Preface:
Continue worthless ......
Mobius function:
\ [\ Mu (x) \]
Algorithm defines:
1. \ [\ mu (1) = 1 \]
2.当\[x=\prod_{i=1}^{k}p[i]\]
And when the p [i] is the number of mutually different prime
\ [\ Mu (x) = (- 1) ^ {k} \]
(The powers of prime factors of less than 2)
3. When all of the powers of prime factors greater than 1
\ [\ Mu (x) = 0 \]
nature:
1. If \ [n = 1 \]
则\[\sum_{x=1}^{n} x\mid n=1\]
Otherwise \ [\ sum_ {x = 1 } ^ {n} x \ mid n = 0 \]
2.\[\sum_{x=1}^{n}x\mid n\frac{\mu(x)}{x}=\frac{\phi(n)}{n}\]
Algorithm:
Sieve 1. While recording power of a prime number
2. Then statistics Mobius function