−9x−11x−1x−5x−7x−9x−11x−13=∑j=0n∏i=17j+511+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11.$
Similarly,
$a13=0⇒1+x−1x−3x−5x−7x−9x−11x−3x−5x−7x−9x−11x−13=∑j=0n∏i=17j+611+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11.$
From 3 and 4,
$1+x−1x−3x−5x−7x−9x−11x−1x−3x−5x−7x−9x−11=∑j=0n∏i=17j11+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11>1+x−1x−3x−5x−7x−9x−11x−1x−3x−5x−7x−9x−13=∑j=0n∏i=17j+111+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11,$
thus, x−13 > x−11. From 4 and 5
$1+x−1x−3x−5x−7x−9x−11x−1x−3x−5x−7x−9x−13=∑j=0n∏i=17j+111+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11>1+x−1x−3x−5x−7x−9x−11x−1x−3x−5x−7x−11x−13=∑j=0n∏i=17j+211+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11,$
thus, x−11 > x−9. From 5 and 6,
$1+x−1x−3x−5x−7x−9x−11x−1x−3x−5x−7x−11x−13=∑j=0n∏i=17j+211+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11>1+x−1x−3x−5x−7x−9x−11x−1x−3x−5x−9x−11x−13=∑j=0n∏i=17j+311+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11,$
thus, x−9 > x−7. From 6 and 7,
$1+x−1x−3x−5x−7x−9x−11x−1x−3x−5x−9x−11x−13=∑j=0n∏i=17j+311+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11>1+x−1x−3x−5x−7x−9x−11x−1x−3x−7x−9x−11x−13=∑j=0n∏i=17j+411+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11$
thus, x−7 > x−5. From 7 and 8,
$1+x−1x−3x−5x−7x−9x−11x−1x−3x−7x−9x−11x−13=∑j=0n∏i=17j+411+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11>1+x−1x−3x−5x−7x−9x−11x−1x−5x−7x−9x−11x−13=∑j=0n∏i=17j+511+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11$
thus, x−5 > x−3. From 8 and 9,
$1+x−1x−3x−5x−7x−9x−11x−1x−5x−7x−9x−11x−13=∑j=0n∏i=17j+511+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11>1+x−1x−3x−5x−7x−9x−11x−3x−5x−7x−9x−11x−13=∑j=0n∏i=17j+611+x2i−1x2i−3x2i−5x2i−7x2i−9x2i−11,$
thus, x−3 > x−1. From here we obtain x−13 > x−11 > x−9 > x−7 > x−5 > x−3 > x−1. Similarly, we can obtain x−12 > x−10 > x−8 > x−6 > x−4 > x−2 > x0. We arrive at a contradiction which completes the proof of theorem.

Example 2

If the initial conditions are selected as follows:

$x−13=0.98, x−12=0.97, x−11=0.96, x−10=0.95, x−9=0.94, x−8=0.93, x−7=0.92,x−6=0.91, x−5=0.9, x−4=0.89, x−3=0.88, x−2=0.87, x−1=0.86, x0=0.85.$

The graph of the solution is given below, xn = {0.62601, 0.634341, 0.701375, 0.701974, 0.737178, 0.734195, 0.753815, 0.748866, 0.759718, 0.753567, 0.759006, 0.752062, 0.753929, 0.746432, 0.535307, 0.545309, 0.621059, 0.622964, 0.664753, 0.66286, 0.687738, 0.683736, 0.698929, 0.693618, 0.702724, 0.696539, 0.701557, 0.694753, 0.487602, 0.498104, 0.576757, 0.579084, 0.623219, 0.621705, 0.64857, 0.644917, 0.661839, 0.656857, 0.667494,...}

## References

• [1]

Amleh A. M., Grove E. A., Georgiou D. A., (1999), On the recursive sequence x n + 1 = α + x n 1 x n {x_{n + 1}} = \alpha + {{{x_{n - 1}}} \over {{x_n}}} , J. Math. Anal. Appl.

• [2]

Amleh A. M., Kirk, V., Ladas G., (2001), On the dynamics of A + α + bx n 1 A + Bx n 2 A + {{\alpha + {bx_{n - 1}}} \over {A + {Bx_{n - 2}}}} , Math Sci. Res. Hot-Line.

• [3]

Cinar C., (2004), On the positive solutions of the difference equation x n + 1 = x n 1 1 + α x n x n 1 {x_{n + 1}} = {{{x_{n - 1}}} \over {1 + \alpha {x_n}{x_{n - 1}}}} , J. Appl. Math.Comput.

• [4]

Cinar C., (2004), On the positive solutions of the difference equation x n + 1 = x n 1 1 α x n x n 1 {x_{n + 1}} = {{{x_{n - 1}}} \over { - 1\alpha {x_n}{x_{n - 1}}}} , Appl. Math. Comput.

• [5]

Cinar C., (2004), On the positive solutions of the difference equation x n + 1 = α x n 1 1 + bx n x n 1 {x_{n + 1}} = {{\alpha {x_{n - 1}}} \over {1 + {bx_n}{x_{n - 1}}}} , Appl. Math. Comput.

• [6]

DeVault R., Ladas G. and Schultz S. W., (1998), On the recursive sequence x n + 1 = A x n + 1 x n 2 {x_{n + 1}} = {A \over {{x_n}}} + {1 \over {{x_{n - 2}}}} , Proc. Amer. Math. Soc. 24 Dagistan Simsek, Burak Ogul and Fahreddin Abdullayev. Applied Mathematics and Nonlinear Sciences 1(2015) 15–24

• [7]

El-Owaidy, H. M., Ahmed, A. M., Mousa, M. S., (2003), On the recursive sequences x n + 1 = α x n 1 β ± x n {x_{n + 1}} = {{\alpha {x_{n - 1}}} \over {\beta \pm {x_n}}} , J. Appl. Math. Comput., 145, 747–753.

• [8]

El-Owaidy, H. M., Ahmed, A. M., Elsady, Z., (2004), Global attractivity of the recursive sequences x n + 1 = α β x n 1 γ + x n {x_{n + 1}} = {{\alpha - \beta {x_{n - 1}}} \over {\gamma + {x_n}}} , J. Appl. Math. Comput., Vol:151, 827–833.

• [9]

Elabbasy E. M., El-Metwally H., Elsayed E. M., (2006), On the difference equation x n + 1 = α x n bx n cx n dx n 1 {x_{n + 1}} = \alpha {x_n} - {{{bx_n}} \over {c{x_n} - {dx_{n - 1}}}} , Advances in Difference Equation, 1–10.

• [10]

Elabbasy E. M., El-Metwally H., Elsayed E. M., (2007), On the difference equation x n + 1 = α x n k β + γ i = 0 k x n i {x_{n + 1}} = {{\alpha {x_{n - k}}} \over {\beta + \gamma \prod\nolimits_{i = 0}^k {x_{n - i}}}} , J. Conc. Appl. Math., 5(2), 101–113.

• [11]

Elabbasy E. M., El-Metwally H., Elsayed E. M., (2007), Qualitative behavior of higher order difference equation, Soochow Journal of Mathematics, 33(4), 861–873.

• [12]

Elabbasy E. M., El-Metwally H., Elsayed E. M., (2007), Global attractivity and periodic character of a fractional difference equation of order three, Yokohama Mathematical Journal, 53, 89–100.

• [13]

Elabbasy E. M. and Elsayed E. M., (2008), Global attractivity of difference equation of higher order, Carpathian Journal of Mathematics, 24(2), 45–53.

• [14]

Elaydi, S., (1996), An Introduction to Difference Equations, Spinger-Verlag, New York.

• [15]

Elsayed E. M., (2008), On the solution of recursive sequence of order two, Fasciculi Mathematici, 40, 5–13.

• [16]

Elsayed E. M., (2009), Dynamics of a rational recursive sequences, International Journal of Difference Equations, 4(2), 185–200.

• [17]

Elsayed E. M., (2009), Dynamics of a Recursive Sequence of Higher Order, Communications on Applied Nonlinear Analysis, 16(2), 37–50.

• [18]

Elsayed E. M., (2011), Solution and atractivity for a rational recursive sequence, Discrete Dynamics in Nature and Society, 17–30.

• [19]

Elsayed E. M., (2011), On the solution of some difference equation, Europan Journal of Pure and Applied Mathematics, 4(3), 287–303.

• [20]

Elsayed E. M., (2012), On the Dynamics of a higher order rational recursive sequence, Communications in Mathematical Analysis, 12(1), 117–133.

• [21]

Elsayed E. M., (2012), Solution of rational difference system of order two, Mathematical and Computer Modelling, 55, 378–384.

• [22]

Elsayed E. M., (2016), Dynamics of a three-dimensional systems of rational difference equations, Mathematical Methods in the Applied Sciences, 39(5), 1026–1038.

• [23]

Elsayed E. M., (2016), Dynamics of behavior of a higher order rational difference equation, The Journal of Nonlinear Science and Applications, 9(4), 1463–1474.

• [24]

Elsayed E. M., (2015), New method to obtain periodic solutions of period two and three of a rational difference equation, Nonlinear Dynamics, 79(1), 241–250.

• [25]

Simsek D., Cinar C., Yalcinkaya I., (2006), On the recursive sequence x n + 1 = x n 3 1 + x n 1 {x_{n + 1}} = {{{x_{n - 3}}} \over {1 + {x_{n - 1}}}} , Internat. J. Contemp. 9(12), 475–480.

• [26]

Simsek D., Cinar C., Karatas R., Yalcinkaya I., (2006), On the recursive sequence x n + 1 = x n 5 1 + x n 2 {x_{n + 1}} = {{{x_{n - 5}}} \over {1 + {x_{n - 2}}}} , Int. J. Pure Appl. Math., 27(4), 501–507.

• [27]

Simsek D., Cinar C., Karatas R., Yalcinkaya I., (2006), On the recursive sequence x n + 1 = x n 5 1 + x n 1 x n 3 {x_{n + 1}} = {{{x_{n - 5}}} \over {1 + {x_{n - 1}}{x_{n - 3}}}} , Int. J. Pure Appl. Math., 28(1), 117–124.

• [28]

Simsek D., Cinar C., Yalcinkaya I., (2008), On the recursive sequence x n + 1 = x n ( 5 k + 9 ) 1 + x n 4 x n 9 x n ( 5 k + 4 ) {x_{n + 1}} = {{{x_{n - (5k + 9)}}} \over {1 + {x_{n - 4}}{x_{n - 9}}{x_{n - (5k + 4)}}}} Taiwanese Journal of Mathematics, 12(5), 1087–1098.

• [29]

Simsek D., Ogul B., Abdullayev F., (2017), Solutions of the rational difference equations x n + 1 = x n 11 1 + x n 2 x n 5 x n 8 {x_{n + 1}} = {{{x_{n - 11}}} \over {1 + {x_{n - 2}}{x_{n - 5}}{x_{n - 8}}}} , AIP Conference Proceedings 1880(1) 040003.

• [30]

Simsek D., Abdullayev F., (2017), On the Recursive Sequence x n + 1 = α x n 4 k + 3 1 + t = 0 2 x n ( k + 1 ) t k {x_{n + 1}} = {{\alpha {x_{n - 4k + 3}}} \over {1 + \prod\nolimits_{t = 0}^2 {x_{n - (k + 1)t - k}}}} , Journal of Mathematics Sciences, 222(6), 762–771.

• [31]

Simsek D., Dogan A., (2014), On A Class of Recursive Sequence, Manas Journal of Engineering, 2(1), 16–22.

• [32]

Simsek D., Ogul B., (2017), Solutions of the Rational Difference Equations x n + 1 = x n ( 2 k + 1 ) 1 + x n k {x_{n + 1}} = {{{x_{n - (2k + 1)}}} \over {1 + {x_{n - k}}}} , Manas Journal of Engineering, 5(3), 57–68.

• [33]

Simsek D., Ogul B., (2018), Solutions of the Rational Difference Equations, Manas Journal of Engineering, 6(1), 56–74.

If the inline PDF is not rendering correctly, you can download the PDF file here.

• [1]

Amleh A. M., Grove E. A., Georgiou D. A., (1999), On the recursive sequence x n + 1 = α + x n 1 x n {x_{n + 1}} = \alpha + {{{x_{n - 1}}} \over {{x_n}}} , J. Math. Anal. Appl.

• [2]

Amleh A. M., Kirk, V., Ladas G., (2001), On the dynamics of A + α + bx n 1 A + Bx n 2 A + {{\alpha + {bx_{n - 1}}} \over {A + {Bx_{n - 2}}}} , Math Sci. Res. Hot-Line.

• [3]

Cinar C., (2004), On the positive solutions of the difference equation x n + 1 = x n 1 1 + α x n x n 1 {x_{n + 1}} = {{{x_{n - 1}}} \over {1 + \alpha {x_n}{x_{n - 1}}}} , J. Appl. Math.Comput.

• [4]

Cinar C., (2004), On the positive solutions of the difference equation x n + 1 = x n 1 1 α x n x n 1 {x_{n + 1}} = {{{x_{n - 1}}} \over { - 1\alpha {x_n}{x_{n - 1}}}} , Appl. Math. Comput.

• [5]

Cinar C., (2004), On the positive solutions of the difference equation x n + 1 = α x n 1 1 + bx n x n 1 {x_{n + 1}} = {{\alpha {x_{n - 1}}} \over {1 + {bx_n}{x_{n - 1}}}} , Appl. Math. Comput.

• [6]

DeVault R., Ladas G. and Schultz S. W., (1998), On the recursive sequence x n + 1 = A x n + 1 x n 2 {x_{n + 1}} = {A \over {{x_n}}} + {1 \over {{x_{n - 2}}}} , Proc. Amer. Math. Soc. 24 Dagistan Simsek, Burak Ogul and Fahreddin Abdullayev. Applied Mathematics and Nonlinear Sciences 1(2015) 15–24

• [7]

El-Owaidy, H. M., Ahmed, A. M., Mousa, M. S., (2003), On the recursive sequences x n + 1 = α x n 1 β ± x n {x_{n + 1}} = {{\alpha {x_{n - 1}}} \over {\beta \pm {x_n}}} , J. Appl. Math. Comput., 145, 747–753.

• [8]

El-Owaidy, H. M., Ahmed, A. M., Elsady, Z., (2004), Global attractivity of the recursive sequences x n + 1 = α β x n 1 γ + x n {x_{n + 1}} = {{\alpha - \beta {x_{n - 1}}} \over {\gamma + {x_n}}} , J. Appl. Math. Comput., Vol:151, 827–833.

• [9]

Elabbasy E. M., El-Metwally H., Elsayed E. M., (2006), On the difference equation x n + 1 = α x n bx n cx n dx n 1 {x_{n + 1}} = \alpha {x_n} - {{{bx_n}} \over {c{x_n} - {dx_{n - 1}}}} , Advances in Difference Equation, 1–10.

• [10]

Elabbasy E. M., El-Metwally H., Elsayed E. M., (2007), On the difference equation x n + 1 = α x n k β + γ i = 0 k x n i {x_{n + 1}} = {{\alpha {x_{n - k}}} \over {\beta + \gamma \prod\nolimits_{i = 0}^k {x_{n - i}}}} , J. Conc. Appl. Math., 5(2), 101–113.

• [11]

Elabbasy E. M., El-Metwally H., Elsayed E. M., (2007), Qualitative behavior of higher order difference equation, Soochow Journal of Mathematics, 33(4), 861–873.

• [12]

Elabbasy E. M., El-Metwally H., Elsayed E. M., (2007), Global attractivity and periodic character of a fractional difference equation of order three, Yokohama Mathematical Journal, 53, 89–100.

• [13]

Elabbasy E. M. and Elsayed E. M., (2008), Global attractivity of difference equation of higher order, Carpathian Journal of Mathematics, 24(2), 45–53.

• [14]

Elaydi, S., (1996), An Introduction to Difference Equations, Spinger-Verlag, New York.

• [15]

Elsayed E. M., (2008), On the solution of recursive sequence of order two, Fasciculi Mathematici, 40, 5–13.

• [16]

Elsayed E. M., (2009), Dynamics of a rational recursive sequences, International Journal of Difference Equations, 4(2), 185–200.

• [17]

Elsayed E. M., (2009), Dynamics of a Recursive Sequence of Higher Order, Communications on Applied Nonlinear Analysis, 16(2), 37–50.

• [18]

Elsayed E. M., (2011), Solution and atractivity for a rational recursive sequence, Discrete Dynamics in Nature and Society, 17–30.

• [19]

Elsayed E. M., (2011), On the solution of some difference equation, Europan Journal of Pure and Applied Mathematics, 4(3), 287–303.

• [20]

Elsayed E. M., (2012), On the Dynamics of a higher order rational recursive sequence, Communications in Mathematical Analysis, 12(1), 117–133.

• [21]

Elsayed E. M., (2012), Solution of rational difference system of order two, Mathematical and Computer Modelling, 55, 378–384.

• [22]

Elsayed E. M., (2016), Dynamics of a three-dimensional systems of rational difference equations, Mathematical Methods in the Applied Sciences, 39(5), 1026–1038.

• [23]

Elsayed E. M., (2016), Dynamics of behavior of a higher order rational difference equation, The Journal of Nonlinear Science and Applications, 9(4), 1463–1474.

• [24]

Elsayed E. M., (2015), New method to obtain periodic solutions of period two and three of a rational difference equation, Nonlinear Dynamics, 79(1), 241–250.

• [25]

Simsek D., Cinar C., Yalcinkaya I., (2006), On the recursive sequence x n + 1 = x n 3 1 + x n 1 {x_{n + 1}} = {{{x_{n - 3}}} \over {1 + {x_{n - 1}}}} , Internat. J. Contemp. 9(12), 475–480.

• [26]

Simsek D., Cinar C., Karatas R., Yalcinkaya I., (2006), On the recursive sequence x n + 1 = x n 5 1 + x n 2 {x_{n + 1}} = {{{x_{n - 5}}} \over {1 + {x_{n - 2}}}} , Int. J. Pure Appl. Math., 27(4), 501–507.

• [27]

Simsek D., Cinar C., Karatas R., Yalcinkaya I., (2006), On the recursive sequence x n + 1 = x n 5 1 + x n 1 x n 3 {x_{n + 1}} = {{{x_{n - 5}}} \over {1 + {x_{n - 1}}{x_{n - 3}}}} , Int. J. Pure Appl. Math., 28(1), 117–124.

• [28]

Simsek D., Cinar C., Yalcinkaya I., (2008), On the recursive sequence x n + 1 = x n ( 5 k + 9 ) 1 + x n 4 x n 9 x n ( 5 k + 4 ) {x_{n + 1}} = {{{x_{n - (5k + 9)}}} \over {1 + {x_{n - 4}}{x_{n - 9}}{x_{n - (5k + 4)}}}} Taiwanese Journal of Mathematics, 12(5), 1087–1098.

• [29]

Simsek D., Ogul B., Abdullayev F., (2017), Solutions of the rational difference equations x n + 1 = x n 11 1 + x n 2 x n 5 x n 8 {x_{n + 1}} = {{{x_{n - 11}}} \over {1 + {x_{n - 2}}{x_{n - 5}}{x_{n - 8}}}} , AIP Conference Proceedings 1880(1) 040003.

• [30]

Simsek D., Abdullayev F., (2017), On the Recursive Sequence x n + 1 = α x n 4 k + 3 1 + t = 0 2 x n ( k + 1 ) t k {x_{n + 1}} = {{\alpha {x_{n - 4k + 3}}} \over {1 + \prod\nolimits_{t = 0}^2 {x_{n - (k + 1)t - k}}}} , Journal of Mathematics Sciences, 222(6), 762–771.

• [31]

Simsek D., Dogan A., (2014), On A Class of Recursive Sequence, Manas Journal of Engineering, 2(1), 16–22.

• [32]

Simsek D., Ogul B., (2017), Solutions of the Rational Difference Equations x n + 1 = x n ( 2 k + 1 ) 1 + x n k {x_{n + 1}} = {{{x_{n - (2k + 1)}}} \over {1 + {x_{n - k}}}} , Manas Journal of Engineering, 5(3), 57–68.

• [33]

Simsek D., Ogul B., (2018), Solutions of the Rational Difference Equations, Manas Journal of Engineering, 6(1), 56–74.

OPEN ACCESS