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Topic: trying to get an upper bound on 1/log(n/m) when (n,m) = 1 and
n >m > 0 , n & m integers

Replies: 19   Last Post: Apr 16, 2012 3:28 PM

 Messages: [ Previous | Next ]
 Pubkeybreaker Posts: 1,291 Registered: 2/12/07
Re: trying to EXAMINE THE DIGITS OF CERTAIN PRIMES FOR PATTERNS
Posted: Apr 16, 2012 8:50 AM

On Apr 16, 12:43 am, quasi <qu...@null.set> wrote:
> On Mon, 16 Apr 2012 04:43:12 +0200, barker
>

> >> >No.
>
> >> >It is the smallest one whose prime factors have not been
> >> >published. There is a difference...

These factorizations get published very quickly. I know everyone
capable of factoring R281 and none of them have done it. It
REMAINS unfactored.

>
> >> The problem with that is how do we know that it wasn't
> >> obtained simply by multiplying two large primes?

>
> >You have contrived to confuse yourself.
>
> >Pubkeybreaker <pubkeybrea...@aol.com> mentioned the repunit 10^281-1,
> >attributing a property incorrectly to it. I never mentioned it until
> >he did. I am not he. Thereafter, I have provided a large factor of
> >this number, and can, if properly motivated, provide all its prime
> >factors (i.e., decompose the number:
> >149757228779144917176204260591184175807079292786053093556159199041982349
> >206415406540332144366189548551670508994053634039049939359298138323011269
> >238802598321754298938863469321281048841531709492264740073694039201 )

You have provided a COMPOSITE COFACTOR of 10^281-1 after dividing out
some (perhaps all, I did not check) of its KNOWN prime factors.

You are a lying troll.

> >Divide 10^281-1 by this number, and you will find what I say is true.
> >The other factor is relatively small and easy to decompose.

>
> >I could not, therefore, have arrived at this 200+ digit number simply by
> >coming up with two primes and multiplying them!

No. You took 10^281-1 and divided out some (perhaps all, I did not
check)
of the known and published prime factors. The remaining cofactor is
composite.
>
>
> Damn it barker, why can't you just be a troll -- that would
> make it so much easier.
>
> I think it's clear you're not a troll (or at least
> you have a non-troll side).
>
> I apologize for calling you a troll.
>
> Bravo for your partial factorization of (10^281 - 1)/9.

Damn it! There are already 4 known and published factors of
10^281-1. The remaining COFACTOR is unfactored. It is within
reach of NFS, but we just haven't gotten to it yet. The
community working on the Cunninghan project is busy with other
numbers. It will be finished.

Before anyone accepts what this troll has been saying about
factoring, you should first check out:

http://homes.cerias.purdue.edu/~ssw/cun/

Date Subject Author
4/13/12 David Bernier
4/13/12 Pubkeybreaker
4/14/12 barker
4/14/12 quasi
4/14/12 Pubkeybreaker
4/15/12 barker
4/15/12 quasi
4/15/12 J. Antonio Perez M.
4/15/12 quasi
4/15/12 quasi
4/15/12 barker
4/15/12 quasi
4/16/12 Pubkeybreaker
4/16/12 Pubkeybreaker
4/16/12 quasi
4/16/12 Pubkeybreaker
4/16/12 quasi
4/16/12 Pubkeybreaker
4/16/12 Pubkeybreaker
4/14/12 Frederick Williams