Konjac tree-lined review - Mobius inversion

Mobius inversion

Multiplicative function: for a function f, if the prime number p, q, such that f (p) f (q) = f (pq), the function f is a multiplicative function

Provided multiplicative function f, and has the function

　

Clearly, F is determined by f, whether this relationship can be reversed?

 　F(1)=f(1)F(1)=f(1)      
  F(2)=f(1)+f(2)F(2)=f(1)+f(2)      
  F(3)=f(1)+f(3)F(3)=f(1)+f(3)      
  F(4)=f(1)+f(2)+f(4)F(4)=f(1)+f(2)+f(4)   
  F(5)=f(1)+f(5)F(5)=f(1)+f(5)      
  F(6)=f(1)+f(2)+f(3)+f(6)F(6)=f(1)+f(2)+f(3)+f(6)      
  F(7)=f(1)+f(7)F(7)=f(1)+f(7)    
  F(8)=f(1)+f(2)+f(4)+f(8)F(8)=f(1)+f(2)+f(4)+f(8)     

Can be found, we can go by the reverse thrust F f

Formal say, f (x) is equal to a number of the form ± F (x / d), and

You may have this formula:

 

Wherein μ is an arithmetic function, if the equality holds, then

m (1) = 1

m (2) = - 1

m (3) = - 1

m (4) = 0

m (5) = - 1

m (6) = 1

When a prime number p, f (p) = F (p) -F (1)

Then we can find, μ (1) = 1, for any prime number, μ (p) = - 1

Since F (p ^ 2) = f (1) + f (p) + (fp ^ 2), f (p ^ 2) = F (p ^ 2) -F (p)

The μ (p ^ 2) = 0

This requires there to primes p p ^ 2 = 0, down recursive seen f (p ^ 3) = F (p ^ 3) -F (p ^ 2)

Found, μ (1) = 1, μ (p) = - 1, μ (p ^ 2) = 0, μ (p ^ 3) = 0;

Found, μ (p ^ k) = 0;

Pledge number p1p2p3 different from each other

 

For f (p1p2) = F (p1p2) -F (p1) -F (p2) + F (1)

则此时μ(p1p2)=1

对于f(p1p2p3)=F(p1p2p3)-F(p1p2)-F(p1p3)-F(p2p3)+F(p1)+F(p2)+F(p3)-F(1)

可以发现，μ(p1p2p3)=-1,μ(p1p2)=1。

同理，μ(p1p2p3p4)=1,μ(p1p2p3p4p5)=-1

 

则可知，对于μ(1)=1,μ(p1p2p3p4...pk)(互为不同质数)=(-1)^k

 

假设我们现在定义一个合数x且x的质因数分解中有相同的质数记作

x=p1pk^2

那么f(p1pk^2)=F(p1pk^2)-F(p1pk)-F(pk^2)+F(pk^2)

可以发现，μ(p1)=-1,μ(pk^2)=0,μ(p1pk^2)=0

则可以推断出，如果x的质因数分解形如x=p1p2p3p4...pk^2,则μ(x)=0

综上，证明得出莫比乌斯函数μ(x)

1. x=1时μ(x)=1
2. x的质因数分解形如x=p1p2p3p4....pk，p1p2p3p4...互不相同，则μ(x)=(-1)^k
3. 如果不属于以上两种情况，即对x进行质因数分解有重复质数，μ(x)=0
4. 莫比乌斯函数是积性函数

莫比乌斯函数的性质：

容易发现,n=1时易证，

n> 1, seen from the definition of the function of the product $F.(n-)$

Wherein n = p1 ^ a1 * p2 ^ a2 ... * pt ^ at.

For the F (p ^ k) with a (k> 2 when)

 

k <= 2, 1 + (- 1) = 0

Thus, n> 1 when F (n) = 0

Mobius inversion:

for:

Have:

 

 

Another form:

for:

Have: