What is most efficient way to find the sum of the series: (N/2^2)*2^2+(N/3^2)*3^2+(N/4^2)*4^2+........
Efficient computation of $\sum_{i=1}^{i=\left \lfloor {\sqrt{N}} \right \rfloor}\left \lfloor \frac{N}{i^{2}} \right \rfloor$