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headache about Kollatz algorythm |
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| Feb6-05, 11:37 AM | #1 |
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headache about Kollatz algorythm
who can prove that if we have x=((3^n)*m-1)/(2^k) where x and m are odd
then x will be: x=(2^(n-1))*m1-1 where m1 is odd? is it possible to prove it? |
| Feb8-05, 01:13 PM | #2 |
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