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Onaclassofrationaldifference...
15
(b)(I1jI2j...jI3k+3jI1jI2j...jI3k+3j...)isasolutionwithperiod(3k+
3)ofEq.(1).
(c)Im·Im+(k+1)·Im+(2k+2)10form11j2j...j(k+1)
(d)Ifthereexistsno∈Nsuchthat
xn1(2k+1)≥xn+1
foralln>no,then
n→∞
lim
xn10.
(e)Thefollowingformulas
x(3k+3)n+b1xb1(3k+3)(1−
1+xb1(k+1)xb1(2k+2)
xb1(k+1)xb1(2k+2)
×
Σ
jlo
n
Π
il1
3j
1+x(k+1)i+b1(k+1)x(k+1)i+b1(2k+2)
1
)
x(3k+3)n+(k+1)+b1xb1(2k+2)(1−
1+xb1(k+1)xb1(2k+2)
xb1(k+1)xb1(3k+3)
×
Σ
jlo
n
3j+1
il1
Π
1+x(k+1)i+b1(k+1)x(k+1)i+b1(2k+2)
1
)
x(3k+3)n+b+(2k+2)1xb1(k+1)(1−
1+xb1(k+1)xb1(2k+2)
xb1(2k+2)xb1(3k+3)
×
Σ
jlo
n
3j+2
Π
il1
1+x(k+1)i+b1(k+1)x(k+1)i+b1(2k+2)
1
)
holdforb11j2j...jk+1.
Proof.(I)Firstly,fromtheEq.(1),weobtain
xn+1(1+xn1kxn1(2k+1))1xn1(3k+2).
Sincexn1kjxn1(2k+1)∈(0j∞),then(1+xn1kxn1(2k+1))∈(1j∞).So,
xn+1<xn1(3k+2).
Thisimpliesthat
0<...<x6k+7<x3k+4<x1<x1(3k+2)
0<...<x6k+8<x3k+5<x2<x1(3k+1)
...
0<...<x9k+8<x6k+5<x3k+2<x11
0<...<x9k+9<x6k+6<x3k+3<xo
