Let $x \in (1,\infty)\setminus \mathbb{Z}$ be any non-integer greater than one. Let $n\in \mathbb{Z}^+$ be any positive integer such that $\lfloor (n-1)x \rfloor +1 < \lfloor nx \rfloor$ (for example, if $x=\sqrt{2}$, then $n$ might be $5$ since $\lfloor 4\sqrt{2} \rfloor +1 = 6 < \lfloor 5\sqrt{2} \rfloor = 7$). How many sequences of integers $\displaystyle (a_k)_{k=1}^n$ satisfy the following properties?

1. $0<a_1<\cdots < a_n < nx$

2. For each $k=1,\ldots, n-1$, $a_k > kx$

In the example above, with $x=\sqrt{2}, n=5$, there are exactly three such sequences: $\{3,4,5,6,7\}$, $\{2,4,5,6,7\}$, and $\{2,3,5,6,7\}$.

If $n=8$, there are 12 such sequences.

Is there a simple way to compute this? I would imagine there is not a closed form for the solution, but I would like a faster way to calculate it than brute forcing it by actually finding all sequences that satisfy the conditions.