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#12 |
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Jun 2005
Near Beetlegeuse
22×97 Posts |
That will teach you for switching your PC off. You should have it running 24/7 with prime95 using it while you are asleep.
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#13 | |
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Feb 2004
France
39416 Posts |
Quote:
Based on Lehmer technics and on my own previous work, I think I have a proof for (LLRT ?): The proof takes 1 page, based on theorems appearing in Williams' book. I'll write it in (I really prefer that test rather than the LLT, because it mimics the form of Mersenne numbers: Tony |
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#14 |
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Feb 2004
France
22×229 Posts |
Hi all,
Here is a draft of the proof of a new (AFAIK) primality test for Mersenne numbers, also based (like LLT) on Lucas Sequences and Lehmer theorems. (Should I name it: LLRT ?) I would appreciate early readers to check it. Do you think it is worth to propose it for publication in a Mathematic Journal ? It is close to the old LLT, but it flies through the Mersenne number space in a different way, producing different intermediate numbers and residues. Stay tuned, I will add some more properties. Also, I have to fix some minor mistakes in the document I refer to in the paper. For those who know nothing about Lucas Sequences and about Lucas-Lehmer theorems, just know that proving this theorem is easy, because Lucas and Lehmer have done the most difficult work. Thanks also to HC Williams for his wonderful book: "Edouard Lucas and Primality Testing". I think the most difficult part was to imagine that there could be a different way for proving primality of Mersenne numbers. Though providing a new primality proof for Mersenne numbers that has the same cost than LLT seems unuseful at first, I think that it may help the search of Mersenne primes and of Mersenne factors. Regards, Tony |
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#15 |
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"Robert Gerbicz"
Oct 2005
Hungary
2×743 Posts |
I think you waste your effort.
For Lucas Lehmer test x[0]=10 is also a good starting value, x[i+1]=x[i]^2-2. See http://primes.utm.edu/mersenne/ Your Lucas sequence for Mersenne numbers is: s[0]=5, s[i+1]=2*s[i]^2-1. It is easy to prove ( by induction ) that s[n]=x[n]/2, so Mq divides s[q-2] if and only if Mq divides x[q-2] ( because Mq is odd) . So it is not a new discovery. |
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#16 | |
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Feb 2004
France
22·229 Posts |
Quote:
(I thought to check that it produces different numbers, but I forgot to look at the other roots of LLT: At least, I've found some mistakes in the second paper. Is this paper (LLT for Fermat numbers) correct ? or did I make another (big) mistake ? Thanks for your help ! Tony |
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