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* Pyupgrade to Python 3.9
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46 lines
1022 B
Python
46 lines
1022 B
Python
# This theorem states that the number of prime factors of n
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# will be approximately log(log(n)) for most natural numbers n
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import math
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def exactPrimeFactorCount(n):
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"""
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>>> exactPrimeFactorCount(51242183)
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3
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"""
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count = 0
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if n % 2 == 0:
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count += 1
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while n % 2 == 0:
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n = int(n / 2)
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# the n input value must be odd so that
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# we can skip one element (ie i += 2)
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i = 3
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while i <= int(math.sqrt(n)):
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if n % i == 0:
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count += 1
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while n % i == 0:
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n = int(n / i)
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i = i + 2
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# this condition checks the prime
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# number n is greater than 2
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if n > 2:
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count += 1
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return count
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if __name__ == "__main__":
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n = 51242183
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print(f"The number of distinct prime factors is/are {exactPrimeFactorCount(n)}")
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print(f"The value of log(log(n)) is {math.log(math.log(n)):.4f}")
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"""
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The number of distinct prime factors is/are 3
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The value of log(log(n)) is 2.8765
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"""
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