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Balkan MO
2020 Balkan MO
4
4
Part of
2020 Balkan MO
Problems
(1)
Problem 4, BMO 2020
Source: Problem 4, BMO 2020
11/1/2020
Let
a
1
=
2
a_1=2
a
1
=
2
and, for every positive integer
n
n
n
, let
a
n
+
1
a_{n+1}
a
n
+
1
be the smallest integer strictly greater than
a
n
a_n
a
n
that has more positive divisors than
a
n
a_n
a
n
. Prove that
2
a
n
+
1
=
3
a
n
2a_{n+1}=3a_n
2
a
n
+
1
=
3
a
n
only for finitely many indicies
n
n
n
. Proposed by Ilija Jovčevski, North Macedonia
number theory
BMO