# Another Case study

$\large a_n= \left \lfloor \dfrac n{\lfloor \sqrt n\rfloor} \right \rfloor$

For each natural number $$n$$, define $$a_n$$ as above. Find the number of $$n \le 2010$$ for which $$a_{n} > a_{n+1}$$.


Notation: $$\lfloor \cdot \rfloor$$ denotes the floor function.

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