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Re: [PrimeNumbers] prime chain,curios!

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  • Jud McCranie
    ... These formulas have been commonly known for a long time. +---------------------------------------------------------+ ...
    Message 1 of 4 , Oct 1, 2001
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      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 2 of 4 , Oct 1, 2001
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        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 3 of 4 , Oct 2, 2001
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          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
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