2022. 1. 23. 21:14ㆍ수학
f(n)=∑d|ng(d)
이라고 하자. 이때 g(n)의 값을 구하고 싶다. 이때 적용되는 식이 뫼비우스 역공식이다.
g(n)=∑d|nf(d)μ(nd)
f가 곱셈적 함수이고 f(n)=∑d|ng(d)이면 g 역시 곱셈적 함수이다.
- Notation
[P=k] : P=k일 때 1 아니면 0
ϵ(n) : [n=1]
∑d|nμ(d)=ϵ(n)
τ(n) : n의 약수의 개수
a⊥b은 gcd라는 뜻
1.\ \ \sum_{i=1}^{n} \sum_{j=1}^{m} \gcd(i, j)
\sum_{d=1}^{\min(n,m)} d \sum_{i=1}^{n} \sum_{j=1}^{m} [\gcd(i, j)=d]
i=dp,\ j=dq이면
\sum_{d=1}^{\min(n,m)} d \sum_{p=1}^{\left[\frac{n}{d}\right]} \sum_{q=1}^{\left[\frac{m}{d}\right]} [\gcd(p,q)=1]
\epsilon(\gcd(p,q))=[\gcd(p,q)=1]=\sum_{s|\gcd(p,q)} \mu(s) = \sum_{s|p,\ s|q} \mu(s)
\sum_{d=1}^{\min(n,m)} d \sum_{p=1}^{\left[\frac{n}{d}\right]} \sum_{q=1}^{\left[\frac{m}{d}\right]} \sum_{s|p,\ s|q} \mu(s)
\sum_{d=1}^{\min(n,m)} d \sum_{p=1}^{\left[\frac{n}{d}\right]} \sum_{q=1}^{\left[\frac{m}{d}\right]} \sum_{s=1}^{\min(\left[\frac{n}{d}\right],\ \left[\frac{m}{d}\right])} \mu(s)[s|p][s|q]
[s|p][s|q]=1이면 [s|p]=[s|q]=1이므로 p=s\alpha,\ q=s\beta
\sum_{d=1}^{\min(n,m)} d \sum_{s=1}^{\min(\left[\frac{n}{d}\right],\ \left[\frac{m}{d}\right])} \mu(s) \sum_{\alpha=1}^{\left[\frac{n}{sd}\right]} \sum_{\beta=1}^{\left[\frac{m}{sd}\right]} 1
\sum_{d=1}^{\min(n,m)} d \sum_{s=1}^{\min(\left[\frac{n}{d}\right],\ \left[\frac{m}{d}\right])} \mu(s) \left[\frac{n}{sd}\right] \left[\frac{m}{sd}\right]
합을 간단히 만들자.
1 \le d \le \min(n,m)
1 \le s \le \min(\left[\frac{n}{d}\right], \left[\frac{m}{d}\right])
d \le ds \le \min(n,m)
1 \le d \le k=ds \le \min(n,m)
d|k가 성립해야 한다.
\sum_{d=1}^{\min(n,m)} \sum_{s=1}^{\min(\left[\frac{n}{d}\right],\ \left[\frac{m}{d}\right])}
는 다음과 같이 쓸 수 있다.
\sum_{k=1}^{\min(n,m)} \sum_{d=1}^{\min(n,m)}
d|k인 조건과 1 \le d \le \min(n,m)으로 인해 아래와 같이 쓰인다.
\sum_{k=1}^{\min(n,m)} \sum_{d|k}
\sum_{k=1}^{\min(n,m)} \left[\frac{n}{k}\right] \left[\frac{m}{k}\right] \sum_{d|k} d\mu(\frac{k}{d})
좀 더 간단히 하면
n=\sum_{d|n} \phi(d)
\phi(n) = \sum_{d|n} d\mu(\frac{k}{d})
\sum_{k=1}^{\min(n,m)} \left[\frac{n}{k}\right] \left[\frac{m}{k}\right] \phi(k)
2.\ \ \sum_{i=1}^{n} \sum_{j=1}^{m} \operatorname{lcm}(i, j)
\operatorname{lcm}(i, j) = \frac{ij}{\gcd(i,j)} 이므로
\sum_{i=1}^{n} \sum_{j=1}^{m} \frac{ij}{\gcd(i,j)}
\sum_{d=1}^{\min(n,m)} \sum_{i=1}^{n} \sum_{j=1}^{m} \frac{ij}{d}[\gcd(i,j)=d]
i=dp,\ j=dq라 두자.
\sum_{d=1}^{\min(n,m)} \sum_{p=1}^{\left[\frac{n}{d}\right]} \sum_{q=1}^{\left[\frac{m}{d}\right]} dpq [\gcd(p,q)=1]
\epsilon(\gcd(p,q))=[\gcd(p,q)=1]=\sum_{s|\gcd(p,q)} \mu(s) = \sum_{s|p,\ s|q} \mu(s)
\sum_{d=1}^{\min(n,m)} \sum_{p=1}^{\left[\frac{n}{d}\right]} \sum_{q=1}^{\left[\frac{m}{d}\right]} dpq \sum_{s|p,\ s|q} \mu(s)
\sum_{d=1}^{\min(n,m)} \sum_{p=1}^{\left[\frac{n}{d}\right]} \sum_{q=1}^{\left[\frac{m}{d}\right]} dpq \sum_{s=1}^{\min(\left[\frac{n}{d}\right],\ \left[\frac{m}{d}\right])} \mu(s)[s|p][s|q]
\sum_{d=1}^{\min(n,m)} d \sum_{s=1}^{\min(\left[\frac{n}{d}\right],\ \left[\frac{m}{d}\right])} \mu(s) \sum_{p=1}^{\left[\frac{n}{d}\right]} \sum_{q=1}^{\left[\frac{m}{d}\right]} pq[s|p][s|q]
[s|p][s|q]=1이면 [s|p]=[s|q]=1이므로 p=s\alpha,\ q=s\beta
\sum_{d=1}^{\min(n,m)} d \sum_{s=1}^{\min(\left[\frac{n}{d}\right],\ \left[\frac{m}{d}\right])} \mu(s) s^{2} \sum_{\alpha=1}^{\left[\frac{n}{sd}\right]} \sum_{\beta=1}^{\left[\frac{m}{sd}\right]} \alpha\beta
\sum_{d=1}^{\min(n,m)} \sum_{s=1}^{\min(\left[\frac{n}{d}\right],\ \left[\frac{m}{d}\right])} d\mu(s) s^{2} \frac{\left[\frac{n}{sd}\right](\left[\frac{n}{sd}\right]+1)}{2} \frac{\left[\frac{m}{sd}\right](\left[\frac{m}{sd}\right]+1)}{2}
합을 간단히 만들면
\sum_{k=1}^{\min(n,m)} \frac{\left[\frac{n}{k}\right](\left[\frac{n}{k}\right]+1)}{2} \frac{\left[\frac{m}{k}\right](\left[\frac{m}{k}\right]+1)}{2} \sum_{d|k} k\mu(\frac{k}{d}) \frac{k}{d}
\sum_{k=1}^{\min(n,m)} \frac{\left[\frac{n}{k}\right](\left[\frac{n}{k}\right]+1)}{2} \frac{\left[\frac{m}{k}\right](\left[\frac{m}{k}\right]+1)}{2} \sum_{d|k} k\mu(d) d
f(n) = \sum_{d|n} n\mu(d) d
\frac{f(n)}{n^{2}} = \sum_{d|n} \frac{1}{d} \mu(\frac{n}{d})
h(n) = \frac{f(n)}{n^{2}} = \sum_{d|n} g(d) \mu(\frac{n}{d}) \ \ \ g(n)=\frac{1}{n}
g가 곱셈적 함수이고 조건을 만족하므로 h 역시 곱셈적 함수이다.
g(p^{k})=\sum_{i=0}^{k} h(p^{k})
g(p^{k})-g(p^{k-1}) = h(p^{k})=\frac{f(p^{k})}{p^{2k}}
f(p^{k})=p^{k}-p^{k+1}
3.\ \ \sum_{i=1}^{n} \sum_{j=1}^{m} \operatorname{lcm}(i, j) |\mu(gcd(x,y))|
이것 역시 위와 같은 방식으로 정리하면 다음 식을 얻는다.
\sum_{k=1}^{\min(n,m)} \frac{\left[\frac{n}{k}\right](\left[\frac{n}{k}\right]+1)}{2} \frac{\left[\frac{m}{k}\right](\left[\frac{m}{k}\right]+1)}{2} \sum_{d|k} k\mu(\frac{k}{d}) \frac{k}{d}|\mu(d)|
h(n) = \frac{f(n)}{n^{2}} = \sum_{d|n} g(d) \mu(\frac{n}{d}) \ \ \ g(n)=\frac{|\mu(n)|}{n}
d,\ \mu(d)가 각각 곱셈적 함수이므로 g(n)역시 곱셈적 함수이다. 따라서 h(n) 역시 곱셈적 함수이다.
f(p^{k})=p^{k}|\mu(p^{k})|-p^{k+1}|\mu(p^{k-1})|
k=1,\ \ f(p)=p-p^{2}
k=2,\ \ f(p^{2})=-p^{3}
k \ge 3,\ \ f(p^{k})=0
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