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Problem:

If $a_0 = 1 , a_1 = 1$ and $a_n = a_{n – 1} a_{n – 2} + 1$ for $n > 1$, then:

(A) $a_{465}$ is odd and $a_{466}$ is even;
(B) $a_{465}$ is odd and $a_{466}$ is odd;
(C) $a_{465}$ is even and $a_{466}$ is even;
(A) $a_{465}$ is even and $a_{466}$ is odd;

Discussion:

First we note a pattern and then we prove that the pattern actually holds.

Note that:

$a_0 = 1$ is odd
$a_1 = 1$ is odd
$a_2 = a_0 a_1 + 1 = 1\times 1 + 1 = 2$ is even
$a_3 = a_1 a_2 + 1 = 1 \times 2 + 1 = 3$ is odd
$a_4 = a_2 a_3 + 1 = 2 \times 3 + 1 = 7$ is odd
$a_5 = a_3 a_4 + 1 = 3 \times 7 + 1 = 22$ is even

So the pattern that we observe is the following order: odd, odd, even, odd, odd, even…

We show this by strong form of induction. Suppose this pattern holds true for all n upto n = 3k+2

(that is $a_{3k+2} = even , a_{3k+1} = odd, a_{3k}= odd$ ).

Our computations show that this is true for k =1 (so for initial value it is true).

Let us show for the next three values:

$a_{3k+3} = a_{3k+2} \times a_{3k+1} + 1 = even \times odd + 1 = odd$
$a_{3k+4} = a_{3k+3} \times a_{3k+2} + 1 = odd \times even + 1 = odd$
$a_{3k+5} = a_{3k+4} \times a_{3k+3} + 1 = odd \times odd + 1 = even$

Thus we showed that whenever the index is of the form 3j+2, the number is even, otherwise if the index is of the form 3j or 3j+1, the term is odd.

Since 465 and 466 are respectively of the form 3j and 3j+1, hence
$a_{465}$ and $a_{466}$ both are odd.

## Chatuspathi:

• What is this topic: Induction
• What are some of the associated concepts: Strong form of induction
• Where can learn these topics: Cheenta I.S.I. & C.M.I. courseMath Olympiad Program discusses these topics in the ‘Induction’ module.
• Book Suggestions: Elementary number theory by David Burton