Mobius Inversion

狄利克雷卷积

定义

若t=f∗g，则t(n)=∑i∣nf(i)g(i)若 \ \mathbf t=\mathbf f\ast\mathbf g，则\\ \mathbf t(n) = \sum_{i | n} \mathbf f(i)\ \mathbf g(i) 若 t=f∗g，则t(n)=i∣n∑f(i) g(i)

一些性质

交换律：$ \mathbf f\ast\mathbf g=\mathbf g\ast\mathbf f $
结合律：(f∗g)∗h=f∗(g∗h)(\mathbf f \ast \mathbf g) \ast \mathbf h= \mathbf f \ast (\mathbf g \ast \mathbf h)(f∗g)∗h=f∗(g∗h)
分配律：(f+g)∗h=f∗h+g∗h(\mathbf f+\mathbf g)\ast \mathbf h=\mathbf f\ast\mathbf h+\mathbf g\ast\mathbf h(f+g)∗h=f∗h+g∗h
$ (x \ \mathbf f)\ast\mathbf g=x \ (\mathbf f\ast\mathbf g)$
ϵ ∗f=f\epsilon \, \ast \mathbf f = \mathbf fϵ∗f=f
逆元：∀f,andf(1)≠0,∃g,sothatf∗g=ϵ\forall \ \mathbf f, and \ \mathbf f(1) \neq 0, \exist \ \mathbf g, \ so \ that\ \mathbf f \ast \mathbf g = \epsilon∀ f,and f(1)̸=0,∃ g, so that f∗g=ϵ

其中构造逆元的方法：令
g(n)=1f(1)([n=1]−∑i∣n,i≠1f(i)g(ni))\mathbf g(n) = \frac{1}{\mathbf f(1)} ([n = 1]- \sum_{i|n, \ i \neq 1} \mathbf f(i) \ \mathbf g(\frac{n}{i})) g(n)=f(1)1([n=1]−i∣n, i̸=1∑f(i) g(in))
这样：
∑i∣nf(i)g(ni)=f(1)g(n)+∑i∣n,i≠1f(i)g(ni)=[n=1]\sum_{i | n} \mathbf f(i) \ \mathbf g(\frac{n}{i}) \\ = \mathbf f(1) \ \mathbf g(n) + \sum_{i|n, \ i \neq 1} \mathbf f(i) \ \mathbf g(\frac{n}{i}) \\ = [n = 1] i∣n∑f(i) g(in)=f(1) g(n)+i∣n, i̸=1∑f(i) g(in)=[n=1]

例题

已知 f(1)…f(n)\mathbf f(1) \dots \mathbf f(n)f(1)…f(n)，定义
fk(n)=∑d1d2…dk=nf(d1)f(d2)…f(dn)\mathbf f_k(n) = \sum \limits _{d_1d_2\dots d_k = n} \mathbf f(d_1) \ \mathbf f(d_2)\dots \mathbf f(d_n) fk(n)=d1d2…dk=n∑f(d1) f(d2)…f(dn)
求 f(1)…f(n),n,k≤105\mathbf f(1) \dots \mathbf f(n), \ n, k \le 10^5f(1)…f(n), n,k≤105

SolSolSol: 发现 fk=f∗f⋯∗f⎵k\mathbf f_k = \begin{matrix}\underbrace{f \ast f \cdots \ast f}\\k\end{matrix}fk=f∗f⋯∗fk

又由于狄利克雷卷积的结合律，快速幂即可。复杂度为 Θ(nlog⁡nlog⁡k)\Theta(n \ \log n \ \log k)Θ(n logn logk)。

积性函数

定义

积性函数：∀gcd⁡(a,b)=1,f(a×b)=f(a)×f(b)\forall \gcd(a, b) = 1, \ \mathbf f(a \times b) = \mathbf f(a) \times \mathbf f(b)∀gcd(a,b)=1, f(a×b)=f(a)×f(b)
完全积性函数：∀a,b,f(a×b)=f(a)×f(b)\forall a, b, \ \mathbf f(a\times b) = \mathbf f(a) \times \mathbf f(b)∀a,b, f(a×b)=f(a)×f(b)

常见积性函数

φ(n)=n∏p∣n(1−1p)\varphi (n) = n \prod \limits_{p | n} (1 - \frac{1}{p})φ(n)=np∣n∏(1−p1)
μ(n)\mu (n)μ(n)
KaTeX parse error: Expected group after '_' at position 20: …ma _k(n) = \sum_̲\limits{d|n} d …
- σ0(x)\sigma _0(x)σ0(x) 约数个数
  - σ0(ij)=∑x∣i∑y∣j[gcd⁡(x,y)=1]\sigma_0(ij) = \sum \limits_{x \mid i} \sum \limits_{y \mid j} [\gcd(x, y) = 1]σ0(ij)=x∣i∑y∣j∑[gcd(x,y)=1]
- σ1(x)\sigma_1 (x)σ1(x) 约数和
ϵ(n)=[n=1]\epsilon(n) = [n = 1]ϵ(n)=[n=1]
idk(x)=nk\mathbf {id} _k(x) = n^kidk(x)=nk
- 1(x)=1\mathbf 1(x) = 11(x)=1
- id(x)=n\mathbf {id} (x) = nid(x)=n

一些性质

两个积性函数的狄利克雷卷积是积性函数

推导：
t(nm)=∑d∣nmf(d)g(nmd)=∑a∣n,b∣mf(ab)g(nmab)=∑a∣n,b∣mf(a)f(b)g(na)g(mb)=(∑a∣nf(a)g(na))(∑b∣mf(b)g(mb))=t(n)t(m)\begin{aligned}\mathbf t(nm)&=\sum_{d\mid nm}\mathbf f(d)\ \mathbf g\left(\frac{nm}d\right)\\&=\sum_{a\mid n,b\mid m}\mathbf f(ab)\ \mathbf g\left(\frac{nm}{ab}\right)\\&=\sum_{a\mid n,b\mid m}\mathbf f(a) \ \mathbf f(b) \ \mathbf g\left(\frac na\right)\mathbf g\left(\frac mb\right)\\&=\left(\sum_{a\mid n}\mathbf f(a) \ \mathbf g\left(\frac na\right)\right)\left(\sum_{b\mid m}\mathbf f(b)\ \mathbf g\left(\frac mb\right)\right)\\&=\mathbf t(n)\ \mathbf t(m)\end{aligned} t(nm)=d∣nm∑f(d) g(dnm)=a∣n,b∣m∑f(ab) g(abnm)=a∣n,b∣m∑f(a) f(b) g(an)g(bm)=⎝⎛a∣n∑f(a) g(an)⎠⎞⎝⎛b∣m∑f(b) g(bm)⎠⎞=t(n) t(m)

莫比乌斯函数

μ(d)\mu(d)μ(d) 的定义

μ(d)={1d=1(−1)kd=∏i=1kpi,pi为互异素数0d有平方因子\mu (d) = \begin {cases} 1&d = 1 \\ (-1) ^ k & d = \prod \limits_{i = 1} ^ kp_i, \ p_i 为互异素数 \\ 0 & d \ 有平方因子 \end{cases} μ(d)=⎩⎪⎪⎪⎨⎪⎪⎪⎧1(−1)k0d=1d=i=1∏kpi, pi为互异素数d 有平方因子

一些性质

∀n,∑d∣nμ(d)=[n=1]\forall n, \ \sum_{d | n} \mu(d) = [n = 1]∀n, ∑d∣nμ(d)=[n=1]
∀n,∑d∣nμ(d)d=φ(n)n\forall n, \ \sum_{d | n}\frac{\mu(d)}{d} = \frac{\varphi(n)}{n}∀n, ∑d∣ndμ(d)=nφ(n)
φ=μ∗id\varphi = \mu \ast \mathbf{id}φ=μ∗id
ϵ=μ∗1\epsilon = \mu \ast \mathbf 1ϵ=μ∗1
σ0(x)=1∗1\sigma _0 (x) = \mathbf 1 \ast \mathbf 1σ0(x)=1∗1
σ1(x)=id∗1\sigma _1(x) = \mathbf{id} \ast \mathbf 1σ1(x)=id∗1
id=φ∗1\mathbf {id} = \varphi \ast \mathbf 1id=φ∗1

线性筛 μ\muμ

inline void get_mu()
{mu[1] = 1;memset(is_prime, true, sizeof(is_prime));for(ri i = 2; i <= MAXN - 5; i++){if(is_prime[i]){mu[i] = -1;prime[++cnt] = i;}for(ri j = 1; j <= cnt && prime[j] * i <= MAXN - 5; j++){is_prime[i * prime[j]] = false;if(i % prime[j] == 0)break;elsemu[prime[j] * i] = -mu[i];}}
}

莫比乌斯反演

由性质 ϵ=μ∗1\epsilon = \mu \ast \mathbf 1ϵ=μ∗1 可进行如下推导：
如果存在F=f∗1就有f=f∗1∗μ=F∗μ如果存在 \ \mathbf F = \mathbf f \ast \mathbf 1 \\ 就有 \ \mathbf f = \mathbf f \ast \mathbf 1 \ast \mu = \mathbf F \ast \mu 如果存在 F=f∗1就有 f=f∗1∗μ=F∗μ
整理一下可以得到：

如果存在数论函数 f\mathbf ff，$\mathbf F $，满足：
F(n)=∑d∣nf(d)\mathbf F(n) = \sum \limits_{d\mid n} \mathbf f(d) F(n)=d∣n∑f(d)
就有：
f(n)=∑d∣nμ(d)F(nd)\mathbf f(n) = \sum \limits_{d\mid n} \mu(d) \ \mathbf F(\frac{n}{d}) f(n)=d∣n∑μ(d) F(dn)
或者：
f(n)=∑n∣dμ(dn)F(d)\mathbf f(n) = \sum \limits_{n \mid d} \mu(\frac{d}{n}) \ \mathbf F(d) f(n)=n∣d∑μ(nd) F(d)