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Re: Polynomials

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  • djbroadhurst
    ... The best HL constant known for x^2+x+a with a 0, seems to be with a= 1762103599631244727222460212500958264206823876231233415836049593411047
    Message 1 of 33 , Jul 26, 2013
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      --- In primenumbers@yahoogroups.com,
      "djbroadhurst" <d.broadhurst@...> wrote:

      > x^2 + x + 3399714628553118047

      The best HL constant known for x^2+x+a with a>0, seems to be with a=
      1762103599631244727222460212500958264206823876231233415836049593411047

      cacr.uwaterloo.ca/techreports/1999/corr99-32.ps

      For a<0, it seems to be with -a=
      33251810980696878103150085257129508857312847751498190349983874538507313

      http://www.ams.org/journals/mcom/2003-72-241/S0025-5718-02-01418-7/

      David
    • 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
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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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