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Re: [PrimeNumbers] Polynomials

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  • yasep16
    Hallo, ... I seem to remember that it was proved that there can be an infinity of polynomials that produce primes but also no polynomial can provide them all.
    Message 1 of 33 , Jul 25, 2013
      Hallo,

      Le 2013-07-25 18:37, Alexander a écrit :
      > Let us have a look at the polynomial P=n²-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)
      >

      I seem to remember that it was proved that there can be an infinity of
      polynomials that produce primes but also no polynomial can provide them
      all.
      Indeed, primes are by nature "anti-linear" (sorry for the neology)
      and those polynomials are essentially linear...

      What are you trying to find ?
    • 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, 2013
        --- 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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