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24 lines
689 B
24 lines
689 B
def primefactors(n):
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""" Return een (natuurlijk) gesorteerde verzameling (list) van priemfactoren van n (int)
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Return [n] als n een priemgetal is, of wanneer n == 1.
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Tip: maak gebruik van de functie 'is_prime(n)' """
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sqrt = n**(1/2.0)
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factors = []
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if n == 1:
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factors.append(1)
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while n % 2 == 0:
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factors.append(2),
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n = n / 2
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for i in range(3,int(sqrt)+1,2):
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# while i divides n , print i ad divide n
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while n % i== 0:
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factors.append(i),
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n = n / i
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if n > 2:
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factors.append(n)
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return sorted(factors)
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print(primefactors(1))
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