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Re: [PrimeNumbers] Re: another goldbachian theme [Puzzle 9]

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  • Phil Carmody
    ... I guess if there s only a few million terms to the sum, then you only need to add half a dozen extra digits of precision to make sure that none of the
    Message 1 of 64 , Oct 30, 2011
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      --- On Sun, 10/30/11, djbroadhurst <d.broadhurst@...> wrote:
      > --- In primenumbers@yahoogroups.com,
      >
      > Phil Carmody <thefatphil@...> wrote:
      >
      > > So we have to be careful that 'floor' doesn't
      > > round to the wrong value.
      >
      > Indeed, as I had previously suggested, in
      > http://tech.groups.yahoo.com/group/primenumbers/message/23544
      >
      > By choosing a base coprime to 6 and by re-checking
      > at much higher precision, I think it very probable,
      > yet have not proven, that these are correct:
      >
      > T(5^4999973) =  10
      > T(5^5001174) = -11
      > T(7^1993768) =  10
      > T(7^2004371) = -10
      > T(7^2999623) =  10
      > T(7^3005373) = -10
      > T(7^3994545) = -10
      > T(7^4001447) = -10

      I guess if there's only a few million terms to the sum, then you only need to add half a dozen extra digits of precision to make sure that none of the terms sees the problem.

      > To prove these, one would need to use exact integer arithmetic
      > within the loop, which might take more time than it's worth.

      Certainly.

      Any view on the correctness of my minimal results? These 6s made themselves known just after my previous post:
      At 3^15912*2^60184-1, predicted 2301016864 got 2301016858, deficit=6
      At 3^37972*2^25220, predicted 2301031561 got 2301031567, excess=6
      I've looked at the first 9 billion 3-smooth numbers so far, in about 90 minutes on my old G5, about a million a second. Clearly I'll never get to the heady bazillions that your random diving yielded, but I'm hoping that at least some 7s will make themselves known before I retire, and perhaps even 8s by tomorrow. (Anyone who wants to run this exhaustive program for a longer period can have a copy of the source, just ask.)

      However, I wonder if there isn't already some known lattice technique for what is being calculated.

      I wish continued fractions (subract the area of the triangle with slope 1, slide everything down, flip axes, repeat) pointed to a shortcut, but it seems to require too much precision or too many iterations. Or does it?
    • Phil Carmody
      ... I guess if there s only a few million terms to the sum, then you only need to add half a dozen extra digits of precision to make sure that none of the
      Message 64 of 64 , Oct 30, 2011
      • 0 Attachment
        --- On Sun, 10/30/11, djbroadhurst <d.broadhurst@...> wrote:
        > --- In primenumbers@yahoogroups.com,
        >
        > Phil Carmody <thefatphil@...> wrote:
        >
        > > So we have to be careful that 'floor' doesn't
        > > round to the wrong value.
        >
        > Indeed, as I had previously suggested, in
        > http://tech.groups.yahoo.com/group/primenumbers/message/23544
        >
        > By choosing a base coprime to 6 and by re-checking
        > at much higher precision, I think it very probable,
        > yet have not proven, that these are correct:
        >
        > T(5^4999973) =  10
        > T(5^5001174) = -11
        > T(7^1993768) =  10
        > T(7^2004371) = -10
        > T(7^2999623) =  10
        > T(7^3005373) = -10
        > T(7^3994545) = -10
        > T(7^4001447) = -10

        I guess if there's only a few million terms to the sum, then you only need to add half a dozen extra digits of precision to make sure that none of the terms sees the problem.

        > To prove these, one would need to use exact integer arithmetic
        > within the loop, which might take more time than it's worth.

        Certainly.

        Any view on the correctness of my minimal results? These 6s made themselves known just after my previous post:
        At 3^15912*2^60184-1, predicted 2301016864 got 2301016858, deficit=6
        At 3^37972*2^25220, predicted 2301031561 got 2301031567, excess=6
        I've looked at the first 9 billion 3-smooth numbers so far, in about 90 minutes on my old G5, about a million a second. Clearly I'll never get to the heady bazillions that your random diving yielded, but I'm hoping that at least some 7s will make themselves known before I retire, and perhaps even 8s by tomorrow. (Anyone who wants to run this exhaustive program for a longer period can have a copy of the source, just ask.)

        However, I wonder if there isn't already some known lattice technique for what is being calculated.

        I wish continued fractions (subract the area of the triangle with slope 1, slide everything down, flip axes, repeat) pointed to a shortcut, but it seems to require too much precision or too many iterations. Or does it?
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