Solution: The answer is yes. There are many ways to denote such a sequence, let me give an example. Denote $p_n$ as the $p$th prime number, then we can let, for all positive integers $n \ge 3$, \[\ a_n=p_{(101-n+1)}*p_{(101-n)}*\prod_{i=1}^{n} p_{m+i} \] and let \[\ a_1=p_{101}, a_2=p_{101}p_{100} \] where $m$ is an arbitrary large positive integer (i.e $500$ ) We have, \[\ (a_k, a_{k-1})=p_{100-k+1} \] and \[\ (a_k, a_{k+1})=p_{100-k} \] thus, $(a_k, a_{k-1}) > (a_k, a_{k+1})$ , and obviously $a_n > a_{n-1}$ due my defintion of $a_n$, which makes the $a_n$ way larger than $a_{n-1}$. $\Box$ Remark: There are much more ways to denote such a sequence, and mine is pretty loose- anyways, at least it still satisfies the problem