Write as a single sum:
Given $\{a_n\}_n$, $a_i \in \mathbb{Q}$, $0 \lt a_i \le a_{i+1}$
$\sum_{i=1}^{n} \sum_{j=i+1}^{n} \lfloor a_j - a_i \rfloor$
I am not sure if this is possible.
I know that if there is no floor, then the sum can be written as:
$\sum_{i=1}^{n} (2i-n-1) a_i$