Simple number coefficient

Question

Here is the question

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.

Now my question

• I start with i = 2 and checking f (x) by incrementing i by 1 every time. is there an effective way to do this?
  2. I also continue the cycle until I get an answer. can this be done with some other final condition, because it can go in an infinite loop? Please, help

Answer 1

Each number can be represented as:

x = a ₁ ^{p ₁} .a ₁ ^{p _{1 sub>}} ... a <sub> psub> ^{p <sub> psub>}

Where:

a _i 1 <= <= n - .

, x 2, x 2 ^ f(x).

x: (p ₁ + 1) (p ₂ + 1)... (p _n + 1), !