数论、莫比乌斯（mobius） 更新中

基本式子

$$n=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdot \cdot \cdot p_m^{{\alpha }_m}(p\in prime,i,{\alpha }_i\in Z_{+})$$

$$lcm(i,j)\ast gcd(i,j)=i\ast j$$

常用数论函数

$ϵ(n)=\{\begin{array}{ll}1& n=1,\\ 0& n>1.\end{array}$

$I(n)=1$

$id(n)=n$

$d(n)因子个数(除数函数也记作\tau (n))$

$\sigma (n)因数和$

$\mu (n)=\{\begin{array}{ll}1& \text{n=1},\\ (-1{)}^k& \prod _{i=1}^m{\alpha }_i=1,\\ 0& \exists {\alpha }_i>1.\end{array}$

$\phi (n)欧拉函数(小于n且与n互质的个数)$

$\omega (n)n的不同素因数的个数,\omega (1)=0$

$\theta (x)=\sum _{p\le x}\ln p$

$\Lambda (n)=\{\begin{array}{ll}\ln p& n=p^{\alpha },p\in prime,\alpha \ge 1,\\ 0& other.\end{array}$

$\psi (x)=\sum _{n<=x}\Lambda (n)$

数论相关公式(定理)

$\phi \ast I=id$

$\mu \ast I=ϵ$

$\mu \ast id=\phi$

$1.\phi (n)=n\prod _{p|n}(1-\frac{1}{p})$

$2.\phi (n)=\sum _{l=1}^n\prod _{p|n}(1-\frac{\sum _{a=1}^p{e}^{\frac{2\pi la}{p}}}{p})$

$3.a^{\phi (n)}\equiv 1(modn),(a,n)=1Fermat小定理$

$4.a^p\equiv 1(modp),p\in primeEuler定理$

$5.(p-1)!\equiv -1(modp)Wilson定理(特殊地)$

$6.\sum _{d|n}\mu (d)\phi (d)=0,2|n$

$7.\sum _{d|n}{\mu }^2(d)=2^{\omega (n)}$

$8.\sum _{d|n}\mu (d)\tau (d)=(-1{)}^{\omega (n)}$

$9.\sum _{d|n}\mu (d)\sigma (d)=(-1{)}^{\omega (n)}=\prod _{p|n}(-p)$

$10.\sum _{d\le x}\mu (d)\lfloor \frac{x}{d}\rfloor =1$

$11.\forall m\ge 1,\exists p,stm＜p\le 2m即\pi (2m)-\pi (m)\ge 1Betrand假设$

$12.\sum _{d|n}\Lambda (d)=\ln n$

$13.x^{p-1}-1\equiv (x-1)...(x-p-1)(modp)$

$14.\frac{(p-1)!}{1}+\frac{(p-1)!}{2}+...+\frac{(p-1)!}{p-1}+\equiv 0(mod{p}^2)$

$15.\sum _{d|n}\frac{{\mu }^2(d)}{\phi (d)}=\frac{n}{\phi (n)}$

$I(n)=\sum _{d|n}\mu (d)$

莫比乌斯反演相关公式

$F(n)=\sum _{d|n}f(d)⟹f(n)=\sum _{d|n}\mu (d)F(\lfloor \frac{n}{d}\rfloor )$
$⟹f(n)=\sum _{d|n}\mu (\lfloor \frac{n}{d}\rfloor )F(d)$

$F(n)|\tau (n)\sigma (n)n\lfloor \frac{1}{n}\rfloor \frac{\phi (n)}{n}\Lambda (n)|$
$f(n)|1n\phi (n)\mu (n)\frac{\mu (n)}{n}\ln n|$

$1.\phi (n)=\sum _{d|n}\mu (d)\ast \lfloor \frac{n}{d}\rfloor$

$2.\sum _{d^2|n}\mu (d)=u^2(n)=|\mu (n)|$

$3.\sum _{d|n}\frac{{\mu }^2(d)}{\phi (d)}=\frac{n}{\phi (n)}$

$4.gcd(i,j)=\sum _{d|i,d|j}\phi (d)$

交换运算顺序

$1.\prod _{i=1}^n(a_i\ast b_i)=\prod _{i=1}^n{a}_i\ast \prod _{i=1}^n{b}_i$

$2.\prod _{i=1}^{n}k\ast a_i=k^n\ast \prod _{i=1}^n{a}_i$

$3.\sum _{i=1}^n(a_i+b_i)=\sum _{i=1}^n{a}_i+\sum _{i=1}^n{b}_i$

$4.\sum _{i=1}^{n}k\ast a_i=k\ast \sum _{i=1}^n{a}_i$

一些结论$(n<m)$

$$\sum _{d|n}\frac{\mu (d)}{d}=\frac{\phi (n)}{n}$$
$1.\sum _{i=1}^{n}lcm(i,n)=\frac{n}{2}(\sum _{d|n}d\phi (d)+1)=n\sum _{d|n}\sum _{i=1}^{d}i[gcd(i,d)==1]$

$2.\sum _{i=1}^n\sum _{j=1}^m[gcd(i,j)==x]=\sum _{d=1}^{n/x}\mu (d)\lfloor \frac{n}{dx}\rfloor \lfloor \frac{m}{dx}\rfloor =2\sum _{d=1}^{n/x}\phi (\frac{n}{x})-1$

$3.\sum _{i=1}^n\sum _{j=1}^{m}gcd(i,j)=\sum _{d=1}^n\phi (d)\lfloor \frac{n}{d}\rfloor \lfloor \frac{m}{d}\rfloor =\sum _{x=1}^n\sum _{d=1}^{n/x}\mu (d)\lfloor \frac{n}{dx}\rfloor \lfloor \frac{m}{dx}\rfloor$

$4.\sum _{i=1}^n\sum _{j=1}^{m}lcm(i,j)=\sum _{d=1}^{n}d\sum _{d^{\prime }=1}^{n/d}{d}^{\prime 2}\mu (d^{\prime })(\sum _{i=1}^{n/d^{\prime }d}i)(\sum _{j=1}^{m/d^{\prime }d}j)$

$5.\prod _{i=1}^n\prod _{j=1}^{m}gcd(i,j)=\prod _{d=1}^n{d}^{\sum _{i=1}^n\sum _{j=1}^m[gcd(i,j)==d]}$

$6.\sum _{i=1}^{n}gcd(i,n)=\sum _{d|n}\frac{n}{d}\phi (d)(Euler反演)$

$7.\sum _{i=1}^n\sum _{j=1}^{n}ij=(\frac{n(n+1)}{2}{)}^2=\sum _{i=1}^n{i}^3$

$8.\prod _{i=1}^n\prod _{j=1}^{n}ij=(n!{)}^{2n}$

$9.\sum _{i=1}^n\sum _{j=1}^{n}ij[gcd(i,j)==1]=\sum _{i=1}^n{i}^2\phi (i)$

$10.\sum _{i=1}^n\sum _{j=1}^{n}ijgcd(i,j)=\sum _{i=1}^n\sum _{j=1}^{n}ij\ast id(gcd(i,j))=\sum _{i=1}^n\sum _{j=1}^{n}ij\sum _{d|i,j|i}\phi (d)$

$11.\phi (ij)=\frac{\phi (i)\phi (j)gcd(i,j)}{\phi (gcd(i,j))}$