Loading ...
Sorry, an error occurred while loading the content.

prime chain,curios!

Expand Messages
  • Norman Luhn
    Hi all! Perhaps,i have found a new methode to generate primes. Here is the largest example,where i found it. We get (41)! primes for this equal px=41+(x^2-x)
    Message 1 of 4 , Oct 1, 2001
    • 0 Attachment
      Hi all!
      Perhaps,i have found a new methode to generate
      primes. Here is the largest example,where i found it.

      We get (41)! primes for this equal px=41+(x^2-x)
      for x=0 to 40. I think it is fantastic.

      A smaller example is 17 primes for
      px=17+(x^2-x), x=0 to 16.

      best wishes

      Norman.


      __________________________________________________________________
      Do You Yahoo!?
      Gesendet von Yahoo! Mail - http://mail.yahoo.de
    • Jud McCranie
      ... These formulas have been commonly known for a long time. +---------------------------------------------------------+ ...
      Message 2 of 4 , Oct 1, 2001
      • 0 Attachment
        At 11:03 PM 10/1/2001 +0200, Norman Luhn wrote:
        >Hi all!
        >Perhaps,i have found a new methode to generate
        >primes. Here is the largest example,where i found it.
        >
        >We get (41)! primes for this equal px=41+(x^2-x)
        >for x=0 to 40. I think it is fantastic.
        >
        >A smaller example is 17 primes for
        >px=17+(x^2-x), x=0 to 16.

        These formulas have been commonly known for a long time.


        +---------------------------------------------------------+
        | Jud McCranie |
        | |
        | Programming Achieved with Structure, Clarity, And Logic |
        +---------------------------------------------------------+
      • d.broadhurst@open.ac.uk
        ... Precisely! There is an informative note by Christian Radoux in http://www.math.niu.edu/~rusin/known-math/98/163 and Ribenboim p.200 completes the link to
        Message 3 of 4 , Oct 1, 2001
        • 0 Attachment
          Jud McCranie asked:
          > The class number problem, right?
          Precisely! There is an informative note by Christian Radoux in
          http://www.math.niu.edu/~rusin/known-math/98/163
          and Ribenboim p.200 completes the link to Euler's 41.
        • d.broadhurst@open.ac.uk
          ... No, it s very much part of the theory: 744=240*3+24 where 240 comes from Rademacher and 24 (as ever) from Ramanujan. [BTW: guess why bosonic string theory
          Message 4 of 4 , Oct 2, 2001
          • 0 Attachment
            Thomas Hadley asked:

            > Is 744 a fudge factor?

            No, it's very much part of the theory:

            744=240*3+24

            where 240 comes from Rademacher and 24 (as ever) from Ramanujan.
            [BTW: guess why bosonic string theory started in 24+2=26 dimensions.]

            Here's a bit of Pari:

            \p1000
            q=-1/exp(Pi*sqrt(4*41-1));
            J(m)=(1+240*sum(k=1,m,sigma(k,3)*q^k))^3/prod(j=1,m,1-q^j)^24/q;
            forstep(m=1,61,10,print((-J(m))^(1/3)));

            You were taking only m=1 term in the Rademacher sum and Ramanujan
            product. If you take m=61 terms, then you get to within better than
            1 part in 10^1000 of the integer
            N=640320=2^6*3*5*23*29,
            where(-N)^3 is the modular invariant.
            To see how Norman's (Leonard's, actually)
            other examples fare, compare the approximations

            (exp(Pi*sqrt(4*41-1))-744)^(1/3) \approx 640320
            (exp(Pi*sqrt(4*17-1))-744)^(1/3) \approx 5280
            (exp(Pi*sqrt(4*11-1))-744)^(1/3) \approx 960
            (exp(Pi*sqrt(4* 5-1))-744)^(1/3) \approx 96
            (exp(Pi*sqrt(4* 3-1))-744)^(1/3) \approx 32
            (exp(Pi*sqrt(4* 2-1))-744)^(1/3) \approx 15

            with the last being as miserable as observing that
            n^2+n+2 is composite with n>0.

            The connection with n^n+n+p is that neither the
            prime generation with p=41 nor the modular trickery with
            -D=4*41-1 can be improved upon, since |D|=163 gives the
            biggest (in magnitude) discriminant of an imaginary
            quadratic number field with unique factorization
            (class number=1).

            David

            PS: Here's a place where folk factorize Laurent coefficients of J:
            http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha135.htm
          Your message has been successfully submitted and would be delivered to recipients shortly.