杜教筛之逆运算1

\[\sum^n_{i=1}\lfloor\frac{n}{i}\rfloor f(i)=\sum^n_{i=1}I*f(i) \]

证明:
设$$F(x)=\sum^x_{i=1}\lfloor\frac{x}{i}\rfloor f(i)$$
则

\[F(x)-F(x-1)=\sum^x_{i=1}\lfloor\frac{x}{i}\rfloor f(i)-\sum^{x-1}_{i=1}\lfloor\frac{x-1}{i}\rfloor f(i) \]

\[=\sum^{x-1}_{i=1}(\lfloor\frac{x}{i}\rfloor-\lfloor\frac{x-1}{i}\rfloor)f(i)+f(x) \]

\[\text{又因为}\lfloor\frac{x}{i}\rfloor-\lfloor\frac{x-1}{i}\rfloor=1\text{当且仅当}i | x \]

所以

\[F(x)-F(x-1)=\sum_{d|x}f(d)=I*f(x) \]

\[=>F(x)=F(1)+\sum_{i=2}^xI*f(i)=\sum^x_{i=1}I*f(i) \]

即

\[\sum^n_{i=1}\lfloor\frac{n}{i}\rfloor f(i)=\sum^n_{i=1}I*f(i) \]

其中\(I\)为恒等函数\(I(x)=1\)

应用举例:

\[1.\sum^n_{i=1}\sum_{j=1}^n\mu(i)\phi(j)\lfloor\frac{n}{ij}\rfloor=\sum_{T=1}^n\lfloor\frac{n}{T}\rfloor\mu*\phi(T)=\sum_{i=1}^n\phi(i) \]

\[2.\sum^n_{i=1}\lfloor\frac{n}{i}\rfloor\phi(i)=\sum_{i=1}^ni=n*(n+1)/2 \]

还有很多式子可以用其化简(基本上可以和\(I\)卷的都可以)

实际上，这就是杜教筛式子特殊化的逆运算

杜教筛式子：$$\sum_{i=1}^{nf(i)\sum_{j=1}}\rfloor}g(i)=\sum_{i=1}^nf*g(i)$$

若把\(g(i)\)改为\(I(i)\)则可得上述式子

杜教筛式子在此就不证明，很好证的