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Topic: More hailstone numbers (Read 1568 times) |
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JocK
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More hailstone numbers
« on: Jun 25th, 2005, 8:42am » |
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Starting from a positive integer k, one can generate a series of integers by iterating the mapping:
k' = k/2 if k even,
k' = (3k+5n)/2 if k odd.
Here, n is an odd positive integer constant.
Can you find a starting value and a n-value for which the iteration fails to yield a value not exceeding 347n?
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| « Last Edit: Jun 25th, 2005, 8:43am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Barukh
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Re: More hailstone numbers
« Reply #1 on: Jun 26th, 2005, 11:43pm » |
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The minimal solution I've found is: n = 29, k = 12655. It generates a 66-long cycle with minimal element 19055.
Of course, this was achieved using computing power. Currently, I have no idea how to tackle this problem more intelligently.
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