Another Case study

Consider $n$ between two perfect squares $k^2$ and $(k+1)^2$ that is $k^2 \le n \le (k+1)^2$ or $k^2 \le n \le k^2+2k+1$ . Then we have $a_{k^2} = \left \lfloor \dfrac {k^2}{\lfloor \sqrt{k^2} \rfloor} \right \rfloor = \left \lfloor \dfrac {k^2}k \right \rfloor = k$ . Similarly, $a_{(k+1)^2} = k+1$ . We note that:

$\begin{aligned} a_{k^2} & = k \\ a_{k^2+1} & = \left \lfloor \frac {k^2+1}{\lfloor \sqrt{k^2+1} \rfloor} \right \rfloor = \left \lfloor \frac {k^2+1}k \right \rfloor = k \\ a_{k^2+2} & = \left \lfloor \frac {k^2+2}k \right \rfloor = k \\ \cdots \ & = \ \cdots \\ a_{k^2+k-1} & = \left \lfloor \frac {k^2+k-1}k \right \rfloor = k \\ a_{k^2+k-1} & = \left \lfloor \frac {k^2+k}k \right \rfloor = \color{#3D99F6} k + 1 \\ a_{k^2+k+1} & = \left \lfloor \frac {k^2+k+1}k \right \rfloor = \color{#3D99F6} k + 1 \\ \cdots \ & = \ \cdots \\ a_{k^2+2k-1} & = \left \lfloor \frac {k^2+2k-1}k \right \rfloor = \color{#3D99F6} k + 1 \\ a_{k^2+2k} & = \left \lfloor \frac {k^2+2k}k \right \rfloor = \color{#D61F06} k + 2 \\ a_{(k+1)^2} & = \color{#3D99F6} k + 1 \end{aligned}$

This implies that $a_n > a_{n+1}$ , whenever, $n = k^2 - 1$ , where $k \ge 2$ . Since $\lfloor \sqrt {2010} \rfloor = 44$ , there are $\boxed{43}$ acceptable $k$ 's.