莫比乌斯反演学习笔记

莫比乌斯反演，之前做过一些题，一直没有太理解，膜了下faebdc学长的姿势，终于搞懂了一些。
首先我们有两个式子：
1:$\sum_{d|n} \phi(d)=n$2:$\sum_{d|n} \mu(d)=e(n)$
1式证明:对于$n$的质因数$x$对$\phi(n)$贡献了$(x-1)*x^{t-1}$
单独对于$x$而言约数可以为$x^{0},x^{1},...,x^{t}$,设约数$x^{t-1}$满足以上式子；
则对于$x^{t}$而言有$x^{t-1}+(x-1)*x^{t-1}=x^t$，同样成立，归纳法得证。
2式证明，这与莫比乌斯函数性质有关。
然后我们就可以推式子了：
upd:$\sum_{i=1}^{n} i*e(gcd(i,n))={\phi(n)*n \over 2}$
1Dgcd
$\sum_{i=1}^{n} gcd(i,n)=\sum_{i=1}^{n}\sum_{d|gcd(i,n)}\phi(d)=\sum_{i=1}^{n}\sum_{d|i,d|n}\phi(d)=\sum_{d|n}\phi(d) \lfloor {n \over d} \rfloor$
2Dgcd
$\sum_{i=1}^{n}\sum_{j=1}^{m}gcd(i,j)=\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{d|i,d|j}\phi(d)=\sum_{d=1}^{min(n,m)}\phi(d)\lfloor {n \over d} \rfloor\lfloor {m \over d} \rfloor$
1D[gcd==1]
$\sum_{i=1}^{n}e(gcd(i,n))=\sum_{i=1}^{n}\sum_{d|i,d|n}\mu(d)=\sum_{d|n}\mu(d)\lfloor{n\over d}\rfloor$
2D[gcd==k]
$\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)==k]=\sum_{i=1}^{\lfloor {n\over k}\rfloor}\sum_{j=1}^{\lfloor {m\over k}\rfloor}e(gcd(i,j))=\sum_{i=1}^{\lfloor {n\over k}\rfloor}\sum_{j=1}^{\lfloor {m\over k}\rfloor}\sum_{d|i,d|j}\mu(d)=\sum_{i=1}^{min(\lfloor {n\over k}\rfloor,\lfloor {m\over k}\rfloor)}\mu(d)\lfloor {n\over kd}\rfloor\lfloor {m\over kd}\rfloor$
1D lcm
kind 1
$\sum_{i=1}^{n} lcm(i,n)=\sum_{i=1}^{n} {i*n \over gcd(i,n)}=\sum_{i=1}^{n} \sum_{d|n}[gcd(i,j)==d] {i*n \over d}=n*\sum_{d|n} \sum_{i=1}^{\lfloor {n \over d} \rfloor} e(gcd(i,{n \over d}))$
$=n*\sum_{d|n}\sum_{i=1}^{\lfloor {n \over d} \rfloor} i \sum_{k|{n \over d},k|i} \mu(k)=n*\sum_{d|n} \sum_{k|{n \over d}} k*\mu(k) {({\lfloor {n\over dk} \rfloor}+1)*{\lfloor {n\over dk} \rfloor} \over 2}$
kind 2
$\sum_{i=1}^{n} lcm(i,n)=\sum_{i=1}^{n} {i*n \over gcd(i,n)}=\sum_{i=1}^{n} \sum_{d|n}[gcd(i,j)==d] {i*n \over d}=n*\sum_{d|n} \sum_{i=1}^{\lfloor {n \over d} \rfloor} e(gcd(i,{n \over d}))$
$=n*\sum_{d|n}\sum_{i=1}^{d}i*e(gcd(i,d))=n*\sum_{d|n} {\phi(d)*d \over 2}$
2D lcm
我们定义$sum(n,m)=\sum_{i=1}^{n}\sum_{j=1}^{m}i*j$
$\sum_{i=1}^{n}\sum_{j=1}^{m} lcm(i,j)=\sum_{i=1}^{n}\sum_{j=1}^{m}{i*j\over gcd(i,j)}$
$=\sum_{i=1}^{\lfloor {n\over d}\rfloor}\sum_{j=1}^{\lfloor {m\over d}\rfloor} {di*dj \over d}e(gcd(i,j))$
$=\sum_{d=1}^{min(n,m)}d\sum_{i=1}^{\lfloor {n\over d}\rfloor}\sum_{j=1}^{\lfloor {m\over d}\rfloor} i*j e(gcd(i,j))$
$=\sum_{d=1}^{min(n,m)}d \sum_{k=1}^{min(\lfloor {n\over d} \rfloor,\lfloor {m\over d} \rfloor)}\mu(k)*k^2*sum(\lfloor {\lfloor {n\over d} \rfloor \over k} \rfloor,\lfloor {\lfloor {m\over d} \rfloor \over k} \rfloor)$