\(前置技能\)
- 學會莫比烏斯函數必須要先知道狄利克雷函數
- 以及什么是逆元（一本正經胡說八道）
\(狄利克雷卷積\)
兩個函數 $f,g$
<font size="3">$\begin{aligned}f*g(n)=\sum_{d|n}^nf(d)g(\frac{n}{d})\end{aligned}$
* 性質
__交換律__$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f*g=g*f$
__結合律__$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f*g*h=f*(g*h)$
__加法分配率__$\ \ \ \ \ \ \ \ \ \ f*(g+h)=f*g+f*h$
\(幾個定理\)
* ① 定義單位函數$I(n)=[n=1]$，即只有當ｎ等於１時，$I(1)=1$，其余情況都為０
* ② $I * f=f$
* ③ 已知函數$f$，定義函數$f$的逆$f^{-1}$，滿足$f * f^{-1}=I$，且積性函數的逆也是積性函數
* ④ 由③可得$f^{-1}(1)=\frac{1}{f(1)}$
* ⑤ 令$\xi(n)=1$
\(莫比烏斯函數\)
- 莫比烏斯函數 \(\mu=\xi^{-1}\)
\(n=p_1^{a_1}·p_2^{a_2}·\cdots p_k^{a_k}\)
\(\mu(n)=\xi^{-1}(n)= \left\{\begin{matrix}1 &n=1 \\(-1)^k &a1=a2=\cdots =ak=1&\\ 0&otherwise \end{matrix}\right.\)
\(如何推導\)
-
\(如何求逆\)
③等價於
\(\begin{aligned}\forall n: \sum_{d|n} f(d)·f^{-1}(\frac nd)=[n=1]\end{aligned}\)
\(\begin{aligned}\sum_{d|n}f(d)·f^{-1}(\frac nd)=I(n) (n=\not 1)=0\end{aligned}\)
\(f^{-1}(n)·f(1)=-\sum_{d|n,d=\not1}f(d)·f^{-1}(\frac nd)\)
\(\begin{aligned}f^{-1}(n)=\frac{-\sum_{d|n,d=\not1}f(d)·f^{-1}(\frac nd)}{f(1)}\end{aligned}\) -
\(莫比烏斯函數\)
令 \(f=\xi\),\(p\)為質數
\(f^{-1}(1)=1\)
\(f^{-1}(p)=\frac{-f(p)·f^{-1}(1)}{f(1)}=-1\)
\(f^{-1}(p^2)=-(f(p^2)·f^{-1}(1)+f(p)·f^{-1}(p))=0\)
\(f^{-1}(p^3)=-(f(p^3)·f^{-1}(1)+f(p^2)·f^{-1}(p)+f(p)·f^{-1}(p^2))=0\)
\(\begin{aligned}f^{-1}(p^k)=-\sum_{i=1}^k f(p^i)·f^{-1}(p^{k-i})\end{aligned}\)
其中只有 \(f(p^k)·f^{-1}(1)=1,f(p^{k-1})·f^{-1}(p)=-1\)，其余項都是0
\(f^{-1}(p^k)=\left\{\begin{matrix}-1 & k=1\\0 & k>1 \end{matrix}\right.\)
\(n=p_1^{a_1}·p_2^{a_2}·\cdots p_k^{a_k}\)
所以莫比烏斯函數可以推出來了
\(\mu(n)=f^{-1}(n)= \left\{\begin{matrix}1 &n=1 \\(-1)^k &a1=a2=\cdots =ak=1&\Leftarrow 積性函數推出 \\ 0&otherwise \end{matrix}\right.\)
\(莫比烏斯函數的性質\)
- \(\mu*\xi=I\)
因為\(\mu=\xi^{-1}\)
所以有\(\begin{aligned}\sum_{d|n}^n\mu(d)=\sum_{d|n}^n\mu·1=\mu*\xi(n)=I(n)=[n=1]\end{aligned}\)
記住這句，下面的證明要用到 - \(I*f=f\)
- \(f*\mu*\xi=f\)
- \(f*\xi=g \Rightarrow f=g*\mu\)
\(線性篩\)
利用積性函數的性質以及\(\mu\)的表達式
int cnt;
int prime[maxn],mu[maxn];
bool vis[maxn];
void mu_sieve (int n)
{
mu[1]=1;
for (int i=2;i<=n;++i){
if (!vis[i]){ prime[++cnt]=i,mu[i]=-1;}
for (int j=1;j<=cnt&&i*prime[j]<=n;++j){
vis[i*prime[j]]=true;
if (i%prime[j]==0){
mu[i*prime[j]]=0;
break;
}
mu[i*prime[j]]=-mu[i];
}
}
}
\(莫比烏斯反演\)
-
若有 \(\begin{aligned}g(n)=\sum_{d|n}f(d)\end{aligned}\)
則有 \(\begin{aligned}f(n)=\sum_{d|n}\mu(d)g(\frac{n}{d})=\sum_{d|n}\mu(\frac{n}{d})g(d)\end{aligned}\)-
\(證明\)
法一 \(\begin{aligned} \sum_{d|n}{\mu(d)g(\frac{n}{d})}=\sum_{d|n}{\mu(d)\sum_{x|\frac{n}{d}}{f(x)}}=\sum_{x|n}{f(x)}\sum_{d|\frac{n}{x}}{\mu(d)}=\sum_{x|n}{f(x)[\frac{n}{x}=1]}=f(n) \end{aligned}\)
法二 \(g=f*\xi,f=g*\mu\)
-
-
若有\(\begin{aligned}g(n)=\sum_{n|d}f(d)\end{aligned}\)
則有\(\begin{aligned}f(n)=\sum_{n|d}\mu(\frac{d}{n})g(d)\end{aligned}\)-
\(證明\)
\(\begin{aligned}\sum_{n|d}\mu(\frac{d}{n})g(d)=\sum_{k=1}^{+\infty}\mu(k)g(kn)=\sum_{k=1}^{+\infty}\mu(k)\sum_{kn|d}f(d)=\sum_{n|d}f(d)\sum_{k|\frac{d}{n}}\mu(k)=\sum_{n|d}f(d)[\frac{d}{n}=1]=f(n) \end{aligned}\)
這一種一般用的比較多
-
\(莫比烏斯反演的應用\)
莫比烏斯函數一般用於\(gcd\)中
\(\begin{aligned} \sum_{i=1}^n\sum_{j=1}^m\gcd(i,j)&=\sum_d^{min(n,m)} d\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)=d] \\ &=\sum_{d}^{min(n,m)}d\sum_{i=1}^{\frac nd}\sum_{j=1}^{\frac md}[gcd(i,j)=1] \\ &=\sum_{d}^{min(n,m)}d\sum_{i=1}^{\frac nd}\sum_{j=1}^{\frac md}\sum_{k\mid gcd(i,j)}\mu(k) \\ &=\sum_d^{min(n,m)} d\sum_k^{min(\frac{n}{d},\frac{m}{d})}\mu(k)\sum_{k\mid i}^{\frac nd}\sum_{k\mid j}^{\frac md} \\ &=\sum_{d}^{min(n,m)}d\sum_k^{min(\frac{n}{d},\frac{m}{d})}\mu(k)\lfloor \frac n{kd}\rfloor \lfloor\frac m{kd}\rfloor \end{aligned}\)
令\(T=kd\)
\(\begin{aligned} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{T}^{min(n,m)}\lfloor\frac nT\rfloor\lfloor\frac mT\rfloor\sum_{k\mid T}\frac Tk\mu(k)\end{aligned}\)
以上是莫比烏斯反演推出來的式子
\(\begin{aligned}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{T}^{min(n,m)}\lfloor\frac nT\rfloor\lfloor\frac mT\rfloor\varphi(T)\end{aligned}\)
這一步是歐拉反演，證明請移步歐拉函數|(擴展)歐拉定理|歐拉反演
當然，這里這個例子不如直接用歐拉反演
\(\begin{aligned}\sum_{i=1}^n\sum_{j=1}^m\gcd(i,j)&=\sum_{i=1}^n\sum_{j=1}^m\sum_{d|gcd(i,j)}\varphi(d)=\sum_{d=1}^{min(n,m)}\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor\varphi(d)\end{aligned}\)
如有哪里講得不是很明白或是有錯誤，歡迎指正
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