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

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

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**Notation:** \( \lfloor \cdot \rfloor \) denotes the floor function.

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