$$\sum_{d=1}^nf(d)\sum_{p=1}^{\lfloor{\frac{n}{d}}\rfloor}g(p)h(dp) = \sum_{T=1}^nh(T)\sum_{d|T}f(d)g(\frac{T}{d})$$
$$g(1)S(n)=\sum_{i=1}^n(f*g)(i)-\sum_{i=2}^ng(i)*S(\frac{n}{d})$$
$$其中有S(n)=\sum_{i=1}^nf(i)$$
$$\mu * 1 = \epsilon$$
$$\varphi * 1 = id$$
$$\mu * id = \varphi$$