Problem:
Let a1,a2,… be a sequence for which
a1=2,a2=3, and an=an−2an−1 for each positive integer n≥3
What is a2006?
Answer Choices:
A. 21
B. 32
C. 23
D. 2
E. 3
Solution:
Note that the first several terms of the sequence are:
2,3,23,21,31,32,2,3,…
so the sequence consists of a repeating cycle of 6 terms. Since 2006=334⋅6+2, we have a2006=a2=(E)3.
The problems on this page are the property of the MAA's American Mathematics Competitions