That's as close as I can get to pinning down the value of . Its integer part (when n is a perfect square) must be either or .
Second attempt at this problem. We are interested in the sum when is a perfect square, say . Then the difference between successive sums of this form is
where . As in the previous comment, this is a lower Riemann sum for
(because ). Therefore .
The aim now is to show by induction that the integer part of is . The base case is k=2, and for that you need to verify that , so that .
For the inductive step, suppose that . The above argument shows that , so that . The reverse inequality comes from the previous comment, where I showed that .
Therefore the integer part of , when n is a perfect square greater than 1, is given by .