1 Answer

  1. Gabriel- Reply

    2019-11-14

    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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