41 lines
1.2 KiB
Python
41 lines
1.2 KiB
Python
######################################################################
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# The following iterative sequence is defined for the set of positive integers:
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#
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# n → n/2 (n is even)
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# n → 3n + 1 (n is odd)
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#
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# Using the rule above and starting with 13, we generate the following sequence:
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#
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# 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
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# It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms.
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# Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
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#
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# Which starting number, under one million, produces the longest chain?
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#
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# NOTE: Once the chain starts the terms are allowed to go above one million.
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######################################################################
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longestChain = []
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currentChain = []
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for i in range(2, 1000000):
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if i not in longestChain:
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print(i)
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currentChain.clear()
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n = i
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currentChain.append(n)
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while n != 1:
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if n % 2 == 0: # even
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n /= 2
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else:
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n = 3*n + 1
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currentChain.append(n)
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if len(currentChain) > len(longestChain):
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longestChain = currentChain.copy()
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print(longestChain[0])
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# Solution: 837799
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