Answer: 486

Small Hint:

Try building a quasi-increasing permutation of $1, \ldots, n$ by inserting $n$ into a quasi-increasing permutation of $1, \ldots, n-1$

Big Hint:

The $n$ can go immediately before $n-1,$ immediately before $n-2,$ or at the very end — always exactly $3$ places, so the count triples with each new $n$

Solution:

Let $S_n$ be the number of quasi-increasing permutations of $1, \ldots, n.$ Insert $n$ into a quasi-increasing permutation of $1, \ldots, n - 1:$ the entry following $n$ must be at least $n - 2,$ so $n$ can go immediately before $n - 1,$ immediately before $n - 2,$ or at the very end — exactly $3$ positions, and each insertion keeps every other adjacent condition intact.

Conversely, deleting $n$ from a quasi-increasing permutation of $1, \ldots, n$ leaves a quasi-increasing permutation of $1, \ldots, n - 1,$ since the entries around the deleted $n$ satisfy $a_{k-1} \le n - 1 \le a_{k+1} + 2$ when $n \ge 3.$ So $S_n = 3S_{n-1}$ for $n \ge 3.$

Since $S_2 = 2,$ we get $S_7 = 2 \cdot 3^5 = 486.$