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

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• 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
Message 1 of 4 , Sep 5, 2005
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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-----
> Sent: 05 September 2005 16:57
> 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.
>
>
>
>
>
>
>
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
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• ... 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
Message 1 of 4 , Sep 5, 2005
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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.
>

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.
Message 1 of 4 , Sep 5, 2005
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Thank you Mr.Steven and Mr.Paul.

--- 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.
> >
>
>
>
> 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
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