With Du teach sieve equation we can do a lot of things
\[\sum_{i=1}^nf*g(i)=\sum_{i=1}^nf(i)\sum_{j=1}^{\lfloor\frac{n}{i}\rfloor}g(j)\]
For this two forms of expression can spike with Du teach sieve formula
\[ 1.\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor f(i)\\ 2.\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor^2 f(i) \]
The first formula
Du will teach sieve \ (g () \) special into \ (I (): I ( x) = 1 \) was
\[\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor f(i)=\sum_{i=1}^nI*f(i)\]
The second formula
To \ (F () \) on a volume \ (id (): id ( x) = x \) to give
\[ \begin{align} \sum_{i=1}^nid*f(i)&=\sum_{i=1}^nf(i)\sum_{j=1}^{\lfloor\frac{n}{i}\rfloor}id(j)\\ &=\sum_{i=1}^nf(i)\frac{\lfloor\frac{n}{i}\rfloor\times(\lfloor\frac{n}{i}\rfloor+1)}{2}\\ &=\frac{\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor^2 f(i)+\sum_{i=1}^nI*f(i)}{2} \end{align} \]
\[\sum_{i=1}^n\lfloor\frac{n}{i}\rfloor^2 f(i)=2\times\sum_{i=1}^nid*f(i)-\sum_{i=1}^nI*f(i)\]