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Factors for 2^n -1, n>400 and n even

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  • jtrjtrjtr2001
    hi, I am looking for a pre-computed table of the prime factorization of 2^n -1,400
    Message 1 of 4 , Sep 5, 2005
      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.
    • Paul Jobling
      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 2 of 4 , Sep 5, 2005
        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.
        >
        >
        >
        >
        >
        >
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      • Steven Whitaker
        ... 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 3 of 4 , Sep 5, 2005
          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
        • jtrjtrjtr2001
          Thank you Mr.Steven and Mr.Paul. Sarad.
          Message 4 of 4 , Sep 5, 2005
            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
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