scramble prime challenge
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02-18-2020, 03:05 PM
Post: #21
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RE: scramble prime challenge
(02-18-2020 12:55 AM)Don Shepherd Wrote: So, only a few scramble primes with 2 or 3 or 4 digits, and none beyond that. I had no idea how many I expected, but I guess I'm not surprised at the lack of scramble primes with more than 3 digits; primes are less elusive than I would have thought. Yeah, I suppose it makes sense intuitively when you think about it. The more digits in the number, the more permutations, and the more chances that at least one will be composite. I wasn't expecting none over 4, though. |
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02-18-2020, 05:26 PM
(This post was last modified: 02-18-2020 05:38 PM by Don Shepherd.)
Post: #22
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RE: scramble prime challenge
(02-18-2020 03:05 PM)Dave Britten Wrote:(02-18-2020 12:55 AM)Don Shepherd Wrote: So, only a few scramble primes with 2 or 3 or 4 digits, and none beyond that. I had no idea how many I expected, but I guess I'm not surprised at the lack of scramble primes with more than 3 digits; primes are less elusive than I would have thought. Yeah, I was hoping that, for example, 917371993771911 might be a scramble prime, but with over 1 trillion permutations of those 15 digits, it would be highly unlikely. |
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02-19-2020, 12:16 AM
Post: #23
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RE: scramble prime challenge
(02-18-2020 05:26 PM)Don Shepherd Wrote: Yeah, I was hoping that, for example, 917371993771911 might be a scramble prime, but with With repeating digits, permutations should use multinomial coefficient For 917371993771911, permutations = \(\large\binom{15}{5,2,4,4} = {15! \over 5! 2! 4! 4!} \) = 9,459,450 I made the same mistake with Python's itertools.permutations, which do n! permutations. With repeated digits, many permutations produce the same patterns, wasting cycles. Then, I googled and found this: Distinct permutations of the string | Set 2 Post #9 unique_permute were "borrowed" from article's Python3 code |
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02-19-2020, 11:04 AM
Post: #24
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RE: scramble prime challenge
Thanks Albert.
Fascinating stuff indeed. |
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