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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)
3
"""
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("The value of log(log(n)) is {0:.4f}".format(math.log(math.log(n))))
"""
The number of distinct prime factors is/are 3
The value of log(log(n)) is 2.8765
"""