# Prove that every element of $a_{n+2013}=\frac{a_{n+1}a_{n+2}…a_{n+2012}+1}{a_n}$ is an integer

Given $\displaystyle a_1=a_2=\cdots=a_{2013}=1$ and $\displaystyle a_{n+2013}=\frac{a_{n+1}a_{n+2}\cdots a_{n+2012}+1}{a_n}$.

Prove that $a_{n+2013}\in\mathbb{N}$ for all $n\in\mathbb{N}$.

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This looks very much like it is from a competition, and if it is, then asking for help here probably violates the rules of said competition. Where did you come by this question? –  Arthur Dec 10 '12 at 13:27
I get this from my friend. –  ᴊ ᴀ s ᴏ ɴ Dec 10 '12 at 13:51
It's so weird that no one have any idea... –  ᴊ ᴀ s ᴏ ɴ Dec 11 '12 at 3:26

For these kind of questions, it is clear that $2013$ is arbitrary. Let us replace $2013$ by $k \geq 2$, so $$a_1, a_2, \ldots , a_k=1, a_{n+k}=\frac{a_{n+1}a_{n+2} \ldots a_{n+k-1}+1}{a_n}, n \geq 1$$

It is clear that $a_{n+k}>0$.

We prove by induction on $n \geq 1$ that $a_{n+k} \in \mathbb{Z}$.

When $1 \leq n \leq k$, $a_n=1$, so clearly $a_{n+k}=\frac{a_{n+1}a_{n+2} \ldots a_{n+k-1}+1}{a_n}=(a_{n+1}a_{n+2} \ldots a_{n+k-1}+1) \in \mathbb{Z}$.

Suppose that the statement holds for $1 \leq n \leq m, m \geq k$. Then taking $\pmod{a_{m+1}}$, \begin{align} & a_{m+2}a_{m+3} \ldots a_{m+k}+1 \\ & \equiv -(a_{m+1}a_{m+1-k}-1)(a_{m+2}a_{m+3} \ldots a_{m+k})+1\\ & \equiv -(a_{m-(k-2)}a_{m-(k-3)} \ldots a_{m})(a_{m+2}a_{m+3} \ldots a_{m+k})+1 \\ & \equiv -\prod_{i=1}^{k-1}{(a_{m+1-k+i}a_{m+1+i})}+1 \\ & \equiv -\prod_{i=1}^{k-1}{(a_{(m+1-k+i)+1}a_{(m+1-k+i)+2} \ldots a_{m+1} \ldots a_{(m+1-k+i)+(k-1)}+1)}+1 \\ & \equiv 0 \pmod{a_{m+1}} \end{align}

Thus $a_{(m+1)+k}=\frac{a_{m+2}a_{m+3} \ldots a_{m+k}+1}{a_{m+1}} \in \mathbb{Z}$, and we are done by induction.

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