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

Re: [PrimeNumbers] Polynomials

Expand Messages
  • Chroma
    ... ad1 n = 0,1,...10000 a primes 247757 5028 595937 4978 1544987 4809 2367767 4767 239621 4724 701147 4713 115721 4691 771581 4670
    Message 1 of 33 , Jul 25, 2013
      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)
      >

      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
    • 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
      Your message has been successfully submitted and would be delivered to recipients shortly.