Mobius function
definition
Set \ (n-= \ prod_ {I =. 1} ^ {K} P_i ^ {C_i} \) , then \ (\ MU (n-) = (-. 1) ^ K \) , in particular \ (\ mu (1 ) = 1 \) .
nature
The most common properties
\(\sum_{d|n}\mu(d)=[n=1]\)
Inversion properties
\(F(n)=\sum_{d|n}f(d) \Longleftrightarrow f(n)=\sum_{d|n}F(d)\mu(\frac{n}{d})\)
\(F(n)=\sum_{n|d}f(d) \Longleftrightarrow f(n)=\sum_{n|d}F(d)\mu(\frac{d}{n})\)
Common property
\(\sum_{i=1}^{n}\sum_{j=1}^{n}[gcd(i,j)=1]=\sum_{d=1}^{n}\mu(d)\lfloor\frac{n}{d}\rfloor \lfloor\frac{n}{d}\rfloor\)
Euler function
definition
设\(n=\prod_{i=1}^{k}p_i^{c_i}\)，有\(\phi(n)=n\prod_{i=1}^{k}\frac{p_i-1}{p_i}\)。
nature
The most common properties
\(\sum_{d|n}\phi(d)=n\)
Some properties piecemeal (did not finish the collection)
\ (\ Phi (ab) = \ frac {\ phi (a) \ phi (b) gcd (a, b)} {\ phi (gcd (a, b))} \)
Contact Mobius function
$\frac{\phi(n)}{n}=\sum_{d|n}\frac{\mu(d)}{d} \Longleftrightarrow \phi(n)=\sum_{d|n}\mu(d)\frac{n}{d}=\sum_{d|n}\mu(\frac{n}{d})d $
Du teach sieve
Thinking
Dirichlet convolution
Two function \ (f, g \) of Dirichlet convolution \ ((f * g) \ ) is defined as follows:
\((f*g)(i)=\sum_{d|i}f(d)g(\frac{i}{d})\)
Prefix and function
令\(S(n)=\sum_{i=1}^{n}f(i)\)
\(\sum_{i=1}^{n}(f*g)(i)\\=\sum_{i=1}^{n}\sum_{d|i}g(d)f(\frac{i}{d})\\=\sum_{d=1}^{n}g(d)\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor }f(i)\\=\sum_{d=1}^{n}g(d)S(\lfloor \frac{n}{d} \rfloor )\)
Transposition was:
\(g(1)S(n)=\sum_{i=1}^{n}(f*g)(i)-\sum_{d=2}^{n}g(d)S(\lfloor \frac{n}{d} \rfloor )\)
So it can be solved recursively.
Mobius function and the prefix
令\(f(i)=\mu(i),g(i)=1\)，有：
\(\sum_{i=1}^{n}(f*g)(i)\\=\sum_{i=1}^{n}\sum_{d|i}g(d)f(\frac{i}{d})\\=\sum_{i=1}^{n}\sum_{d|i}\mu(\frac{i}{d})\\=\sum_{i=1}^{n}[i=1]\\=1\)
So \ (S (n) = 1- \ sum_ {d = 2} ^ {n} g (d) S (\ lfloor \ frac {n} {d} \ rfloor) \)
Euler function and the prefix
Order \ (F (I) = \ Phi (I), G (I) =. 1 \) , are:
\(\sum_{i=1}^{n}(f*g)(i)\\=\sum_{i=1}^{n}\sum_{d|i}g(d)f(\frac{i}{d})\\=\sum_{i=1}^{n}\sum_{d|i}\phi(\frac{i}{d})\\=\sum_{i=1}^{n}i\\=\frac{n(n+1)}{2}\)
所以\(S(n)=\frac{n(n+1)}{2}-\sum_{d=2}^{n}g(d)S(\lfloor \frac{n}{d} \rfloor )\)
Prefix and a number of other functions
若\(f(i)=\mu(i)*i\)，则\(g(i)=i\)。
\(\sum_{i=1}^{n}(f*g)(i)\\=\sum_{i=1}^{n}\sum_{d|i}g(d)f(\frac{i}{d})\\=\sum_{i=1}^{n}\sum_{d|i}d*\frac{i}{d}*\mu(\frac{i}{d})\\=\sum_{i=1}^{n}i\sum_{d|i}\mu(\frac{i}{d})\\=\sum_{i=1}^{n}[i=1]\\=1\)
So \ (S (n) = 1 - \ sum_ {d = 2} ^ {n} g (d) S (\ lfloor \ frac {n} {d} \ rfloor) \)
Leave a pit first. . . . .