对于任意整数 \(x\in[1,n]\) 设 \(g(x)=\lfloor\frac{k}{\lfloor\frac{k}{x}\rfloor}\rfloor\)
\(\because f(x)=k/x\) 单调递减
又 \(g(x)=\lfloor\frac{k}{\lfloor\frac{k}{x}\rfloor}\rfloor\geq\lfloor\frac{k}{(\frac{k}{x})}\rfloor=x\)
\(\therefore \lfloor\frac{k}{g(x)}\rfloor\leq\lfloor\frac{k}{x}\rfloor\)
\(\because \lfloor\frac{k}{g(x)}\rfloor\geq\lfloor\frac{k}{(\frac{k}{\lfloor\frac{k}{x}\rfloor})}\rfloor=\lfloor\frac{k}{k}\lfloor\frac{k}{x}\rfloor\rfloor=\lfloor\frac{k}{x}\rfloor\)
\(\therefore \lfloor\frac{k}{g(x)}\rfloor=\lfloor\frac{k}{x}\rfloor\)
综上得,对于 \(i\in[x,\lfloor\frac{k}{\lfloor\frac{k}{x}\rfloor}\rfloor]\),\(\lfloor\frac{k}{i}\rfloor\)的值相等
\(\mu(n) = \begin{cases} 0, & \exists i \in [1,m],c_i>1 \\ 1, & m \equiv 0 \pmod 2,\forall i \in [1,m],c_i=1 \\ -1, & m \equiv 1 \pmod 2,\forall i \in [1,m],c_i=1 \end{cases}\)
Code:
mu[1]=1;
for(re int i=2;i<N;i++){
if(!mark[i]){
p[++cnt]=i;
mu[i]=-1;
}
for(re int j=1;j<=cnt&&i*p[j]<N;j++){
mark[i*p[j]]=1;
if(!(i%p[j])){
mu[i*p[j]]=0;
break;
}else mu[i*p[j]]=-mu[i];
}
}
对于 \(h(n)=\sum_{d|n} f(d)*g(\frac{n}{d})\)
记作 \(h=f\otimes g\) ,若 \(f,g\) 为积性函数,则 \(h\) 也为积性函数
对于给定数论函数 \(f,g\)
\(f(n)=\sum_{d|n} g(d) \iff g(n)=\sum_{d|n} \mu(d)*f(\frac{n}{d})\)
性质:
\(\sum_{d|n} \mu(d) = [n=1]\)
\(\sum_{d|n} \phi(d) =n\)
原文:https://www.cnblogs.com/wwlwQWQ/p/11272819.html