Möbius transformation easy to get started
I wrote a few questions Mobius inversion, less likely to inversion routine, but feel directly Möbius transformation would be easier to use a little ...
(Shenkaoziliao: "On the basis of the number of" pan chengdong a)
Some functions
- Mobius function: \ [\ MU (n-) = \ Cases the begin {}. 1. 1 & n-\\ = (-. 1) ^ n-S & P_1 = \ \\ 0 & cdots P_s other \ end {cases} \] i.e. containing function function of the square value of the number of prime factors is 0,1 is 1
- Convolution unit element: \ [the I (n-) = [n-==. 1] \]
- Multiplicative identity element: \ (U (the n-) \ equiv1 \)
- Identity function: \ (E (n-) = n-\)
- Euler function: \ (\ varphi (n) = \ SUM \ limits_. 1} ^ {n = I [(I, n) ==. 1] \) , and from 1 ~ n n number of prime numbers.
Dirichlet product
Like addition, subtraction, is a convolution operation, to be familiar with its use.
If \ (F (n-) \) , \ (G (n-) \) is the two arithmetic functions, the
\[h(n)=\sum\limits_{d|n}f(d)g(\frac{n}{d})\]
Called \ (F (n-) \) , \ (G (n-) \) a \ (the Dirichlet \) product (convolution).
// is written in general: \ (H = f * G \)
This operation, which satisfies the commutative and associative law !
- Commutative: \ (F * F * G = G \)
- Associativity: \ ((F * G) = F * H * (G * H) \)
Proved to be another form of convolution: \ ((F * G) (n-) = \ SUM \ n-limits_} = F {ab & (A) G (B) \)
- Commutative is obvious
- 结合律：\[(f*(g*h))(n)=\sum\limits_{ad=n}f(a)(g*h)(d)=\sum\limits_{ad=n}f(a)\sum\limits_{bc=d}g(b)h(c)=\sum\limits_{abc=n}f(a)g(b)h(c)=((f*g)*h)(n)\]
Useful conclusions
\(I=\mu *u\)，即\(I(n)=\sum\limits_{d|n}\mu(d)=\begin{cases}1 & n=1\\0& n≠1\end{cases}\)
- \ (n-=. 1 \) , clearly
\(n=p_1^{\alpha_1}\cdots p_1^{\alpha_1}\)，\(\sum\limits_{d|n}=1+C_s^1(-1)+C_s^2(-1)^2+\cdots+C_s^s(-1)^s=(1-1)^n=0\)
Based on the above conclusions, see similar \ ([\ gcd (x, y) == 1] \) can be substituted into the equation to the
Möbius transformation
If \ (F = f * u \ ) then \ (F. \) Called \ (F \) is converted Möbius
That \ [F (n) = \ sum \ limits_ {d | n} f (d) \]
By the way make up about inversion is inverse transform, is to use \ (f \) represents \ (F \)
- 若\(F=f*u\)，则\(f=F*\mu\)
Volume on both sides at the same time it \ (\ MU \) : \ (F * \ MU = f * U * \ MU the I = f = f * \)
// This is the \ (I \) reason is called convolution units yuan
- Familiar expression: If \ (F. (N-) = \ SUM \ limits_ {D | n-} F (D) \) , then the \ (f (n) = \ sum \ limits_ {d | n} \ mu (d) \)
Basic Operations
- Replacing enumeration summed number, for example, \ (\ SUM \ limits_. 1} ^ {n-I = \ SUM \ limits_ {D | I. 1} = \ SUM \ _ Limits. 1} ^ {n-D = \ lfloor \ frac {n} {d} \ rfloor \)
- Enumeration and change the way, like, \ (\ SUM \ limits_. 1} = X ^ {m \ SUM \ limits_. 1} = ^ {n-Y [\ GCD (X, Y) == K] = \ SUM \ limits_ {x = 1} ^ { m / k} \ sum \ limits_ {y = 1} ^ {n / k} [\ gcd (x, y) == 1] \)
- Enumeration of common factor, such as, \ (\ SUM \ limits_. 1 = {X} ^ {m / K} \ SUM \ limits_. 1} = {Y {n-^ / K} [\ GCD (X, Y) == 1] = \ sum \ limits_ { x = 1} ^ {m / k} \ sum \ limits_ {y = 1} ^ {n / k} \ sum \ limits_ {d = 1} ^ {\ min (m / k , n / k)} [d | \ gcd (x, y)] \)
- The common factor is the factor common divisor of two numbers
- \([d|\gcd(x,y)]=[d|x][d|y]\)
- There formula: \ (D (ij of) = \ SUM \ limits_ {X | I} \ SUM \ limits_ {Y | J} [GCD (X, Y) ==. 1] \)