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International Contests
IMO Longlists
1988 IMO Longlists
45
45
Part of
1988 IMO Longlists
Problems
(1)
g(n+2) = g(n) + g(n+1) + 1
Source: IMO LongList 1988, Ireland 4, Problem 45 of ILL
11/3/2005
Let
g
(
n
)
g(n)
g
(
n
)
be defined as follows:
g
(
1
)
=
0
,
g
(
2
)
=
1
g(1) = 0, g(2) = 1
g
(
1
)
=
0
,
g
(
2
)
=
1
and
g
(
n
+
2
)
=
g
(
n
)
+
g
(
n
+
1
)
+
1
,
n
≥
1.
g(n+2) = g(n) + g(n+1) + 1, n \geq 1.
g
(
n
+
2
)
=
g
(
n
)
+
g
(
n
+
1
)
+
1
,
n
≥
1.
Prove that if
n
>
5
n > 5
n
>
5
is a prime, then
n
n
n
divides
g
(
n
)
⋅
(
g
(
n
)
+
1
)
.
g(n) \cdot (g(n) + 1).
g
(
n
)
⋅
(
g
(
n
)
+
1
)
.
quadratics
number theory unsolved
number theory