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Resbog
searching PlanetScale…
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3 ms
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by
Resbog
7y ago
Since we're multiplying by 3 and adding 1 only if n is odd, i.e. for some n = 2k + 1, the result will be even 100% of the time, i.e. 3n + 1 = 3(2k + 1) + 1 = 6k + 4 = 2(3k + 2). Since this result is even, we'll immediately divide
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by
Resbog
7y ago
No. If you could prove that you'll always reach a previously seen number, then you could simply say "Let n be the smallest number for which the Collatz Conjecture is false, we have shown that we'll always reach a number <