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All-Russian Olympiad Regional Round
2000 All-Russian Olympiad Regional Round
10.6
10.6
Part of
2000 All-Russian Olympiad Regional Round
Problems
(1)
a_{n+1} = a^2_n-5 for odd a_n - All-Russian MO 2000 Regional (R4) 10.6
Source:
9/26/2024
Given a natural number
a
0
a_0
a
0
, we construct the sequence
{
a
n
}
\{a_n\}
{
a
n
}
as follows
a
n
+
1
=
a
n
2
−
5
a_{n+1} = a^2_n-5
a
n
+
1
=
a
n
2
−
5
if
a
n
a_n
a
n
is odd, and
a
n
2
\frac{a_n}{2}
2
a
n
if
a
n
a_n
a
n
is even. Prove that for any odd
a
0
>
5
a_0 > 5
a
0
>
5
in the sequence
{
a
n
}
\{a_n\}
{
a
n
}
arbitrarily large numbers will occur.
recurrence relation
number theory