Problem 1859

For every $k \geqslant 2$ define the quotient $b_{k} = 2(a_{1} + \cdots + a_{k-1}) / a_{k}$, which must be a positive integer. We first prove the following properties of the sequence $(b_{k})$:

Claim 1. We have $b_{k+1} \leqslant b_{k} + 1$ for all $k \geqslant 2$.

Proof. By subtracting $b_{k} a_{k} = 2(a_{1} + \cdots + a_{k-1})$ from $b_{k+1} a_{k+1} = 2(a_{1} + \cdots + a_{k})$, we find that $b_{k+1} a_{k+1} = b_{k} a_{k} + 2 a_{k} = (b_{k} + 2) a_{k}$. From $a_{k} < a_{k+1}$ it follows that $b_{k} + 2 > b_{k+1}$.

Claim 2. The sequence $(b_{k})$ is unbounded.

Proof. We start by rewriting $b_{k+1} a_{k+1} = (b_{k} + 2) a_{k}$ as
$$a_{k+1} = a_{k} \cdot \frac{b_{k} + 2}{b_{k+1}} \Longrightarrow a_{k+1} \mid a_{k}(b_{k} + 2)$$
If the sequence $(b_{k})$ were bounded, say by some positive integer $B$, then the prime factors of the terms of the sequence $(a_{k})$ could only be primes less than or equal to $B + 2$ or those dividing $a_{1}$ or $a_{2}$, which contradicts the property in the statement of the problem.

Consider now an arbitrary positive integer $n$. We assume $n > b_{2}$, otherwise we replace $n$ by an arbitrary multiple of $n$ that is bigger than $b_{2}$. By Claim 2, there exists $k$ such that $b_{k+1} \geqslant n$. Consider the smallest such $k$. From Claim 1, it follows that we must have $b_{k} = n - 1$ and $b_{k+1} = n$ (we assumed $n > b_{2}$ to ensure that $k \geqslant 2$). We now find that
$$a_{k+1} = a_{k} \cdot \frac{b_{k} + 2}{b_{k+1}} = a_{k} \cdot \frac{n + 1}{n}.$$
Because $n$ and $n + 1$ are coprime, this immediately implies that $a_{k}$ is divisible by $n$.