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Old 2011-01-27, 01:17   #1
wblipp
 
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Default Review of Odd Perfect Paper

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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Old 2011-01-27, 01:24   #2
R. Gerbicz
 
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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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Old 2011-01-27, 01:25   #3
fivemack
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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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Old 2011-01-27, 01:29   #4
R. Gerbicz
 
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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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Old 2011-01-27, 01:35   #5
R. Gerbicz
 
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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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Old 2011-01-27, 02:51   #6
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Quote:
Originally Posted by wblipp View Post
Pascal Ochem and Michael Rao asked for comments on the draft of their paper about Odd Perfect Numbers [...]

The current draft of the paper is available here:

http://www.lri.fr/~ochem/opn/opn.pdf
Typo in the introduction's first paragraph:
Replace \sigma(n)_{-1} by \sigma_{-1}(n).
(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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Old 2011-01-27, 04:21   #7
Pascal Ochem
 
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Quote:
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.
The new player sigma(p_(j,e)^e) is sigma(p_(j,e')^e') when e=e'. It is explicitely written now.

Quote:
what happens if p_(j,e') is not an exact divisor of sigma(p_(i,e)^e) ?
The exact same thing happens. A typo transformed | into ||. Very sorry for that.
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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Old 2011-01-27, 04:32   #8
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This is a fantastic piece of work!
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Old 2011-01-27, 12:29   #9
R. Gerbicz
 
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Quote:
Originally Posted by Pascal Ochem View Post
The new player sigma(p_(j,e)^e) is sigma(p_(j,e')^e') when e=e'. It is explicitely written now.

The exact same thing happens. A typo transformed | into ||. Very sorry for that.
I hope the remark sentence makes more sense now. And no, I don't feel like the professional guy.
The "then" part in the sentence begining by "Remark that" is still without any proof. It took me about 10-15 minutes to prove it.

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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Old 2011-01-27, 16:18   #10
R. Gerbicz
 
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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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Old 2011-01-27, 17:32   #11
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Quote:
Originally Posted by R. Gerbicz View Post
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.
The paper fails to give credit to the people who wrote the software
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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