$$\sum_{d=1}^nf(d)\sum_{p=1}^{\lfloor{\frac{n}{d}}\rfloor}g(p)h(dp) = \sum_{T=1}^nh(T)\sum_{d|T}f(d)g(\frac{T}{d})$$

$$g(1)S(n)=\sum_{i=1}^n(f*g)(i)-\sum_{i=2}^ng(i)*S(\frac{n}{d})$$

$$其中有S(n)=\sum_{i=1}^nf(i)$$

$$\mu * 1 = \epsilon$$

$$\varphi * 1 = id$$

$$\mu * id = \varphi$$