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Re: [PrimeNumbers] Polynomials

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  • Jack Brennen
    The answer to both (1) and (2) is yes, if you believe the first Hardy-Littlewood conjecture. There is undoubtedly a value of (a) which will produce 100
    Message 1 of 33 , Jul 25, 2013
      The answer to both (1) and (2) is yes, if you believe
      the first Hardy-Littlewood conjecture.

      There is undoubtedly a value of (a) which will produce
      100 different prime numbers for n=1..100. You're
      basically describing an admissible 100-tuple. (It's
      provably admissible because n^2-n cannot cover all
      residues for any prime modulus.)

      Naming such a value for (a) is currently out of reach;
      we're not even ready to systematically search for
      50-tuples with any chance of success, much less 100-tuples.



      On 7/25/2013 9:37 AM, Alexander wrote:
      >
      > 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)
      >
      >
      >
      > [Non-text portions of this message have been removed]
      >
      >
      >
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    • 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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