莫比乌斯反演

随便整理下，其实我也不是很懂

整除分块

(lfloor frac{n}{i} floor leq frac{n}{i} implies i leq lfloor frac{n}{lfloor frac{n}{i} floor} floor)
即(lfloor frac{n}{i} floor = k)，(i)能取到的最大值为(lfloor frac{n}{lfloor frac{n}{i} floor} floor)

狄利克雷卷积

((f*g)(n)=sum_limits{d|n}f(d)g(frac{n}{d}))
满足交换律、结合律

常用积性函数

欧拉函数：(varphi(n)=largesum_{i=1}^{n}[gcd(i,n)=1])
莫比乌斯函数：(mu(n) = egin{cases} 1 & n=1\ 0 &n=p_1^{a_1}p_2^{a_2}...p_k^{a_n} (exists a_i>1,即n含有平方因子)\(-1)^k & n=p_1p_2...p_k end{cases})
单位函数：(epsilon(n) = [n = 1])
恒等函数：(id_(n) = n)
常数函数：(I(n) = 1)
除数函数：(sigma(n)或d(n) = sum limits_{d|n}1（即因数个数）)

线性筛莫比乌斯函数

void get_mu(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) break;
            else mu[i*prime[j]] = -mu[i];
        }
    }
}

性质

(mu*I=epsilon)
(varphi*I=id)
(id*mu=varphi)
(I*I=d)
(I*id=sigma)

莫比乌斯反演

设(F(n)=sum_{d|n}f(d))，则
(egin{array}{l} f(n) = sumlimits_{d|n} {mu(d)F(lfloorfrac{n}{d} floor)} \ f(n) = sumlimits_{n|d} {mu(lfloorfrac{d}{n} floor)F(d)} end{array})

常用结论

(ecause sumlimits_{d∣n}mu(d)=[n=1])
( herefore [gcd(i,j)=1]=sumlimits_{d∣gcd(i,j)}​μ(d))

(sumlimits_{d|n}frac{mu(d)}{d}=frac{varphi(n)}{n})

(ecause varphi(n) = sum_{d|n}mu(d)frac{n}{d})
由莫比乌斯反演得
( herefore n = sum_{d|n}varphi(d))