The outer loop runs $n$ times. The inner loop runs $\lfloor \frac{n}{i} \rfloor$ times. So it would be $O(n \cdot \lfloor \frac{n}{i} \rfloor)$. I do not know how to transform that into a proper expression involving Big Oh and n. Maybe it could be first expressed as a sum:
$\qquad\displaystyle \sum_{i=1}^n \sum_{j=i}^n \frac{n}{j}$.
for i = 1 to n:
j = i
while j ≤ n:
A[j] = A[j] + 1
j = j + i
Could you please indicate, apart from the actual running time of the algorithm, how to show the said running time?
