On the recursive sequence x(n+1)=(x(n-(4k+3)))/(1+∏(t=0-2(x(n-(k+1)t-k))))
In this paper a solution of some difference equation was investigated.
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Інститут прикладної математики і механіки НАН України
2016
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irk-123456789-1450792019-01-15T01:23:36Z On the recursive sequence x(n+1)=(x(n-(4k+3)))/(1+∏(t=0-2(x(n-(k+1)t-k)))) Simsek, D. Abdullayev, F. In this paper a solution of some difference equation was investigated. 2016 Article On the recursive sequence x(n+1)=(x(n-(4k+3)))/(1+∏(t=0-2(x(n-(k+1)t-k)))) / D. Simsek, F. Abdullayev // Український математичний вісник. — 2016. — Т. 13, № 3. — С. 376-387. — Бібліогр.: 26 назв. — англ. 1810-3200 2010 MSC. 39A10, 39A14, 39A20. http://dspace.nbuv.gov.ua/handle/123456789/145079 en Український математичний вісник Інститут прикладної математики і механіки НАН України |
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In this paper a solution of some difference equation was investigated. |
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Simsek, D. Abdullayev, F. |
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Simsek, D. Abdullayev, F. On the recursive sequence x(n+1)=(x(n-(4k+3)))/(1+∏(t=0-2(x(n-(k+1)t-k)))) Український математичний вісник |
| author_facet |
Simsek, D. Abdullayev, F. |
| author_sort |
Simsek, D. |
| title |
On the recursive sequence x(n+1)=(x(n-(4k+3)))/(1+∏(t=0-2(x(n-(k+1)t-k)))) |
| title_short |
On the recursive sequence x(n+1)=(x(n-(4k+3)))/(1+∏(t=0-2(x(n-(k+1)t-k)))) |
| title_full |
On the recursive sequence x(n+1)=(x(n-(4k+3)))/(1+∏(t=0-2(x(n-(k+1)t-k)))) |
| title_fullStr |
On the recursive sequence x(n+1)=(x(n-(4k+3)))/(1+∏(t=0-2(x(n-(k+1)t-k)))) |
| title_full_unstemmed |
On the recursive sequence x(n+1)=(x(n-(4k+3)))/(1+∏(t=0-2(x(n-(k+1)t-k)))) |
| title_sort |
on the recursive sequence x(n+1)=(x(n-(4k+3)))/(1+∏(t=0-2(x(n-(k+1)t-k)))) |
| publisher |
Інститут прикладної математики і механіки НАН України |
| publishDate |
2016 |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/145079 |
| citation_txt |
On the recursive sequence x(n+1)=(x(n-(4k+3)))/(1+∏(t=0-2(x(n-(k+1)t-k)))) / D. Simsek, F. Abdullayev // Український математичний вісник. — 2016. — Т. 13, № 3. — С. 376-387. — Бібліогр.: 26 назв. — англ. |
| series |
Український математичний вісник |
| work_keys_str_mv |
AT simsekd ontherecursivesequencexn1xn4k31t02xnk1tk AT abdullayevf ontherecursivesequencexn1xn4k31t02xnk1tk |
| first_indexed |
2025-07-10T20:48:08Z |
| last_indexed |
2025-07-10T20:48:08Z |
| _version_ |
1837294416507699200 |
| fulltext |
Український математичний вiсник
Том 13 (2016), № 3, 376 – 387
On the recursive sequence xn+1 =
xn−(4k+3)
1+
2∏
t=0
xn−(k+1)t−k
Dağıstan Simsek and Fahreddin Abdullayev
Abstract. In this paper a solution of the following difference equation
was investigated
xn+1 =
xn−(4k+3)
1 +
2∏
t=0
xn−(k+1)t−k
, n = 0, 1, 2, . . .
where x−(4k+3), x−(4k+2), . . . , x−1, x0 ∈ (0,∞) and k = 0, 1, . . . .
2010 MSC. 39A10, 39A14, 39A20.
Key words and phrases. Difference equation, period 4k+4 solution.
1. Introduction
Difference equations appear naturally as discrete analogues and as
numerical solutions of differential and delay differential equations having
applications in biology, ecology, physics, etc [26].
Recently there has been a lot of interest in studying the periodic na-
ture of nonlinear difference equations. For some recent result concerning
among other problems, the periodic nature of scalar nonlinear difference
equations see, for examples [1–26].
Cinar, studied the following problems with positive initial values
xn+1 =
xn−1
1 + axnxn−1
xn+1 =
xn−1
−1 + axnxn−1
xn+1 =
axn−1
1 + bxnxn−1
Received 28.08.2016
ISSN 1810 – 3200. c⃝ Iнститут математики НАН України
D. Simsek, F. Abdullayev 377
for n = 0, 1, 2, . . . , in [2, 3, 4], respectively.
In [18] Stevic assumed that β = 1 and solved the following problem
xn+1 =
xn−1
1 + xn
for n = 0, 1, 2, . . .
where x−1, x0 ∈ (0,∞). Also, this result was generalized to the equation
of the following form:
xn+1 =
xn−1
g(xn)
for n = 0, 1, 2, . . .
where x−1, x0 ∈ (0,∞).
In [19, 20, 21] studied the following problems with positive initial
values:
xn+1 =
xn−3
1 + xn−1
,
xn+1 =
xn−5
1 + xn−2
,
xn+1 =
xn−5
1 + xn−1xn−3
for n = 0, 1, 2, . . . , respectively.
In this paper we investigate the following nonlinear difference equation
xn+1 =
xn−(4k+3)
1 +
2∏
t=0
xn−(k+1)t−k
for n = 0, 1, 2, . . . (1)
where x−(4k+3), x−(4k+2), . . . , x−1, x0 and k = 0, 1, . . . .
2. Main result
Theorem 2.1. Consider the difference equation (1). Then the following
statements are true.
a) The sequences (x(4k+4)n−(4k+3)), (x(4k+4)n−(4k+2)), . . . , (x(4k+4)n)
are decreasing and there exists a1, a2, . . . , a4k+4 > 0 such that
lim
n→∞
x(4k+4)n−(4k+3)+t = a1+t, t = 0, 1, . . . , (4k + 3).
b) (a1, a2, . . . , a4k+4, a1, a2, . . . , a4k+4, . . . ) is a solution of equation
(1) of period 4k+4.
c)
3∏
t=0
lim
n→∞
x(4k+4)n−(k+1)t−k = 0, . . . ,
3∏
t=0
lim
n→∞
x(4k+4)n−(k+1)t = 0 or
378 On the recursive sequence. . .
3∏
t=0
a(k+1)t+1 = 0, . . . ,
4∏
t=1
a(k+1)t = 0.
d) If there exists n0 ∈ N such that
2∏
t=0
xn−(k+1)t−k ≥ xn+1
1∏
t=0
xn−(k+1)t−k
for all n > n0, then
lim
n→∞
xn = 0.
e) The following formulas
x(4k+4)n+r(k+1)+s+1
= x(r−4)(k+1)+s+1
1 −
(
4∏
m=1
x−(mk+m−1)+s
)
/ x(r−4)(k+1)+s+1
1 + (
3∏
m=1
x−(mk+m−1)+s)
×
n∑
j=0
4j+r∏
i=1
1
1 +
3∏
m=1
x(k+1)i−(mk+m−1)+s
,
where r = 0, 1, 2, 3 and s = 0, 1, . . . , k hold.
f) If x(4k+4)n+(t−1)k+t → at ̸= 0, (t = 1, 2, 3) then x(4k+4)n+3k+4 →
0 as n → ∞, . . . . If x(4k+4)n+tk+t → atk+t ̸= 0, (t = 1, 2, 3) then
x(4k+4)n+4k+4 → 0 as n→ ∞.
Proof. a) Firstly, from the equation (1), we obtain
xn+1(1 + xn−kxn−(2k+1)xn−(3k+2)) = xn−(4k+3).
If xn−k, xn−(2k+1), xn−(3k+2) ∈ (0,+∞), then
(1 + xn−kxn−(2k+1)xn−(3k+2)) ∈ (1,+∞).
Since xn+1 < xn−(4k+3), n ∈ N , we obtain that there exist
lim
n→∞
x(4k+4)n−(4k+3)+t = a1+t, t = 0, 1, . . . , (4k + 3).
b) (a1, a2, . . . , a4k+4, a1, a2, . . . , a4k+4, . . . ) is a solution of equation
(1) of period 4k + 4.
D. Simsek, F. Abdullayev 379
c) In view of the equation (1), we obtain
x(4k+4)n+1 =
x(4k+4)n−(4k+3)
1 +
2∏
t=0
x(4k+4)n−(k+1)t−k
.
Taking limits as n→ ∞ on both sides of the above equality, we get
lim
n→∞
x(4k+4)n+1 = lim
n→∞
x(4k+4)n−(4k+3)
1 +
2∏
t=0
x(4k+4)n−(k+1)t−k
.
Then
3∏
t=0
lim
n→∞
x(4k+4)n−(k+1)t−k = 0 or
3∏
t=0
a(k+1)t+1 = 0.
Similarly,
3∏
t=0
lim
n→∞
x(4k+4)n−(k+1)t = 0 or
4∏
t=1
a(k+1)t = 0.
d) If there exists n0 ∈ N such that xn−kxn−(2k+1)xn−(3k+2) ≥
xn+1xn−kxn−(2k+1) for all n > n0, then a1 ≤ ak+2 ≤ a2k+3 ≤ a3k+4 ≤
a1, . . . , ak+1 ≤ a2k+2 ≤ a3k+3 ≤ a4k+4 ≤ ak+1. Since
3∏
t=0
a(k+1)t+1 = 0, . . . ,
4∏
t=1
a(k+1)t = 0,
we obtain the result.
e) Subracting xn−(4k+3) from the left and right-hand sides in equation
(1) we obtain
xn+1 − xn−(4k+3) =
1
1 + xn−kxn−(2k+1)xn−(3k+2)
(xn−k − xn−(5k+4)),
and the following formula
for n ≥ k + 1
x(k+1)n−((k+1)2−1) − x(k+1)n−[(k+2)2+2k]
= (x1 − x−(4k+3))
n−(k+1)∏
i=1
1
1+
2∏
t=0
x(k+1)i−(k+1)t−k
.
.
.
x(k+1)n−((k+1)2−(k+1)) − x(k+1)n−[(k+2)2+k]
= (xk+1 − x−(3k+3))
n−(k+1)∏
i=1
1
1+
2∏
t=0
x(k+1)i−(k+1)t
(2)
380 On the recursive sequence. . .
holds. Replacing n by 4j in (2) and summing from j = 0 to j = n we
obtain
x(4k+4)n+1 − x−(4k+3) = (x1 − x−(4k+3))
n∑
j=0
4j∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
(3)
.
.
.
x(4k+4)n+k+1 − x−(3k+3) = (xk+1 − x−(3k+3))
n∑
j=0
4j∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t
.
Also, replacing n by 4j + 1 in (2) and summing from j = 0 to j = n
we obtain
x(4k+4)n+k+2−x−(3k+2) = (x1−x−(4k+3))
n∑
j=0
4j+1∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
,
(4)
.
.
.
x(4k+4)n+2k+2 − x−(2k+2) = (xk+1 − x−(3k+3))
n∑
j=0
4j+1∏
i=1
1
1+
2∏
t=0
x(k+1)i−(k+1)t
.
Also, replacing n by 4j + 2 in (2) and summing from j = 0 to j = n
we obtain
x(4k+4)n+2k+3 − x−(2k+1) = (x1 − x−(4k+3))
n∑
j=0
4j+2∏
i=1
1
1+
2∏
t=0
x(k+1)i−(k+1)t−k
,
(5)
.
.
.
x(4k+4)n+3k+3 − x−(k+1) = (xk+1 − x−(3k+3))
n∑
j=0
4j+2∏
i=1
1
1+
2∏
t=0
x(k+1)i−(k+1)t
.
D. Simsek, F. Abdullayev 381
Also, replacing n by 4j + 3 in (2) and summing from j = 0 to j = n
we obtain
x(4k+4)n+3k+4 − x−(k) = (x1 − x−(4k+3))
n∑
j=0
4j+3∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
,
(6)
.
.
.
x(4k+4)n+4k+4 − x0 = (xk+1 − x−(3k+3))
n∑
j=0
4j+3∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t
.
Now, we obtained of the above formulas;
x(4k+4)n+r(k+1)+s+1
= x(r−4)(k+1)+s+1
1 −
(
4∏
m=1
x−(mk+m−1)+s
)
/ x(r−4)(k+1)+s+1
1 + (
3∏
m=1
x−(mk+m−1)+s)
×
n∑
j=0
4j+r∏
i=1
1
1 +
3∏
m=1
x(k+1)i−(mk+m−1)+s
where r = 0, 1, 2, 3 and s = 0, 1, . . . , k hold.
f) Suppose that a1 = ak+2 = a2k+3 = a3k+4 = 0. By (e) we have
lim
n→∞
x(4k+4)n+1 = lim
n→∞
x−(4k+3)
1 −
2∏
t=0
x−(k+1)t−k
1 +
2∏
t=0
x−(k+1)t−k
×
n∑
j=0
4j∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
.
382 On the recursive sequence. . .
a1 = x−(4k+3)
1 −
2∏
t=0
x−(k+1)t−k
1 +
2∏
t=0
x−(k+1)t−k
∞∑
j=0
4j∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
,
a1 = 0 ⇒
1 +
2∏
t=0
x−(k+1)t−k
2∏
t=0
x−(k+1)t−k
=
∞∑
j=0
4j∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
. (7)
Similarly,
lim
n→∞
x(4k+4)n+k+2 = lim
n→∞
x−(3k+2)
1 −
(
3∏
t=0
x−(k+1)t−k)/x−(3k+2)
1 +
2∏
t=0
x−(k+1)t−k
n
×
∑
j=0
4j+1∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
.
ak+2 = 0 ⇒
1+
2∏
t=0
x−(k+1)t−k
(
3∏
t=0
x−(k+1)t−k)/x−(3k+2)
=
∞∑
j=0
4j+1∏
i=1
1
1+
2∏
t=0
x(k+1)i−(k+1)t−k
. (8)
Similarly,
a2k+3 = x−(2k+1)
1 −
(
3∏
t=0
x−(k+1)t−k)/x−(2k+1)
1 +
2∏
t=0
x−(k+1)t−k
∞
×
∑
j=0
4j+2∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
,
a2k+3 = 0 ⇒
1 +
2∏
t=0
x−(k+1)t−k
(
3∏
t=0
x−(k+1)t−k)/x−(2k+1)
D. Simsek, F. Abdullayev 383
=
∞∑
j=0
4j+2∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
. (9)
Similarly,
lim
n→∞
x(4k+4)n+3k+4 = lim
n→∞
x−(k)
1 −
(
3∏
t=0
x−(k+1)t−k)/x−k
1 +
2∏
t=0
x−(k+1)t−k
n
×
∑
j=0
4j+3∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
a3k+4 = x−(k)
1 −
(
3∏
t=0
x−(k+1)t−k)/x−k
1 +
2∏
t=0
x−(k+1)t−k
×
∞∑
j=0
4j+3∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
,
a3k+4 = 0 ⇒
1 +
2∏
t=0
x−(k+1)t−k
(
3∏
t=0
x−(k+1)t−k)/x−k
=
∞∑
j=0
4j+3∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
. (10)
From the equation (7) and (8),
1 +
2∏
t=0
x−(k+1)t−k
2∏
t=0
x−(k+1)t−k
=
∞∑
j=0
4j∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
(11)
384 On the recursive sequence. . .
>
1 +
2∏
t=0
x−(k+1)t−k
(
3∏
t=0
x−(k+1)t−k)/x−(3k+2)
=
∞∑
j=0
4j+1∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
thus x−(4k+3) > x−(3k+2).
From the equation (8) and (9),
1 +
2∏
t=0
x−(k+1)t−k
(
3∏
t=0
x−(k+1)t−k)/x−(3k+2)
=
∞∑
j=0
4j+1∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
>
1 +
2∏
t=0
x−(k+1)t−k
(
3∏
t=0
x−(k+1)t−k)/x−(2k+1)
=
∞∑
j=0
4j+2∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
, (12)
thus, x−(3k+2) > x−(2k+1).
From the equation (9) and (10),
1 +
2∏
t=0
x−(k+1)t−k
(
3∏
t=0
x−(k+1)t−k)/x−(2k+1)
=
∞∑
j=0
4j+2∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
>
1 +
2∏
t=0
x−(k+1)t−k
(
3∏
t=0
x−(k+1)t−k)/x−k
=
∞∑
j=0
4j+3∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t−k
(13)
thus, x−(2k+1) > x−k, x−(4k+3) > x−(3k+2) > x−(2k+1) > x−k. We arrive
at a conradiction.
Suppose that ak+1 = a2k+2 = a3k+3 = a4k+4 = 0.
From that the equation (14) in (e) follows, Proof of the equation (11)
is similar and will be omitted.
1 +
2∏
t=0
x−(k+1)t
2∏
t=0
x−(k+1)t
=
∞∑
j=0
4j∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t
D. Simsek, F. Abdullayev 385
>
1 +
2∏
t=0
x−(k+1)t
(
3∏
t=0
x−(k+1)t)/x−(2k+2)
=
∞∑
j=0
4j+1∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t
, (14)
thus, x−(3k+3) > x−(2k+2).
From that the equation (15) in (e) follows, the proof of the equation
(12) is similar and will be omitted.
1 +
2∏
t=0
x−(k+1)t
(
3∏
t=0
x−(k+1)t)/x−(2k+2)
=
∞∑
j=0
4j+1∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t
>
1 +
2∏
t=0
x−(k+1)t
(
3∏
t=0
x−(k+1)t)/x−(k+1)
=
∞∑
j=0
4j+2∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t
, (15)
thus, x−(2k+2) > x−(k+1).
From that the equation (16) in (e) follows, the proof of the equation
(13) is similar and will be omitted.
1 +
2∏
t=0
x−(k+1)t
(
3∏
t=0
x−(k+1)t)/x−(k+1)
=
∞∑
j=0
4j+2∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t
>
1 +
2∏
t=0
x−(k+1)t
(
3∏
t=0
x−(k+1)t)/x0
=
∞∑
j=0
4j+3∏
i=1
1
1 +
2∏
t=0
x(k+1)i−(k+1)t
(16)
thus, x−(k+1) > x0, x−(3k+3) > x−(2k+2) > x−(k+1) > x0. We arrive at a
conradiction which completes the proof of theorem.
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Contact information
Dağıstan Simsek Kyrgyz-Turkish Manas University,
Bishkek, Kyrgyzstan
&
Selcuk University
Konya, Turkey
E-Mail: dagistan.simsek@manas.edu.kg
dsimsek@selcuk.edu.tr
Fahreddin
Abdullayev
Kyrgyz-Turkish Manas University,
Bishkek, Kyrgyzstan
&
Mersin University
Mersin, Turkey
E-Mail: fabdul@mersin.edu.tr
fahreddin.abdullayev@manas.edu.kg
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