Time Complexity Analysis

When a function goes in the form of

for(int i=1;i<=n;++i)
{
    for(int j=2*i;j<=n;j+=i)
    cout<<(i,j);
}

$\sum_{i=0}^n \frac{n}{i} \implies n*\sum_{i=0}^n \frac{1}{i}$

The Summation part takes O(logN) hence the total time is O(N*logN)