, 2015

The answer is in the affirmative. The idea is to consider a sequence of pairwise distinct primes $p_1, p_2, p_3, \dots$, cover the positive integers by a sequence of finite non-empty sets $I_n$ such that $I_m$ and $I_n$ are disjoint if and only if $m$ and $n$ are one unit apart, and set $a_n = \prod_{i \in I_n} p_i$, $n = 1, 2, 3, \dots$. For instance, we may take
$$I_n = \{2n - 4k - 1: k = 0, 1, \dots, \lfloor(n-1)/2\rfloor\} \cup \{2n - 4k - 2: k = 1, 2, \dots, \lfloor n/2 \rfloor - 1\} \cup \{2n\},$$
$n = 1, 2, 3, \dots$, where the set in the middle is understood to be empty for $n = 1, 2, 3$. It is readily checked that $I_n$ and $I_{n+1}$ are disjoint for every index $n$, and, if the indices $m$ and $n$ are at least two units apart, then $I_m$ and $I_n$ both contain $m + n - |m-n| - e$, where $e = 0$ if $m$ and $n$ have opposite parities, and $e = 1$ otherwise. This ends the proof.