1. The function of number theory
Defines two arithmetic functions \ (f (n) \) and \ (g (n) \)
The \ ((f + g) ( n) = f (n) + g (n) \)
2. Dirichlet convolution
The definition of Dirichlet convolution of two arithmetic functions \ (* \)
Definition of arithmetical function \ (t = f * g \ )
则\(\mathbf t(n)=\sum_{ij=n}\mathbf f(i)\mathbf g(j)\)
显然,\(f*g=g*f,f*(g*k)=(f*g)*k,f*(g+k)=f*g+f*k\)
Definition of arithmetical function \ (F \) of the unit element \ ([epsilon] \) , so \ (\ epsilon \ ast \ mathbf f = \ mathbf f * ε \)
Easy to see, when the \ (I \) is equal to \ (1 \) when, \ ([epsilon] (I) = 1 \) , otherwise, \ ([epsilon] (I) = 0 \)
Definition of arithmetical function \ (G \) as a function \ (F \) of the inverse element, i.e. \ (f * g = ε \ )
则\(g(n)=\dfrac{1}{f(1)}\left([n==1]−\sum\limits_{i|n,i\ne1}f(i)g(\dfrac{n}{i})\right)\)
3. multiplicative function
If a number theoretic function \ (F \) when \ (n⊥m \) when satisfying \ (F (nm) = F (n-) F (m) \) , this function is called multiplicative function
Common multiplicative function: \ ([sigma] * * _0 \) ( \ (n-\) number factor), \ (\ varphi (n-) \) ( \ ([. 1, n-] \) with \ (n- \) number of prime numbers)
in conclusion:
1. Dirichlet convolution of two functions is the product of a function of the product
Certify as follows:
设\(t=f*g\),则
\[ \begin{aligned}\mathbf t(nm)&=\sum_{d\mid nm}\mathbf f(d)\mathbf g\left(\frac{nm}d\right)\\&=\sum_{a\mid n,b\mid m}\mathbf f(ab)\mathbf g\left(\frac{nm}{ab}\right)\\&=\sum_{a\mid n,b\mid m}\mathbf f(a)\mathbf f(b)\mathbf g\left(\frac na\right)\mathbf g\left(\frac mb\right)\\&=\left(\sum_{a\mid n}\mathbf f(a)\mathbf g\left(\frac na\right)\right)\left(\sum_{b\mid m}\mathbf f(b)\mathbf g\left(\frac mb\right)\right)\\&=\mathbf t(n)\mathbf t(m)\end{aligned} \]
2.一个积性函数的逆也是积性函数
Certify as follows:
Set \ (g \) is \ (f \) inverse element
1. When \ (nm = 1 \) , then \ (G (. 1). 1 = \) , the conclusion holds
2.当\(nm>1\)时,则有
\[ \begin{aligned}\mathbf g(nm)&=-\sum_{d\mid nm,d\neq1}\mathbf f(d)\mathbf g\left(\frac{nm}d\right)\\&=-\sum_{a\mid n,b\mid m,ab\neq1}\mathbf f(ab)\mathbf g\left(\frac{nm}{ab}\right)\\&=-\sum_{a\mid n,b\mid m,ab\neq1}\mathbf f(a)\mathbf f(b)\mathbf g\left(\frac na\right)\mathbf g\left(\frac mb\right)\\&=\mathbf f(1)\mathbf f(1)\mathbf g(n)\mathbf g(m)-\sum_{a\mid n,b\mid m}\mathbf f(a)\mathbf f(b)\mathbf g\left(\frac na\right)\mathbf g\left(\frac mb\right)\\&=\mathbf g(n)\mathbf g(m)-\left(\sum_{a\mid n}\mathbf f(a)\mathbf g\left(\frac na\right)\right)\left(\sum_{b\mid m}\mathbf f(b)\mathbf g\left(\frac mb\right)\right)\\&=\mathbf g(n)\mathbf g(m)-\epsilon(n)\epsilon(m)\\&=\mathbf g(n)\mathbf g(m)\end{aligned} \]
Tips:对于一个积性函数,Obviously ,\ (f (1) = 1 \)\ (f (1) \ ne1 \) situation temporarily without discussion
use:
For a multiplicative function may be linear recursive method to quickly screen to give
We can put a multiplicative function \ (F \) , can be \ (f (n) \) decomposed \ (\ mathbf f (n) = \ prod_ {i = 1} ^ t \ mathbf f (p_i ^ { } K_i) \) , i.e. it is the prime factorization of
Since the mesh number of linear prime way we can determine the minimum number of each prime factor \ (P_1 \) , the minimum number of prime factors \ (k_1 \) and \ (n-/ k_1 P_1 ^ {} \) , can be recursive formula \ (\ mathbf f (n) = \ mathbf f (p_1 ^ {k_1}) \ mathbf f (n / p_1 ^ {k_1}) \) calculated directly \ (\ mathbf f \) of
Because $ σ_0 $ and (\ varphi \) \ readily obtained value prime power at, \ (K> 0 \) when, \ (\ sigma_0 (P ^ K) = K +. 1, \ varphi (P ^ K ) = p ^ {k-1 } (p-1) \)
从而推出公式:
\[ \sigma_0(n)=\prod_{i=1}^t(k_i+1), \varphi(n)=\prod_{i=1}^np_i^{k_i-1}(p_i-1)=n\prod_{i=1}^t\left(1-\frac1{p_i}\right) \]
4. Mobius inversion
The definition of \ (\ mathbf1 \) the inverse is \ (\ mu \)
Thus for \ (\ mathbf G = \ mathbf F \ AST \ mathbf1 \) , there \ (\ mathbf f = \ mathbf f \ ast \ mathbf1 \ ast \ mu = \ mathbf g \ ast \ mu \)
That is, if \ (\ mathbf G (n-) = \ sum_ {D \ MID n-} \ mathbf F (D) \) , then \ (\ mathbf f (n) = \ sum_ {d \ mid n} \ mu \ left (\ frac nd \ right) \ mathbf g (d) \)
This proves that the Möbius inversion
For \ (\ mu \) method of calculating
Obviously, because \ (\ mathbf1 \) is the product of nature, and \ (\ mu \) is \ (\ mathbf 1 \) inverse, so the \ (\ mu \) is the product of
Can be obtained:
\ [\ MU (P ^ K) = \ the begin {Cases}. 1 & K = 0 \\ -. 1 & K =. 1 \\ 0 & K>. 1 \ End {Cases} \]
to arrive:
\ [\ MU (n-) = \ begin {cases} (- 1) ^ t & n = p_1p_2 \ dots p_t \ text { and $ P_i $ mutually different} \\ 0 & n \ text {condition is not satisfied} \ End {Cases} \]
Tips: for \ (n-=. 1 \) , then \ (t = 0 \)
Inversion inference:
\[ [n==1]=\sum_{d|n}{\mu(d)} \]
From \ (\ mu \) is the inverse of an apparently well understood