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Re: [PrimeNumbers] Polynomials

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  • Jose Ramón Brox
    A few years ago there was a programming contest from Al Zimmermann looking for the best prime-generating polynomials. The best results are essentially found in
    Message 1 of 33 , Jul 25 2:56 PM
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      A few years ago there was a programming contest from Al Zimmermann looking
      for the best prime-generating polynomials. The best results are essentially
      found in this page:

      http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html

      The best one is [image: 1/4(n^5-133n^4+6729n^3-158379n^2+1720294n-6823316)].

      Regards,
      Jose Brox


      2013/7/25 Chroma <chromatella@...>

      > **
      >
      >
      > Alexander wrote:
      > >
      > > Let us have a look at the polynomial P=n�-n+a. a=41 results in the
      > > well-known Euler-formula, which produces 86 different prime numbers for
      > > n=1..100. The second-best a seems to be 27941 with 77 different primes.
      > > Two questions:
      > >
      > > 1) Is there a more efficient a (except 41)?
      > >
      > > 2) Is there another polynomial form which is more efficient? (no
      > > repeats, no negative results)
      > >
      >
      > ad1
      > n = 0,1,...10000
      >
      > a primes
      > 247757 5028
      > 595937 4978
      > 1544987 4809
      > 2367767 4767
      > 239621 4724
      > 701147 4713
      > 115721 4691
      > 771581 4670
      > 333491 4669
      > 765197 4666
      > 1974881 4644
      > 361637 4634
      > 383681 4613
      > 722231 4610
      > 136517 4609
      > 601037 4605
      > 206807 4602
      > 310691 4591
      > 579431 4590
      > 1272851 4570
      >
      > ..........
      > 41 4179 ( 201's position)
      >
      > --
      > marian otremba
      >
      >
      >



      --
      La verdad (blog de raciocinio pol�tico e informaci�n
      social)<http://josebrox.blogspot.com/>


      [Non-text portions of this message have been removed]
    • djbroadhurst
      ... Bad models. ... No. Rather it is that n1, the begining of the sampling interval, needs to be substantially larger than sqrt(a), for the HL heuristic to win
      Message 33 of 33 , Jul 28 8:25 AM
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        --- In primenumbers@yahoogroups.com,
        "WarrenS" <warren.wds@...> wrote:

        > > Let N(a,n1,n2) be the number of primes of the form
        > > n^2+n+a with n in [n1,n1+n2]. Then the data
        > >
        > > N(247757,0,10^6) = 324001
        > > N(3399714628553118047,0,10^6) = 251841
        > >
        > > seem to favour the smaller value of a. Yet these data
        > >
        > > N(247757,10^12,10^6) = 148817
        > > N(3399714628553118047,10^12,10^6) = 193947
        > >
        > > indicate that the larger value of a is better, in the long run.
        >
        > --these numbers seem to be in vast violation of naive statistical
        > models.

        Bad models.

        > Is the reason, that the length n2 of the sampling interval,
        > needs to be substantially larger than a, in order for naive
        > statistical models to become reasonably valid?

        No. Rather it is that n1, the begining of the sampling
        interval, needs to be substantially larger than sqrt(a),
        for the HL heuristic to win out. Clearly when
        n1 < sqrt(3399714628553118047), Marion was comparing apples
        and oranges, since log(n^2+n+a) was dominated by "a".

        All I did was to level the playing field, here:

        > N(247757,10^12,10^6) = 148817
        > N(3399714628553118047,10^12,10^6) = 193947

        to allow the HL heuristic to show through.

        It's a simple as that. No shock-horror for statisticians;
        Just a trivial observation by a log-lover :-)

        David
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