java - 数的质因数

Question

问题来了

f(x)=sum of power of prime factor of x
now if f(x) is given then find the least value of x which also setisfy the condition
(No of divisor of x)-1=f(x).
Eg: f(x)=2 given and i need to find x 
Step 1:check for x=2 then f(x)=1
Step 2:check for x=3 then f(x)=1
Step 3: check for x=4 then f(x)=2 (i.e-x=2^2,and we know that f(x) is some of power of prime factor so f(x) here is 2)
And divisor of 4 is 1,2 & 4 so ,no of divisor -1=f(x)

so the Answer is x=4.

Now the approach i followed
Step 1-start for i=2 till we get the answer
step 2-find the prime factor for i and calculate the sum of power
Step 3-check if calculated sum is equal to the f(x) or not if not then increment i and repeat from step -1.

现在我的问题是

1. 我从 i=2 开始，每次将 i 递增 1 检查 f(x)。有什么有效的方法吗？
   1. 我也在继续循环，直到我得到答案。是否可以使用其他一些结束条件来完成，因为它可能会进入无限循环？ 请帮忙

Answer 1

每个数字都可以表示为:

x = a₁^p₁.a₁^p₁...a_n^p_n

地点:

a_i 1 <= i <= n 是质因数。

因此，要使 x 尽可能小，所有因子都应为 2，因此 x 将为 2 ^ f( x).

x 的约数是:(p₁ + 1)(p₂ + 1)...(p_n + 1) 所以你添加的条件也将得到满足!