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National and Regional Contests
China Contests
China Team Selection Test
1991 China Team Selection Test
2
f(n+2) = 23 * f(n+1) + f(n)
f(n+2) = 23 * f(n+1) + f(n)
Source: China TST 1991, problem 5
June 27, 2005
function
number theory unsolved
number theory
Problem Statement
Let
f
f
f
be a function
f
:
N
∪
{
0
}
↦
N
,
f: \mathbb{N} \cup \{0\} \mapsto \mathbb{N},
f
:
N
∪
{
0
}
↦
N
,
and satisfies the following conditions: (1)
f
(
0
)
=
0
,
f
(
1
)
=
1
,
f(0) = 0, f(1) = 1,
f
(
0
)
=
0
,
f
(
1
)
=
1
,
(2)
f
(
n
+
2
)
=
23
⋅
f
(
n
+
1
)
+
f
(
n
)
,
n
=
0
,
1
,
…
.
f(n+2) = 23 \cdot f(n+1) + f(n), n = 0,1, \ldots.
f
(
n
+
2
)
=
23
⋅
f
(
n
+
1
)
+
f
(
n
)
,
n
=
0
,
1
,
…
.
Prove that for any
m
∈
N
m \in \mathbb{N}
m
∈
N
, there exist a
d
∈
N
d \in \mathbb{N}
d
∈
N
such that
m
∣
f
(
f
(
n
)
)
⇔
d
∣
n
.
m | f(f(n)) \Leftrightarrow d | n.
m
∣
f
(
f
(
n
))
⇔
d
∣
n
.
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