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:∑d|nϕ(d)=n\sum_{d|n} \phi(d)=n2:∑d|nμ(d)=e(n)\sum_{d|n} \mu(d)=e(n)

1式证明:对于nn的质因数xx对ϕ(n)\phi(n)贡献了(x−1)∗xt−1(x-1)*x^{t-1}

单独对于xx而言约数可以为x0,x1,...,xtx^{0},x^{1},...,x^{t},设约数xt−1x^{t-1}满足以上式子；

则对于xtx^{t}而言有xt−1+(x−1)∗xt−1=xtx^{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:∑ni=1i∗e(gcd(i,n))=ϕ(n)∗n2\sum_{i=1}^{n} i*e(gcd(i,n))={\phi(n)*n \over 2}

1Dgcd

∑ni=1gcd(i,n)=∑ni=1∑d|gcd(i,n)ϕ(d)=∑ni=1∑d|i,d|nϕ(d)=∑d|nϕ(d)⌊nd⌋\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

∑ni=1∑mj=1gcd(i,j)=∑ni=1∑mj=1∑d|i,d|jϕ(d)=∑min(n,m)d=1ϕ(d)⌊nd⌋⌊md⌋\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]

∑ni=1e(gcd(i,n))=∑ni=1∑d|i,d|nμ(d)=∑d|nμ(d)⌊nd⌋\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]

∑ni=1∑mj=1[gcd(i,j)==k]=∑⌊nk⌋i=1∑⌊mk⌋j=1e(gcd(i,j))=∑⌊nk⌋i=1∑⌊mk⌋j=1∑d|i,d|jμ(d)=∑min(⌊nk⌋,⌊mk⌋)i=1μ(d)⌊nkd⌋⌊mkd⌋\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

∑ni=1lcm(i,n)=∑ni=1i∗ngcd(i,n)=∑ni=1∑d|n[gcd(i,j)==d]i∗nd=n∗∑d|n∑⌊nd⌋i=1e(gcd(i,nd))\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∗∑d|n∑⌊nd⌋i=1i∑k|nd,k|iμ(k)=n∗∑d|n∑k|ndk∗μ(k)(⌊ndk⌋+1)∗⌊ndk⌋2=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

∑ni=1lcm(i,n)=∑ni=1i∗ngcd(i,n)=∑ni=1∑d|n[gcd(i,j)==d]i∗nd=n∗∑d|n∑⌊nd⌋i=1e(gcd(i,nd))\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∗∑d|n∑di=1i∗e(gcd(i,d))=n∗∑d|nϕ(d)∗d2=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)=∑ni=1∑mj=1i∗jsum(n,m)=\sum_{i=1}^{n}\sum_{j=1}^{m}i*j

∑ni=1∑mj=1lcm(i,j)=∑ni=1∑mj=1i∗jgcd(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)}

=∑⌊nd⌋i=1∑⌊md⌋j=1di∗djde(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))

=∑min(n,m)d=1d∑⌊nd⌋i=1∑⌊md⌋j=1i∗je(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))

=∑min(n,m)d=1d∑min(⌊nd⌋,⌊md⌋)k=1μ(k)∗k2∗sum(⌊⌊nd⌋k⌋,⌊⌊md⌋k⌋)=\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)