莫比乌斯反演学习笔记 II
如无特殊说明，$n\le m$。
狄利克雷卷积 $(f*g)=\displaystyle\sum_{d|n}f(d)g(\frac n d)$
- $f*\epsilon=f$
- $\mu*I=\epsilon$
- $\varphi*I=id$
- $\mu*id=\varphi$
莫比乌斯反演 $f(n)=\displaystyle\sum_{d|n}g(d)\to g(n)=\displaystyle\sum_{d|n}\mu(d)f(\frac n d)$
$\displaystyle\sum_{d|n}\mu(d)=[n=1]$
$\displaystyle\sum_{i=1}^n\displaystyle\sum_{j=1}^m[\gcd(i,j)=1]=\displaystyle\sum_{d=1}^{n}\mu(d)\left\lfloor\frac{n}{d}\right\rfloor\left\lfloor\frac{m}{d}\right\rfloor$
$\displaystyle\sum_{i=1}^n\sum_{j=1}^mf(\gcd(i,j))=\sum_{s=1}^n\left\lfloor\frac n{s}\right\rfloor\left\lfloor\frac m{s}\right\rfloor\sum_{d|s}f(d)\mu(\frac s d)$
11.求$$\sum_{i=1}^n\sum_{j=1}^nij\gcd(i,j)$$ $\text{Link}$$$\sum_{d=1}^nd^3\sum_{i=1}^{\lfloor\frac n d\rfloor}\sum_{j=1}^{\lfloor\frac n d\rfloor}ij[\gcd(i,j)=1]$$$$\sum_{d=1}^nd^3\sum_{i=1}^{\lfloor\frac n d\rfloor}\sum_{j=1}^{\lfloor\frac n d\rfloor}ij\sum_{q|i}\sum_{q|j}\mu(q)$$$$\sum_{d=1}^nd^3\sum_{q=1}^{\lfloor\frac n d\rfloor}\mu(q)q^2\sum_{i=1}^{\lfloor\frac n {dq}\rfloor}\sum_{j=1}^{\lfloor\frac n {dq}\rfloor}ij$$ 例 $5$ 中有$$g(n)=\sum_{i=1}^n\sum_{j=1}^nij=\frac{n^2(n+1)^2}4$$$$\sum_{d=1}^nd^3\sum_{q=1}^{\lfloor\frac n d\rfloor}\mu(q)q^2g(\lfloor\frac n {dq}\rfloor)$$ 令 $t=dq$
$$\sum_{t=1}^ng(\lfloor\frac n {t}\rfloor)\sum_{d|t}d^3\mu(\frac t d)(\frac t d)^2$$$$\sum_{t=1}^ng(\lfloor\frac n {t}\rfloor)\sum_{d|t}d\mu(\frac t d) t^2$$$$\sum_{t=1}^ng(\lfloor\frac n {t}\rfloor) t^2\sum_{d|t}d\mu(\frac t d)$$ 可以看出，后面是狄利克雷卷积的形式，即有 $\mu*id=\varphi$
$$\sum_{t=1}^ng(\lfloor\frac n {t}\rfloor) t^2\varphi(t)$$ 记 $f(x)=x^2\varphi(x)$，显然为积性函数，于是考虑 $h(x)$ 使得 $(f*h)(x)$ 可以快速计算$$(f*h)(x)=\sum_{d|x}d^2\varphi(d)h(\frac x d)$$ 想要约去 $d^2$，则可设计函数 $h(x)=x^2$
$$(f*h)(x)=x^2\sum_{d|x}\varphi(d)$$ 欧拉反演得$$(f*h)(x)=x^3$$ 接下来考虑杜教筛，记$$S(n)=\sum_{i=1}^nf(i)$$$$h(1)S(n)=\sum_{i=1}^nh(i)S(\lfloor\frac n i\rfloor)-\sum_{i=2}^nh(i)S(\lfloor\frac n i\rfloor)$$$$S(n)=\sum_{i=1}^nh(i)S(\lfloor\frac n i\rfloor)-\sum_{i=2}^nh(i)S(\lfloor\frac n i\rfloor)$$ 先算前面的$$\sum_{i=1}^nh(i)S(\lfloor\frac n i\rfloor)$$$$\sum_{i=1}^nh(i)\sum_{j=1}^{\lfloor\frac n i\rfloor}f(j)$$$$\sum_{j=1}^n\sum_{i|j}f(j)h(\frac j i)$$$$\sum_{j=1}^n(f*h)(j)$$
$$\sum_{j=1}^nj^3$$$$g(n)$$$$S(n)=g(n)-\sum_{i=2}^nh(i)S(\lfloor\frac n i\rfloor)$$ 为了达到 $O(n^{\frac 2 3})$ 的时间复杂度，需要预处理 $S(1)$ 至 $S(\lfloor n^{\frac 2 3}\rfloor)$ 的值，可以用 $S(n)=S(n-1)+n^2\varphi(n)$。
12.求$$\sum_{i=1}^n\sum_{j=1}^n(i+j)^k\gcd(i,j)\mu^2(\gcd(i,j)))$$ $\text{Link}$$$\sum_{d=1}^n\mu^2(d)d^{k+1}\sum_{i=1}^{\lfloor\frac n d\rfloor}\sum_{j=1}^{\lfloor\frac n d\rfloor}(i+j)^k[\gcd(i,j)=1]$$$$\sum_{d=1}^n\mu^2(d)d^{k+1}\sum_{q=1}^{\lfloor\frac n {d}\rfloor}\mu(q)q^k\sum_{i=1}^{\lfloor\frac n {dq}\rfloor}\sum_{j=1}^{\lfloor\frac n {dq}\rfloor}(i+j)^k$$ 记 $\displaystyle f_k(x)=\sum_{i=1}^{x}\sum_{j=1}^{x}(i+j)^k,s=dq$
$$\sum_{d=1}^n\mu^2(d)d^{k+1}\sum_{q=1}^{\lfloor\frac n {d}\rfloor}\mu(q)q^kf_k(\lfloor\frac n {s}\rfloor)$$$$\sum_{s=1}^nf_k(\lfloor\frac n {s}\rfloor)\sum_{d|s}\mu^2(d)d^{k+1}\mu(\frac s d)(\frac s d)^k$$$$\sum_{s=1}^nf_k(\lfloor\frac n {s}\rfloor)s^k\sum_{d|s}\mu^2(d)d\mu(\frac s d)$$ 令$$g(x)=\sum_{i=1}^xi^k$$$$h(x)=\sum_{i=1}^xg(i)$$$$f_k(n)=\sum_{i=n+1}^{2n}\sum_{j=1}^i j^k-\sum_{i=1}^n\sum_{j=1}^i j^k$$$$f_k(n)=\sum_{i=n+1}^{2n}g(i)-\sum_{i=1}^ng(i)$$$$f_k(n)=h(2n)-2h(n)$$ 就可以线性筛预处理了。
$$F(x)=\sum_{d|x}\mu^2(d)\mu(\frac x d)d$$ 是 $(x)=\mu^2(x)x$ 与 $(x)=\mu(x)$ 的狄利克雷卷积，为积性函数。
于是只需考虑 $F(p^k)$ 即可。
- $k=0$，$F(1)=1$；
- $k=1$，$F(p)=\mu(p)+\mu^2(p)p=p-1$；
- $k=2$，$F(p^2)=\mu^3(p)p=-p$；
- $k\ge3$，$\mu(p^c)=0\text{ or }\mu(p^{k-c})=0,F(p^k)=0$。
推完了，时间复杂度 $O(n+T\sqrt n)$。
13.求$$\sum_{i=1}^n\sum_{j=1}^m\gcd(i,j)^k$$ $\text{Link}$
$$\sum_{d=1}^nd^k\sum_{i=1}^{\lfloor\frac n d\rfloor}\sum_{j=1}^{\lfloor\frac m d\rfloor}\sum_{q|\gcd(i,j)}\mu(q)$$$$\sum_{d=1}^nd^k\sum_{q=1}^{\lfloor\frac n d\rfloor}\mu(q)\lfloor\frac n {dq}\rfloor\lfloor\frac m {dq}\rfloor$$ 令 $s=dq$
$$\sum_{s=1}^n\lfloor\frac n {s}\rfloor\lfloor\frac m {s}\rfloor\sum_{d|s}d^k\mu(\frac s d)$$ 只需求 $f(x)=\displaystyle\sum_{d|x}d^k\mu(\frac x d)$ 的前缀和即可，注意到 $f$ 是积性函数，设$$x=\prod_{i=1}^yp_i^{q_i}$$$$f(x)=\prod_{i=1}^yf(p_i^{q_i})$$$$\prod_{i=1}^yp_i^{k(q_i-1)}\mu(p_i)+p_i^{kq_i}\mu(1)$$$$\prod_{i=1}^yp_i^{k(q_i-1)}(p_i^{k}-1)$$ 就可以线性筛了，时间复杂度 $O(n+T\sqrt n)$。
14.求$$\sum_{i=1}^n\sum_{j=1}^n\gcd(i,j)^{i+j}$$ $\text{Link}$
$$\sum_{d=1}^n\sum_{i=1}^{\lfloor\frac n d\rfloor}\sum_{j=1}^{\lfloor\frac n d\rfloor}[\gcd(i,j)=1]d^{id+jd}$$$$\sum_{d=1}^n\sum_{i=1}^{\lfloor\frac n d\rfloor}\sum_{j=1}^{\lfloor\frac n d\rfloor}d^{id+jd}\sum_{q|\gcd(i,j)}\mu(q)$$$$\sum_{d=1}^n\sum_{q=1}^{\lfloor\frac n d\rfloor}\mu(q)\sum_{i=1}^{\lfloor\frac n {dq}\rfloor}\sum_{j=1}^{\lfloor\frac n {dq}\rfloor}d^{idq+jdq}$$ 令 $s=dq$
$$\sum_{d=1}^n\sum_{q=1}^{\lfloor\frac n d\rfloor}\mu(q)\sum_{i=1}^{\lfloor\frac n {s}\rfloor}\sum_{j=1}^{\lfloor\frac n {s}\rfloor}d^{is+js}$$$$\sum_{d=1}^n\sum_{q=1}^{\lfloor\frac n d\rfloor}\mu(q)\sum_{i=1}^{\lfloor\frac n {s}\rfloor}d^{is}\sum_{j=1}^{\lfloor\frac n {s}\rfloor}d^{js}$$$$\sum_{d=1}^n\sum_{q=1}^{\lfloor\frac n d\rfloor}\mu(q){\left(\sum_{i=1}^{\lfloor\frac n {s}\rfloor}{(d^{s})}^i\right)}^2$$
设 $S_n$ 为等比数列前 $n$ 项和，$a$ 为共商。
$$S_n=S_{\lfloor\frac 2 n\rfloor}(1+a^{\lfloor\frac 2 n\rfloor})+a^n[2\nmid n]$$
即可 $O(\log n)$ 求解。
最终时间复杂度为 $O(n\log n)$。
15.设 $f_x$ 表示斐波那契数列第 $x$ 项，求$$\prod_{i=1}^n\prod_{j=1}^mf_{\gcd(i,j)}$$
$\text{Link}$$$\prod_{d=1}^nf_d^{\sum_{i=1}^{\lfloor\frac n d\rfloor}\sum_{j=1}^{\lfloor\frac m d\rfloor}[\gcd(i,j)=1]}$$ 考虑指数$$\sum_{i=1}^{\lfloor\frac n d\rfloor}\sum_{j=1}^{\lfloor\frac m d\rfloor}[\gcd(i,j)=1]$$$$\sum_{q=1}^{\lfloor\frac n d\rfloor}\mu(q){\lfloor\frac n {dq}\rfloor}{\lfloor\frac m {dq}\rfloor}$$ 令 $s=dq$
$$\prod_{d=1}^nf_d^{\sum_{q=1}^{\lfloor\frac n d\rfloor}\mu(q){\lfloor\frac n {s}\rfloor}{\lfloor\frac m {s}\rfloor}}$$$$\prod_{s=1}^n\prod_{d|s}f_d^{\mu(\frac s d){\lfloor\frac n {s}\rfloor}{\lfloor\frac m {s}\rfloor}}$$$$\prod_{s=1}^n{\left(\prod_{d|s}f_d^{\mu(\frac s d)}\right)}^{{\lfloor\frac n {s}\rfloor}{\lfloor\frac m {s}\rfloor}}$$
时间复杂度 $O(n\ln n+n\log\bmod+T\sqrt n\log\bmod)$。
16.$$\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=1][\gcd(j,k)=1]$$ $\text{Link}$$$\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=1][\gcd(j,k)=1]$$$$\sum_{i=1}^n\sum_{j=1}^m\sum_{d|\gcd(i,j)}\mu(d)[\gcd(j,k)=1]$$$$\sum_{d=1}^n\mu(d)\sum_{d|i}^n\sum_{d|j}^m[\gcd(j,k)=1]$$$$\sum_{d=1}^n\mu(d)\sum_{i=1}^{\lfloor\frac n d\rfloor}\sum_{j=1}^{\lfloor\frac m d\rfloor}[\gcd(jd,k)=1]$$$$\sum_{d=1}^n\mu(d)[\gcd(d,k)=1]\lfloor\frac n d\rfloor\sum_{j=1}^{\lfloor\frac m d\rfloor}[\gcd(j,k)=1]$$$$\begin{aligned}g(n)&=\sum_{i=1}^n[\gcd(i,k)=1]\\ &=\lfloor\frac n k\rfloor\sum_{i=1}^k[\gcd(i,k)=1]+g(n\bmod k)\\ &=\lfloor\frac n k\rfloor\varphi(k)+g(n\bmod k)\end{aligned}$$$$\sum_{d=1}^n\mu(d)[\gcd(d,k)=1]\lfloor\frac n d\rfloor g\left(\lfloor\frac m d\rfloor\right)$$$$\begin{aligned}f(n)&=\sum_{i=1}^n\mu(i)[\gcd(i,k)=1]\\ &=\sum_{i=1}^n[\gcd(i,k)=1]f\left(\lfloor\frac n d\rfloor\right)-\sum_{i=2}^n[\gcd(i,k)=1]f\left(\lfloor\frac n d\rfloor\right)\\ &=\sum_{i=1}^n\sum_{j=1}^{\lfloor\frac n i\rfloor}\mu(j)[\gcd(ij,k)=1]-\sum_{i=2}^n[\gcd(i,k)=1]f\left(\lfloor\frac n d\rfloor\right)\\ &=\sum_{q=1}^n[\gcd(q,k)=1]\sum_{d|q}\mu(d)-\sum_{i=2}^n[\gcd(i,k)=1]f\left(\lfloor\frac n d\rfloor\right)\\ &=\sum_{q=1}^n[\gcd(q,k)=1][q=1]-\sum_{i=2}^n[\gcd(i,k)=1]f\left(\lfloor\frac n d\rfloor\right)\\ &=1-\sum_{i=2}^n[\gcd(i,k)=1]f\left(\lfloor\frac n d\rfloor\right)\end{aligned}$$ 时间复杂度 $O(n^{\frac 2 3}+k\log k)$。
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