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#1 |
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"William"
May 2003
Near Grandkid
1001010001112 Posts |
Pascal Ochem and Michael Rao asked for comments on the draft of their paper about Odd Perfect Numbers in a different thread. The Forum Trolls were supposed to move those posts to a new thread here. One of the trolls was drunk, and deleted the posts instead of moving them. Please repost here.
The current draft of the paper is available here: http://www.lri.fr/~ochem/opn/opn.pdf |
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#2 |
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"Robert Gerbicz"
Oct 2005
Hungary
3×547 Posts |
On page 7 I see a big hole in the proof. The sentence beginning with "Remark that": I see at the end of the sentence sigma(p_(j,e)^e), but this term is a new player, before it I see only _(j,e') and _(i,e) indexes.
Further: what happens if p_(j,e') is not an exact divisor of sigma(p_(i,e)^e) ? That situation makes (2) invalid, by the following "semi-counter example": (this can't happen because sigma(p^2)!=square, but the proof currently doesn't handle this case) Code:
sigma(p1^2)=p2^2*q1^2 sigma(p2^2)=p3^2*q2^2 sigma(p3^2)=p4^2*q3^2 sigma(p4^2)=p1^2*q4^2 n_(2,3)=4, produced 8 factors for sigma(n) not cancelled by p_i^2 But you write 3*n_(2,3)=12 factors for the other side of sigma(n)=2n equation, so 12<=8, contradicition. Last fiddled with by R. Gerbicz on 2011-01-27 at 01:25 |
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#3 |
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(loop (#_fork))
Feb 2006
Cambridge, England
193616 Posts |
I think my main comment was that it's not terribly clear to use 'roadblock' both to describe a number whose complete factorisation is unknown, and a number whose factorisation is only as a product of primes already used in that part of the tree. Maybe a paragraph to describe exactly what branching means would be helpful, to avoid the paper being entirely dependent on already understanding the field.
In section 4, I don't see why you can start off by assuming that all p are greater than 10^8, I wonder whether it might be possible to get a more precise contradiction than the result than a positive number must be less than -10^14, and I'd like some idea of what other computations would be required to increase the bound on the largest prime factor of an odd perfect number. |
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#4 |
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"Robert Gerbicz"
Oct 2005
Hungary
31518 Posts |
Moreover: n_2=n_(2,2)+n_(2,3)+{0,1} // add 0 or 1.
is totally trivial if the above proofs are true on page 7. This makes a little interesting why (1) is so important. |
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#5 |
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"Robert Gerbicz"
Oct 2005
Hungary
164110 Posts |
Final note (before sleep), I've spent so far about one and half an hour to understand page 7, without any success. The remark sentence is very interesting, currently I don't understand that by "since" you prove the "then" or the "so" part, for this I would need two proofs, not? Or is it trivial? Probably you are much more professional than me on odd perfect numbers but for me it is a very compressed proof.
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#6 | |
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Apr 2010
22·3·13 Posts |
Quote:
Replace (And in the same sentence, make n an $n$ so that it's math italic.) Use \DeclareMathOperator{\gcd}{gcd} or something like that. $gcd$ looks like g times c times d. If you want to avoid the page break after Remark 2, try \beginparpenalty=10000. Last fiddled with by ccorn on 2011-01-27 at 03:00 |
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#7 | ||
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Apr 2006
109 Posts |
Quote:
Quote:
I hope the remark sentence makes more sense now. And no, I don't feel like the professional guy. n_2=n_(2,2)+n_(2,3)+{0,1} is trivial indeed. We want to keep a basic system of inequalities, and the useful one is n_2<=n_(2,2)+n_(2,3)+1, not n_2>=n_(2,2)+n_(2,3). Thinking again about the notion of roadblock and writing a paragraph on the meaning of branching is on the todo list. I assume that all primes are > 10^8 because we have ruled out by computation all primes < 10^8. We forbid {127, 19, 7, 11, 331, 31, 97, 61, 13, 398581, 1093, 3, 5, 307, 17} in section 2, {3, 5, 7, 1} in section 3, and { all primes < 10^8 } in section 4. We could be more tight using bounds on pi(x) instead of known values for some particular x and have a sharp contradiction, but the obtained bound wouldn't be much better. We would not reach 2*10^62. I think we can extract from the system of inequalities a bound of C*B^8 (for some small constant C) if we forbid all primes up to B. So this gives about 10^62 for B=10^8. The bound on the largest prime doesn't seem to fit in the common framework we have for the other 3 types of bound. We would need a bound on k such that k' >= k implies that sigma(p^k') has a prime factor greater than 10^10. I know that k=10^10 works but it is not good enough. Thank you very much for the comments. |
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#8 |
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Aug 2006
135448 Posts |
This is a fantastic piece of work!
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#9 | |
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"Robert Gerbicz"
Oct 2005
Hungary
31518 Posts |
Quote:
On page 8: "Also, notice that at least (n_(2,2)-1)...", Ok I can notice but WHY? I see why and you also (see page 7). |
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#10 |
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"Robert Gerbicz"
Oct 2005
Hungary
3×547 Posts |
I'm really curious, how much time have you spent on factoring (when there is a small p<10^8 divisor) to prove the theorem 5 ? The same question for the other two proofs.
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#11 | |
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"Bob Silverman"
Nov 2003
North of Boston
11101100011012 Posts |
Quote:
that performed the factorizations! They should also acknowledge that many factorizations were performed by many different people and mention the major outside contributors. The paper should also say something about the level of computational effort needed to gather their results. Nor do they say where the factoring data can be found so that others might confirm their calculations. |
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