Program to find probability that any proper divisor of n would be an even perfect square number in Python

Suppose we have a number n, we have to find out the probability that any proper divisor of n would be an even perfect square.

So, if the input is like n = 36, then the output will be 1/8 because there are eight proper divisors of 36, these are {1,2,3,4,6,9,12,18} and among them only one number (4) is perfect square and even.

To solve this, we will follow these steps −

• if n mod 4 is not same as 0, then
  • return 0
• otherwise,
  • nc := n, ptr := 2
  • l := a new list
  • while ptr <= square root of nc , do
    • a := 0
    • while nc mod ptr is same as 0, do
      • a := a + 1
      • nc := floor of (nc / ptr)
    • if a > 0, then
      • append a into the list l
    • ptr := ptr + 1
  • if nc > 1, then append 1 into the list l
  • k := l[0]
  • d := k + 1
  • no := floor of (k / 2)
  • for each i in l[from index 1 to end], do
    • d := d *(i + 1)
    • no := no * floor of (i / 2) + 1
  • d := d - 1
  • if n is a perfect square, then
    • no := no - 1
  • g := gcd of d and no
  • d := floor of d / g
  • no := floor of no / g
  • if no is same as 0, then
    • return 0
  • otherwise,
    • return a fraction no/d

Example

Let us see the following implementation to get better understanding −

from math import gcd

def solve(n):
   if n % 4 != 0:
      return 0
   else:
      nc = n
      ptr = 2
      l = []
      while ptr <= nc ** 0.5:
         a = 0
         while nc % ptr == 0:
            a += 1
            nc = nc / ptr
         if a > 0:
            l += [a]
         ptr += 1
      if nc > 1:
         l += [1]
      k = l[0]
      d = k + 1
      no = int(k / 2)
      for i in l[1:]:
         d = d * (i + 1)
         no *= int(i / 2) + 1
      d = d - 1
      if int(n ** 0.5) ** 2 == n:
         no -= 1
      g = gcd(d, no)
      d = d // g
      no = no // g
      if no == 0:
         return 0
      else:
         return str(no) + '/' + str(d)

n = 36
print(solve(n))

Input

4, 27

Output

1/8