莫比乌斯函数&莫比乌斯反演

莫比乌斯函数:
( 1 ) d = 1 , μ ( d ) = 1 (1)若d=1,则\mu(d)=1
( 2 ) d = p 1 p 2 p k , p i ( 1 i k ) μ ( d ) = ( 1 ) k (2)若d=p_1*p_2*\cdots*p_k,其中p_i(1\leq i \leq k)为互异的素数, 则\mu(d)=(-1)^k
( 3 ) μ ( d ) = 0 (3)其余情况\mu(d)=0

莫比乌斯反演:
f ( k ) = d k g ( d ) g ( k ) = d k f ( k d ) μ ( d ) f(k)=\sum^{}_{d|k}g(d)\Rightarrow g(k)=\sum^{}_{d|k}f(\frac{k}{d})\mu(d)
f ( k ) = k d d n f ( d ) g ( k ) = k d d n f ( d k ) μ ( d ) f(k)=\sum^{d\leq n}_{k|d}f(d)\Rightarrow g(k)=\sum^{d\leq n}_{k|d}f(\frac{d}{k})\mu(d)
莫比乌斯函数性质:
( 1 ) n d n μ ( d ) = { 1 n = 1 0 n > 1 (1)对于任意正整数n,有:\sum^{}_{d|n}\mu(d)=\left\{\begin{array}{cc}1,若n=1\\0,若n>1\end{array}\right.
( 2 ) n d n μ ( d ) d = φ ( n ) n (2)对于任意正整数n,有\sum^{}_{d|n}\frac{\mu(d)}{d}=\frac{\varphi(n)}{n}
( 3 ) μ ( n ) (3)莫比乌斯函数\mu(n)是积性函数

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转载自blog.csdn.net/qq_15647239/article/details/89854237