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#78 | |
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Aug 2002
Buenos Aires, Argentina
2×683 Posts |
Quote:
Code:
10 prime factors known for 2 Mersenne numbers 9 prime factors known for 9 Mersenne numbers 8 prime factors known for 86 Mersenne numbers 7 prime factors known for 346 Mersenne numbers 6 prime factors known for 1,788 Mersenne numbers 5 prime factors known for 16,262 Mersenne numbers 4 prime factors known for 139,199 Mersenne numbers Last fiddled with by alpertron on 2014-08-23 at 00:19 Reason: prime factors, not factors |
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#79 |
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May 2013
East. Always East.
11×157 Posts |
Careful... Just because we haven't found many doesn't mean that there aren't. For most of these, we've stopped looking because with the exception of a bit of hobbying, nobody cares. A select few might want a factor for EVERY candidate, regardless of LL result.
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#80 |
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Aug 2002
Buenos Aires, Argentina
2·683 Posts |
OK. Let me rephrase my thinking differently. Suppose that we know not 10 but 9 prime factors of M7508981. The probability of finding the tenth prime factor would be so extremely tiny according to Bob that it would be nearly impossible that we found the 10th prime factor.
Then we can continue backtracking, and finally we should conclude that it would be impossible with the current bounds to find more, than say 5 prime factors. This is not what occurs according to my previous post. So it appears that something in Bob's argument is not entirely correct. |
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#81 |
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Apr 2014
7×17 Posts |
@Silverman
No reason to be rude, your comments are being less than helpful. I will be ignoring your inquiries until you choose to be civil. |
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#82 |
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Aug 2002
Buenos Aires, Argentina
2·683 Posts |
Given that the factor bound is 25 digits, I'd estimate the probability at little less than 50%, but the problem is that you will need a lot of people running ECM in a lot of computers each in order to complete the 45-digit level. Notice that adding 5 digits is about 10 times slower, so adding 20 digits is about ... slower (fill in the blanks).
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#83 |
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"William"
May 2003
New Haven
2·7·132 Posts |
I was going to ask why this is larger than the heuristic in post 59, which is p^2, but then I realized that you have the mean and post 59 has the median. The method in post 59 also handles the case where trial factoring has cleared beyond a known prime - does your method say n^e for that case?.
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#84 |
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May 2013
East. Always East.
32778 Posts |
Well, Silverman is suggesting a breadth-first search for factors. In other words, give each exponent the same amount of work until we can find a reason to focus a lot of work onto a specific one.
The counter-argument is that given that the remainder of M7508981 is not PRP and that there are already 10 known primes, it follows that there ARE 12 factors and we're looking for the eleventh (to do more PRP on the leftover, etc etc). This IS the reason to put so much work onto a specific exponent. The reason certain exponents have two or three factors and that certain others have eight or nine or ten are because the distribution isn't fixed in stone, but made in a statistical way. The occasional exponent will give 10 small factors like our friend M7508981 which is just an outlier (hence why we're bothering to look at it at all). While he is very direct with our massive stupidity he certainly is not with the mental exercise he wants us to perform. I believe what he is suggesting is that we ought to show that we won't be more successful finding another outlier with many (as in even more than 10) factors versus finding the actual eleventh and twelfth factors of this specific one. I don't think he actually knows which path would be more successful. Instead he wants us to justify every little thing we're doing instead of just doing something for the hell of it. Last fiddled with by TheMawn on 2014-08-23 at 06:51 |
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#85 | ||
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Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
22×5×72×11 Posts |
Quote:
Not saying there's anything wrong with that, only trying to make it explicit. AIU Bob's earlier statements here and in other threads suggests he agrees (the nothing wrong bit) but he wants people to understand why they are doing what they are doing and not to mislead (inadvertently) others into thinking that an alternative justification is present. Quote:
Paul Last fiddled with by xilman on 2014-08-23 at 08:02 Reason: Must try not to (inadvertently) split infinitives. |
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#86 | |
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Nov 2003
11101001001002 Posts |
Quote:
that rather than "will be ignoring until you choose to be civil", you have been IGNORING THEM ALL ALONG. Furthermore, I tell you something and then you, despite not knowing any math ARGUE ABOUT IT. Last fiddled with by R.D. Silverman on 2014-08-23 at 08:37 |
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#87 | |
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Nov 2003
22×5×373 Posts |
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This is correct. I do not know which would be more successful. I strongly suspect (based on some experience with similar things) that a breadth-first search will succeed with less expected work. I have not crunched through the numbers, Note that the work being performed here is a random variable. Even if one has success with a depth-first search, it does not prove that such a search will succeeed on average with less work. I do not have the tools or the time (or the inclination) to crunch the numbers to find out which method would be better. I can only suggest the methodology to the problem. DAMN IT! I'm trying to get people to apply some analysis before blindly plunging into a computation. Otherwise it is just "try it and hope". I am, of course, assuming that the goal is to find an M_p with at least 13 prime factors in the most efficient manner possible so that this group effort succeeds in as little elapsed time as possible.. |
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#88 | |
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Aug 2002
Buenos Aires, Argentina
2×683 Posts |
Quote:
Code:
10 prime factors known for 1 Mersenne number 9 prime factors known for 1 Mersenne number 8 prime factors known for 5 Mersenne numbers 7 prime factors known for 16 Mersenne numbers -> 0.003 % 6 prime factors known for 114 Mersenne numbers -> 0.019 % 5 prime factors known for 682 Mersenne numbers -> 0.116 % 4 prime factors known for 4,337 Mersenne numbers -> 0.740 % Last fiddled with by alpertron on 2014-08-23 at 13:56 |
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