莫比乌斯反演一些式子的推导(附题目)

一、一些符号和等式

$1.a|b$意为$a$整除$b$
$2.[a=b]=\{\begin{array}{l}1,a=b\\ 0,a\ne b\end{array}$
$3.\lfloor a\rfloor$意为对$a$向下取整
$4.\lfloor \frac{\lfloor n/d\rfloor }{k}\rfloor =\lfloor \frac{n}{dk}\rfloor$

二、狄利克雷卷积

关于狄利克雷卷积，请戳这里
然后莫比乌斯反演中需要知道的是
$\phi$为欧拉函数，$\mu$为莫比乌斯函数，$id(n)=n$，$\epsilon (n)=[n=1]$，$I(n)=1$
①$\phi \ast I=id$
②$\mu \ast I=\epsilon$
在①等式两边同时$\ast \mu$，我们可以得到$\phi \ast I\ast \mu =\mu \ast id$
再根据②，我们有$\phi \ast \epsilon =\mu \ast id$
对于任意数论函数都有$f=f\ast \epsilon$，故最终我们有：
③$\mu \ast id=\phi$
这里面③是非常重要的，它揭示了欧拉函数和莫比乌斯函数的关系，在有些式子的化简当中起到至关重要的作用

三、莫比乌斯反演

对于莫比乌斯反演的题目，我们只要记住上面的①式就行，写开之后就是这样
$$\sum _{d|n}\mu (d)=[n=1]$$
那么$[gcd(i,j)=1]$就有
$$\sum _{d|gcd(i,j)}\mu (d)=[gcd(i,j)=1]$$
亦可以写作
$$\sum _{d|i,d|j}\mu (d)=[gcd(i,j)=1]$$
再做一下拓展求$[gcd(i,j)=k]$
因$[gcd(i,j)=k]\iff [gcd(\frac{i}{k},\frac{j}{k})=1]$，故有
$$\sum _{d|\frac{i}{k}，d|\frac{j}{k}}\mu (d)=[gcd(i,j)=k]$$

四、经典式子的推导

1. $\sum _{i=1}^n\sum _{j=1}^n[gcd(i,j)=1]$

推导过程

$$\begin{array}{rl}\sum _{i=1}^n\sum _{j=1}^n[gcd(i,j)=1]& =\sum _{i=1}^n\sum _{j=1}^n\sum _{d|i,d|j}\mu (d)\\ & =\sum _{d=1}^n\mu (d)\sum _{d|i}\sum _{d|j}1\\ & =\sum _{d=1}^n\mu (d)\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=1}^{\lfloor n/d\rfloor }1\\ & =\sum _{d=1}^n\mu (d)\lfloor \frac{n}{d}{\rfloor }^2\end{array}$$

其中的第二步，先枚举$d$
再看下一步，因为$d|i,d|j$，因此$i$和$j$一定是$d$的整数倍
于是我们枚举$i=1,2,...,\lfloor n/d\rfloor$，那么$i\ast d$就是原来的$i$值，这样就能保证$d|i$
代码实现比较简单，求一个$\mu (i)$的前缀和，加一个数论分块就行

2.$\sum _{i=1}^n\sum _{j=1}^n[gcd(i,j)=k]$

推导过程

$$\begin{array}{rl}\sum _{i=1}^n\sum _{j=1}^n[gcd(i,j)=k]& =\sum _{i=1}^n\sum _{j=1}^n\sum _{d|\frac{i}{k},d|\frac{j}{k}}\mu (d)\\ & =\sum _{i=1}^{\lfloor n/k\rfloor }\sum _{j=1}^{\lfloor n/k\rfloor }\sum _{d|i,d|j}\mu (d)\\ & =\sum _{d=1}^n\mu (d)\sum _{i=1}^{\lfloor n/dk\rfloor }\sum _{j=1}^{\lfloor n/dk\rfloor }1\\ & =\sum _{d=1}^n\mu (d)\lfloor \frac{n}{dk}{\rfloor }^2\end{array}$$
其中第二步我们将$i$和$j$同时缩小了$k$倍，于是有$\sum _{i=1}^{\lfloor n/k\rfloor }\sum _{j=1}^{\lfloor n/k\rfloor }\sum _{d|i,d|j}\mu (d)$
其余部分和上一条一样的

练习题

3.$\sum _{i=1}^n\sum _{j=1}^{n}gcd(i,j)$

推导过程

$$\begin{array}{rl}\sum _{i=1}^n\sum _{j=1}^{n}gcd(i,j)& =\sum _{d=1}^n\sum _{i=1}^n\sum _{j=1}^{n}d[gcd(i,j)=d]\\ & =\sum _{d=1}^{n}d\sum _{i=1}^n\sum _{j=1}^n[gcd(i,j)=d]\\ & =\sum _{d=1}^{n}d\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=1}^{\lfloor n/d\rfloor }[gcd(i,j)=1]\\ & =\sum _{d=1}d\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=1}^{\lfloor n/d\rfloor }\sum _{t|i,t|j}\mu (t)\\ & =\sum _{d=1}d\sum _{t=1}^{\lfloor n/d\rfloor }\mu (t)\lfloor \frac{n}{dt}{\rfloor }^2\end{array}$$
到这里，我们已经能够在$O(n)$的时间内计算，但是还不够，我们仍可以继续优化下去
令$T=td$，要使$\lfloor \frac{n}{dt}\rfloor >0$，于是我们枚举$T=td=1,2,...,n$
$$\sum _{i=1}^n\sum _{j=1}^{n}gcd(i,j)=\sum _{T=1}^n\lfloor \frac{n}{T}{\rfloor }^2\sum _{t|T}\mu (d)(\frac{T}{d})$$
注意，$\sum _{t|T}\mu (d)(\frac{T}{d})=(\mu \ast id)(T)=\phi (T)$
所以最终我们有
$$\sum _{i=1}^n\sum _{j=1}^{n}gcd(i,j)=\sum _{T=1}^n\lfloor \frac{n}{T}{\rfloor }^2\phi (T)$$
这样就能够$O(\sqrt{n})$计算出我们的答案
记住这里的优化方式，以后会经常碰到，当你发现优化不下去的时候，考虑代换

练习题

4.$\sum _{i=1}^n\sum _{j=1}^{n}lcm(i,j)$

推导过程

首先$lcm(i,j)=\frac{ij}{gcd(i,j)}$
$$\begin{array}{rl}\sum _{i=1}^n\sum _{j=1}^{n}lcm(i,j)& =\sum _{i=1}^n\sum _{j=1}^n\frac{ij}{gcd(i,j)}\\ & =\sum _{d=1}^n\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=1}^{\lfloor n/d\rfloor }\frac{idjd}{d}[gcd(i,j)=1]\\ & =\sum _{d=1}^{n}d\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=1}^{\lfloor n/d\rfloor }ij\sum _{t|i,t|j}\mu (t)\\ & =\sum _{d=1}^{n}d\sum _{t=1}^{\lfloor n/d\rfloor }\mu (t)\sum _{i=1}^{\lfloor n/dt\rfloor }it\sum _{j=1}^{\lfloor n/dt\rfloor }jt\\ & =\sum _{d=1}^{n}d\sum _{t=1}^{\lfloor n/d\rfloor }\mu (t)t^2\sum _{i=1}^{\lfloor n/dt\rfloor }i\sum _{j=1}^{\lfloor n/dt\rfloor }j\end{array}$$

我们记$su{m}_1(n)=1+2+...+n=n(n+1)/2$
其中$\sum _{i=1}^{\lfloor n/dt\rfloor }i=1+2+...+\lfloor n/dt\rfloor =su{m}_1(\lfloor n/dt\rfloor )$
$$\sum _{i=1}^n\sum _{j=1}^{n}lcm(i,j)=\sum _{d=1}^{n}d\sum _{t=1}^{\lfloor n/d\rfloor }\mu (t)t^{2}su{m}_1(\lfloor n/dt\rfloor {)}^2$$
记$f(n)=\sum _{t=1}^n\mu (t)t^{2}su{m}_1(\lfloor n/t\rfloor {)}^2$，那么$f(n)$可以$O(\sqrt{n})$求，于是
$$\sum _{i=1}^n\sum _{j=1}^{n}lcm(i,j)=\sum _{d=1}^{n}df(\lfloor n/d\rfloor )$$
最后总体就是$O(\sqrt{n})O(\sqrt{n})=O(n)$

练习题

5.$\sum _{i=1}^n\sum _{j=1}^{n}ijgcd(i,j)$

推导过程

$$\begin{array}{rl}\sum _{i=1}^n\sum _{j=1}^{n}ijgcd(i,j)& =\sum _{d=1}^{n}d\sum _{i=1}^n\sum _{j=1}^{n}ij[gcd(i,j)=d]\\ & =\sum _{d=1}^{n}d\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=1}^{\lfloor n/d\rfloor }idjd\sum _{t|i,t|j}\mu (t)\\ & =\sum _{d=1}^n{d}^3\sum _{t=1}^{\lfloor n/d\rfloor }\mu (t)\sum _{i=1}^{\lfloor n/dt\rfloor }it\sum _{j=1}^{\lfloor n/dt\rfloor }jt\\ & =\sum _{d=1}^n{d}^3\sum _{t=1}^{\lfloor n/d\rfloor }\mu (t)t^2\sum _{i=1}^{\lfloor n/dt\rfloor }i\sum _{j=1}^{\lfloor n/dt\rfloor }j\\ & =\sum _{d=1}^n{d}^3\sum _{t=1}^{\lfloor n/d\rfloor }\mu (t)t^{2}su{m}_1(\lfloor n/dt\rfloor {)}^2\end{array}$$
到这里，我们已经能够$O(n)$求得答案，但是别慌，我们仍然可以继续优化
又是熟悉的代换，令$T=dt$，于是
$$\begin{array}{rl}\sum _{i=1}^n\sum _{j=1}^{n}ijgcd(i,j)& =\sum _{T=1}^{n}su{m}_1(\lfloor n/T\rfloor {)}^2\sum _{t|T}\mu (t)t^2(\frac{T}{t}{)}^3\\ & =\sum _{T=1}^{n}su{m}_1(\lfloor n/T\rfloor {)}^2{T}^2\sum _{t|T}\mu (t)(\frac{T}{t})\end{array}$$
又是熟悉的$\mu \ast id=\phi$
于是最终
$$\sum _{i=1}^n\sum _{j=1}^{n}ijgcd(i,j)=\sum _{T=1}^{n}su{m}_1(\lfloor n/T\rfloor {)}^2{T}^2\phi (T)$$
复杂度$O(\sqrt{n})$

练习题

这道练习题，因为数据比较大，需要用到杜教筛

五、拓展提高

学完了上面的内容之后，相信你应该已经基本掌握了比较经典的莫比乌斯反演
写了一些题，你会发现，莫比乌斯反演难，一是难在推导，二是难在一些预处理，比如要你预先处理好一些不常见的积性函数
下面的题目就按照难度逐个击破

1. $\sum _{i=1}^N\sum _{j=1}^{N}lcm(a_i,a_j)$

原题链接

给你$N$个数$a_1,a_2,...,a_N$，其中$1\le a_i\le 50000$，求$\sum _{i=1}^N\sum _{j=1}^{N}lcm(a_i,a_j)$

推导过程

是不是看着很熟悉呢？经典推导中第四条
先对式子做基本的变形
$$\sum _{i=1}^N\sum _{j=1}^{N}lcm(a_i,a_j)=\sum _{i=1}^N\sum _{j=1}^N\frac{a_i{a}_j}{gcd(a_i,a_j)}$$
分析一下，光有$a_i$是没用的，我们要将式子中$a_i$转化为数字才行
所以我们不妨直接枚举$i,j=1,2,3,...,n=50000$，这样就有
$$\sum _{i=1}^N\sum _{j=1}^{N}lcm(a_i,a_j)=\sum _{i=1}^n\sum _{j=1}^{n}cn{t}_{i}cn{t}_j\frac{ij}{gcd(i,j)}$$
其中$cn{t}_i,cn{t}_j$为数$i,j$出现次数，于是我们继续
$$\begin{array}{rl}\sum _{i=1}^N\sum _{j=1}^{N}lcm(a_i,a_j)& =\sum _{d=1}^n\sum _{i=1}^n\sum _{j=1}^{n}cn{t}_{i}cn{t}_j\frac{ij}{d}[gcd(i,j)=d]\\ & =\sum _{d=1}^n\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=1}^{\lfloor n/d\rfloor }cn{t}_{id}cn{t}_{jd}\frac{ij{d}^2}{d}[gcd(i,j)=1]\\ & =\sum _{d=1}^{n}d\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=1}^{\lfloor n/d\rfloor }ijcn{t}_{id}cn{t}_{jd}\sum _{t|i,t|j}\mu (t)\\ & =\sum _{d=1}^{n}d\sum _{t=1}^{\lfloor n/d\rfloor }\mu (t)t^2\sum _{i=1}^{\lfloor n/dt\rfloor }icn{t}_{idt}\sum _{j=1}^{\lfloor n/dt\rfloor }jcn{t}_{jdt}\\ & =\sum _{d=1}^{n}d\sum _{t=1}^{\lfloor n/d\rfloor }\mu (t)t^2{\left(\sum _{i=1}^{\lfloor n/dt\rfloor }icn{t}_{idt}\right)}^2\end{array}$$
到这里我们令$T=dt$
$$\begin{array}{rl}\sum _{i=1}^N\sum _{j=1}^{N}lcm(a_i,a_j)& =\sum _{T=1}^n{\left(\sum _{i=1}^{\lfloor n/T\rfloor }icn{t}_{iT}\right)}^2\sum _{t|T}\mu (t)t^2(\frac{T}{t})\\ & =\sum _{T=1}^{n}T{\left(\sum _{i=1}^{\lfloor n/T\rfloor }icn{t}_{iT}\right)}^2\sum _{t|T}\mu (t)t\end{array}$$
先看后面的$\sum _{t|T}\mu (t)t$，我们记做$f(T)$，我们可以$O(nlogn)$预处理：

for (int t = 1; t <= n; t++)
        for (int T = t; T <= n; T += t)
            f[T] += mu[t] * t;

关于复杂度，这个是一个调和级数，即$\frac{n}{1}+\frac{n}{2}+\frac{n}{3}+...+1$，是$nlogn$级别的
再看$\sum _{i=1}^{\lfloor n/T\rfloor }icn{t}_{iT}$，显然它也是是一个关于$T$的函数
我们记$s(T)=\sum _{i=1}^{\lfloor n/T\rfloor }icn{t}_{iT}=\sum _{T|t}\frac{t}{T}cn{t}_t=\frac{1}{T}\sum _{T|t}tcn{t}_t$
这样也能$O(nlogn)$预处理

for (int T = 1; T <= n; T++) {
        for (int t = T; t <= n; t += T)
            s[T] += t * cnt[t];
        s[T] /= T;
    }

于是这题就能轻松搞定了

AC代码

2.$\sum _{i=1}^{n}lcm(i,n)$

原题链接

推导过程

$$\begin{array}{rl}\sum _{i=1}^{n}lcm(i,n)& =\sum _{i=1}^n\frac{in}{gcd(i,n)}\\ & =n\sum _{i=1}^n\frac{i}{gcd(i,n)}\\ & =n\sum _{d|n}\sum _{i=1}^{n/d}\frac{id}{d}[gcd(i,\frac{n}{d})=1]\\ & =n\sum _{d|n}\sum _{i=1}^{n/d}i[gcd(i,\frac{n}{d})=1]\\ & =n\sum _{d|n}\sum _{i=1}^{d}i[gcd(i,d)=1]\\ & =n\sum _{d|n}\sum _{i=1}^{d}i\sum _{t|i,t|d}\mu (t)\\ & =n\sum _{d|n}\sum _{t|d}\mu (t)\sum _{i=1}^{d/t}it\\ & =n\sum _{d|n}\sum _{t|d}\mu (t)t\sum _{i=1}^{d/t}i\\ & =n\sum _{d|n}\sum _{t|d}\mu (t)t\frac{\frac{d}{t}(\frac{d}{t}+1)}{2}\\ & =\frac{n}{2}\sum _{d|n}\left[\sum _{t|d}\mu (t)\frac{d^2}{t}+\sum _{t|d}\mu (t)d\right]\\ & =\frac{n}{2}\sum _{d|n}d\left[\sum _{t|d}\mu (t)\frac{d}{t}+\sum _{t|d}\mu (t)\right]\\ & =\frac{n}{2}\sum _{d|n}d\left[\left(\mu \ast id\right)\left(d\right)+\left(\mu \ast I\right)\left(d\right)\right]\\ & =\frac{n}{2}\sum _{d|n}d\left[\phi \left(d\right)+\epsilon \left(d\right)\right]\end{array}$$
显然所有答案可以像上一题那样$O(nlogn)$处理

AC代码

3. $\sum _{i=1}^{N}gcd\left(\lfloor \sqrt[3]{i}\rfloor ,i\right)$

原题链接(2019年杭电多校第一场)

推导过程

记$n=\sqrt[3]{N}$

$$\begin{array}{rl}\sum _{i=1}^{N}gcd\left(\lfloor \sqrt[3]{i}\rfloor ,i\right)& =\sum _{i=1}^n\sum _{j=i^3}^{\min ({\left(i+1\right)}^3-1,N)}gcd(i,j)\\ & =\sum _{i=1}^{n-1}\sum _{j=i^3}^{{\left(i+1\right)}^3-1}gcd(i,j)+\sum _{j=n^3}^{N}gcd(n,j)\end{array}$$
先求前半部分
不妨记$f\left(n\right)=\sum _{i=1}^n\sum _{j=i^3}^{{\left(i+1\right)}^3-1}gcd\left(i,j\right)$，那么原式就是要求$f\left(n-1\right)$
$$\begin{array}{rl}f\left(n\right)& =\sum _{i=1}^n\sum _{j=i^3}^{{\left(i+1\right)}^3-1}gcd\left(i,j\right)\\ & =\sum _{d=1}^{n}d\sum _{i=1}^n\sum _{j=i^3}^{{\left(i+1\right)}^3-1}\left[gcd\left(i,j\right)=d\right]\\ & =\sum _{d=1}^{n}d\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=\lfloor (id{)}^3/d\rfloor }^{\lfloor \left[{\left(id+1\right)}^3-1\right]/d\rfloor }\left[gcd\left(i,j\right)=1\right]\\ & =\sum _{d=1}^{n}d\sum _{i=1}^{\lfloor n/d\rfloor }\sum _{j=\lfloor (id{)}^3/d\rfloor }^{\lfloor \left[{\left(id+1\right)}^3-1\right]/d\rfloor }\sum _{t|i,t|j}\mu (t)\\ & =\sum _{d=1}^{n}d\sum _{t=1}^{\lfloor n/d\rfloor }\mu (t)\left(\sum _{i=1}^{\lfloor n/dt\rfloor }\sum _{j=\lfloor (idt{)}^3/dt\rfloor }^{\lfloor \left[{\left(idt+1\right)}^3-1\right]/dt\rfloor }\right)\end{array}$$
接下来老方法，令$T=dt$
$$\begin{array}{rl}f\left(n\right)& =\sum _{T=1}^n\left(\sum _{i=1}^{\lfloor n/T\rfloor }\sum _{j=\lfloor (iT{)}^3/T\rfloor }^{\lfloor \left[{\left(iT+1\right)}^3-1\right]/T\rfloor }\right)\sum _{t|T}\mu (t)\left(\frac{T}{t}\right)\\ & =\sum _{T=1}^n\left(\sum _{i=1}^{\lfloor n/T\rfloor }\sum _{j=\lfloor (iT{)}^3/T\rfloor }^{\lfloor \left[{\left(iT+1\right)}^3-1\right]/T\rfloor }\right)\phi (T)\end{array}$$
再看括号里面的部分
$$\begin{array}{rl}g(n,T)& =\sum _{i=1}^{\lfloor n/T\rfloor }\sum _{j=\lfloor (iT{)}^3/T\rfloor }^{\lfloor \left[{\left(iT+1\right)}^3-1\right]/T\rfloor }\\ & =\sum _{i=1}^{\lfloor n/T\rfloor }\left(\lfloor \frac{{\left(iT+1\right)}^3-1}{T}\rfloor -\lfloor \frac{(iT{)}^3}{T}\rfloor +1\right)\\ & =\sum _{i=1}^{\lfloor n/T\rfloor }\left(\lfloor \frac{{\left(iT+1\right)}^3-(iT{)}^3-1}{T}\rfloor +1\right)\\ & =\sum _{i=1}^{\lfloor n/T\rfloor }\left(3i^{2}T+3i+1\right)\\ & =3T\sum _{i=1}^{\lfloor n/T\rfloor }{i}^2+3\sum _{i=1}^{\lfloor n/T\rfloor }i+\lfloor n/T\rfloor \end{array}$$
可以预处理前缀和后$O(1)$得到
那么$f\left(n\right)=\sum _{T=1}^{n}g\left(n,T\right)\phi (T)$，就可以$O(n)$得到

最后来看原式的后半部分$\sum _{j=n^3}^{N}gcd(n,j)$
$$\begin{array}{rl}\sum _{i=n^3}^{N}gcd(n,i)& =\sum _{d|n}d\sum _{i=\lfloor n^3/d\rfloor }^{\lfloor N/d\rfloor }\left[gcd\left(\frac{n}{d},j\right)=1\right]\\ & =\sum _{d|n}d\sum _{i=\lfloor n^3/d\rfloor }^{\lfloor N/d\rfloor }\sum _{t|\frac{n}{d},t|i}\mu (t)\\ & =\sum _{d|n}d\sum _{td|n}\mu (t)\left(\sum _{i=\lfloor n^3/dt\rfloor }^{\lfloor N/dt\rfloor }\right)\Leftarrow T=dt\\ & =\sum _{T|n}\left(\sum _{i=\lfloor n^3/T\rfloor }^{\lfloor N/T\rfloor }\right)\sum _{d|T}\mu (\frac{T}{d})d\\ & =\sum _{T|n}\left(\sum _{i=\lfloor n^3/T\rfloor }^{\lfloor N/T\rfloor }\right)\phi (T)\\ & =\sum _{T|n}\left(\lfloor N/T\rfloor -\lfloor n^3/T\rfloor +1\right)\phi (T)\end{array}$$
于是我们后半部分可以$O\left(\sqrt{n}\right)$求得
总体复杂度就是$O\left(n\right)$

AC代码