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1986 IMO Longlists
22
Common divisors of two terms of the sequence
Common divisors of two terms of the sequence
Source:
August 29, 2010
induction
algebra proposed
algebra
Problem Statement
Let
(
a
n
)
n
≥
0
(a_n)_{n \geq 0}
(
a
n
)
n
≥
0
be the sequence of integers defined recursively by
a
0
=
0
,
a
1
=
1
,
a
n
+
2
=
4
a
n
+
1
+
a
n
a_0 = 0, a_1 = 1, a_{n+2} = 4a_{n+1} + a_n
a
0
=
0
,
a
1
=
1
,
a
n
+
2
=
4
a
n
+
1
+
a
n
for
n
≥
0.
n \geq 0.
n
≥
0.
Find the common divisors of
a
1986
a_{1986}
a
1986
and
a
6891
.
a_{6891}.
a
6891
.
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