鲨币猜想

$设m=2^k(k为正整数) 则\varphi(m)=\dfrac{1}{2}\times m$

$设m=3^k(k为正整数) 则\varphi(m)=\dfrac{2}{3}\times m$

$设m=5^k(k为正整数) 则\varphi(m)=\dfrac{4}{5}\times m$

$设m=7^k(k为正整数) 则\varphi(m)=\dfrac{6}{7}\times m$

$......$

根据此代码得出猜想

long long phi[100000010],isnotp[100000010];
vector<long long> p;
void ouler(){
	phi[1]=1;
	for(int i=2;i<=n;i++){
		if(!isnotp[i]){
			p.push_back(i);
			phi[i]=i-1;
		}
		for(int j=0;j<p.size();j++){
			int x=p[j];
			if(i*x>n) break;
      		isnotp[i*x]=1;
      		if(i%x==0) {
      			phi[i * x] = phi[i] * x;
        		break;
      		}
      		phi[i * x] = phi[i] * phi[x];
		}
	}
}

得出普遍结论

$设m=p^k(k为正整数) 则\varphi(m)=\dfrac{p-1}{p} \times m(p为质数)$