Floor & floor

Let natural number $n = \lfloor x \lfloor x \rfloor \rfloor$ , then we have:

$\begin{aligned} n & = \lfloor x \lfloor x \rfloor \rfloor \\ & = (\lfloor x \rfloor + {\color{#3D99F6}\{x\}}) \lfloor x \rfloor & \small \color{#3D99F6} \text{where } 0 \le \{x\} < 1 \text{ is the fractional part of }x. \\ \implies \lfloor x \rfloor ^2 \le n & {\color{#D61F06}<} (\lfloor x \rfloor + 1) \lfloor x \rfloor \\ \lfloor x \rfloor ^2 \le n & \le \lfloor x \rfloor ^2 + \lfloor x \rfloor - 1 \end{aligned}$

Therefore, if natural number $k = \lfloor x \rfloor$ , then for a $k$ , the valid $n$ 's are $k^2, k^2+1, k^2+2, ... k^2+k-1$ , that is a total $k$ valid $n$ 's. While the smallest $k_{min}=1$ , the largest $k_{max} = \lfloor \sqrt{2016} \rfloor = 44$ . Then the number of $n$ 's between 1 and 2016 inclusive is as follows:

$N = \displaystyle \sum_{k=1}^{44} k = \frac {44(44+1)}2 = \boxed{990}$

By definition, $\lfloor x \rfloor ≤ x < \lfloor x \rfloor + 1$ . Therefore, $\lfloor x \rfloor ^ 2 ≤ x \lfloor x \rfloor < (\lfloor x \rfloor + 1) \lfloor x \rfloor$ . As both $\lfloor x \rfloor ^ 2$ and $(\lfloor x \rfloor + 1) \lfloor x \rfloor$ are integers, $\lfloor x \rfloor ^ 2 ≤ \lfloor x \lfloor x \rfloor \rfloor < (\lfloor x \rfloor + 1) \lfloor x \rfloor$ . Numbers between $n^2$ and $n^2+n-1$ , inclusive, are the answer, while $n \in [1,44]$ , for $44 \times 45 < 2016 < 45^2$ . That gives $1+2+3+…+44=\frac{45 \times 44}{2}=990$ numbers.