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# This theorem states that the number of prime factors of n

# will be approximately log(log(n)) for most natural numbers n

import math

def exactPrimeFactorCount(n):

"""

>>> exactPrimeFactorCount(51242183)

"""

count = 0

if n % 2 == 0:

count += 1

while n % 2 == 0:

n = int(n / 2)

# the n input value must be odd so that

# we can skip one element (ie i += 2)

i = 3

while i <= int(math.sqrt(n)):

if n % i == 0:

count += 1

while n % i == 0:

n = int(n / i)

i = i + 2

# this condition checks the prime

# number n is greater than 2

if n > 2:

count += 1

return count

if __name__ == "__main__":

n = 51242183

print(f"The number of distinct prime factors is/are {exactPrimeFactorCount(n)}")

print(f"The value of log(log(n)) is {math.log(math.log(n)):.4f}")

"""

The number of distinct prime factors is/are 3

The value of log(log(n)) is 2.8765

"""

Sunbelt Computer Software