- You can work out the factorisations of the even values from the tables of

factorisations of 2^n+1 and 2^n-1 given in that document - all of the

required information is present.

Clue: 2^402-1 = (2^201+1)*(2^201-1)

> -----Original Message-----

__________________________________________________

> From: primenumbers@yahoogroups.com

> [mailto:primenumbers@yahoogroups.com]On Behalf Of jtrjtrjtr2001

> Sent: 05 September 2005 16:57

> To: primenumbers@yahoogroups.com

> Subject: [PrimeNumbers] Factors for 2^n -1, n>400 and n even

>

>

> hi,

>

> I am looking for a pre-computed table of the prime factorization of

> 2^n -1,400<n<500 and n is even.

>

> I am only able to find the prime factorization for all odd powers in

> the above range from the table below.

> http://www.ams.org/online_bks/conm22/conm22-whole.pdf

>

> I have been using pari to compute the factorisation without much

> success. Any help is appreciated.

>

> Sarad.

>

>

>

>

>

>

> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com

> The Prime Pages : http://www.primepages.org/

>

>

> Yahoo! Groups Links

>

>

>

>

>

>

>

> __________________________________________________

> Virus checked by MessageLabs Virus Control Centre.

>

Virus checked by MessageLabs Virus Control Centre. - On Mon, 05 Sep 2005 15:56:45 -0000, you wrote:

>hi,

Sarad

>

>I am looking for a pre-computed table of the prime factorization of

>2^n -1,400<n<500 and n is even.

>

>I am only able to find the prime factorization for all odd powers in

>the above range from the table below.

>http://www.ams.org/online_bks/conm22/conm22-whole.pdf

>

>I have been using pari to compute the factorisation without much

>success. Any help is appreciated.

>

>Sarad.

There is an algebraic factorisation for even exponents:

2^(2n)-1 = (2^n-1)(2^n+1)

So there's no point in recording them separately. (Read the introduction

to the short tables in that paper).

Incidentally, there is a more up-to-date list at the Cunningham project

website:

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

Regards

Steve - Thank you Mr.Steven and Mr.Paul.

Sarad.

--- In primenumbers@yahoogroups.com, Steven Whitaker <steven@w...> wrote:

> On Mon, 05 Sep 2005 15:56:45 -0000, you wrote:

>

> >hi,

> >

> >I am looking for a pre-computed table of the prime factorization of

> >2^n -1,400<n<500 and n is even.

> >

> >I am only able to find the prime factorization for all odd powers in

> >the above range from the table below.

> >http://www.ams.org/online_bks/conm22/conm22-whole.pdf

> >

> >I have been using pari to compute the factorisation without much

> >success. Any help is appreciated.

> >

> >Sarad.

>

>

> Sarad

>

> There is an algebraic factorisation for even exponents:

>

> 2^(2n)-1 = (2^n-1)(2^n+1)

>

> So there's no point in recording them separately. (Read the introduction

> to the short tables in that paper).

>

> Incidentally, there is a more up-to-date list at the Cunningham project

> website:

>

> http://homes.cerias.purdue.edu/~ssw/cun/index.html

>

> Regards

> Steve