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Re: About fourteenorbust

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  • Jack Brennen
    ... Yes, Mr. Helm has erred in his use of the formulae put forth by both Yves and myself (which are essentially the same -- although Yves added some small
    Message 1 of 11 , Dec 8, 2002
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      "David Broadhurst " wrote:
      > I noted that Louis Helm gave this advice to his SoB users in
      >
      > http://www.free-dc.org/forum/showthread.php?
      > s=3e927d470bb3654c86b16785f8e932fc&threadid=2087
      >
      > > so yeah, don't waste your time reading these estimates.
      > > they are wrong. -Louie
      >
      > despite the fact that Yves' heuristic seems to be in fair shape
      > when compared with the recent SoB progress.

      Yes, Mr. Helm has erred in his use of the formulae put forth by
      both Yves and myself (which are essentially the same -- although
      Yves added some small adjustments based on experimental evidence,
      the effect of those adjustments is much less than an order of
      magnitude).

      First, a little background on Proth weights -- the first Proth
      weights I ever saw or heard of were those sometimes referred to
      as Nash weights, and calculated by a program called PSIEVE,
      I believe written by Chris Nash and Paul Jobling. The Nash
      weight was useful for comparing two Proth coefficients, but not
      "calibrated" -- it was just a number without context.

      Then I introduced my Proth Weight applet, which was basically
      the same concept, but the weight was calibrated (divided by
      a constant) so that the "average" weight over all odd k would
      be equal to 1.

      Yves Gallot in his paper "On the number of primes in a sequence"
      used the same concept of Proth weights, but he chose to calibrate
      so that members of a sequence with weight 1 would be prime about
      as often as random numbers of the same magnitude.

      It turns out that the "average" weight over all odd k is 2 by Yves'
      reckoning; two because k*2^n+1 is already known to be odd, and so
      twice as likely to be prime as a "random" number.

      So it should be clear that Yves' weights are equal to twice the
      weights given by my Proth Weight applet. Neither weight is wrong,
      they are just different scales of measuring the same thing.


      When I wrote the formula which Mr. Helm has referenced, I used my
      own calibrated Proth weights. Look closely at the formula:

      Product {over all relevant i} of:

      1-Exp(-ProthWeight(k[i])*Log(N/n[i])/Log(2)*2)

      You see the multiplication of the exponent by 2 -- that "adjusts"
      my Proth weight to correspond to Yves' Proth weight.


      Apparently, what Mr. Helm has done is used my formula with the
      Proth weights published in Yves' paper. As I mentioned, Yves'
      weights are calibrated differently. If you use his weights,
      you have to remove the multiplication by 2 from my formula. By
      using Yves' weights with my formula, he was "correcting" Yves'
      weights, which are already "corrected" -- this resulted in his
      exponents being twice as large as they should have been.

      If Mr. Helm would rerun his numbers with the Proth weights
      given by my applet instead of those published in Yves' paper,
      or if he would rerun his numbers with Yves' weights but with
      the "times 2" removed from the exponent, he will see that
      the analysis of the Sierpinski problem hasn't changed nearly
      as dramatically as he says, and in fact, that the results of
      Seventeen or Bust have done nothing to disparage the analyses
      done by Yves or myself. In fact, I would say that their work
      has strengthened the case for the correctness of our analyses.

      The original estimates (from my original post in November 2000)
      and the updated estimates (like Mr. Helm, I blindly assumed that
      all Sierpinski candidates have been checked up to 1M and no further):

      N : 10^7 --> chance to solve : 0.00000263, now 0.00000308
      N : 10^8 --> chance to solve : 0.000949, now 0.00168
      N : 10^9 --> chance to solve : 0.0151, now 0.0245
      N : 10^10 --> chance to solve : 0.0681, now 0.0983
      N : 10^11 --> chance to solve : 0.166, now 0.218
      N : 10^12 --> chance to solve : 0.291, now 0.353
      N : 10^13 --> chance to solve : 0.420, now 0.482
      N : 10^14 --> chance to solve : 0.538, now 0.593
      N : 10^15 --> chance to solve : 0.639, now 0.684
      N : 10^16 --> chance to solve : 0.720, now 0.756
      N : 10^17 --> chance to solve : 0.785, now 0.812
      N : 10^18 --> chance to solve : 0.836, now 0.856
      N : 10^19 --> chance to solve : 0.875, now 0.889
      N : 10^20 --> chance to solve : 0.905, now 0.915
      N : 10^25 --> chance to solve : 0.976, now 0.977
      N : 10^30 --> chance to solve : 0.994, now 0.994


      This new data, based on the work of Seventeen or Bust (and using
      Mr. Helm's own assumptions), is strongly supportive of the
      predictions made by Yves and myself. It is unfortunate that
      Mr. Helm disparaged the predictions so strongly due to his own
      unfortunate misunderstanding of the scaling factor between my
      Proth weights and those given by Yves, but I didn't take any
      offense personally -- he truly believed that the predicted
      probabilities had risen dramatically after these new discoveries.
      I'm sure that now that his mistake has been pointed out, he'll
      be more inclined to take the analysis seriously.


      Jack Brennen
    • Yves Gallot
      ... The probabilities (that were originally computed in November 2000 by Jack Brennen) are correct (as indicated by Jack in the previous message). In my paper,
      Message 2 of 11 , Dec 9, 2002
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        > > so yeah, don't waste your time reading these estimates.
        > > they are wrong. -Louie
        >
        > despite the fact that Yves' heuristic seems to be in fair shape
        > when compared with the recent SoB progress.

        The probabilities (that were originally computed in November 2000 by Jack
        Brennen) are correct (as indicated by Jack in the previous message).

        In my paper, I tried to compute the "frequencies" and then the expected
        number of remaining candidates at n. The computation is mainly the summation
        of Jack's probabilities but surprisingly it works well (maybe because the
        original set was large enough). Because of the summation, there is a degree
        of freedom, a "translation" that can change a bit the numbers of primes
        found in a range. But it doesn't change the chance to solve the problem at n
        that is today strongly verified by the expected and actual frequencies.

        Don't forgot that the variance V(n) is approximately E(n). Then if 3 primes
        are expected in a range, we have about the same chance to find 2, 3, 4 or 5
        primes : it's difficult to estimate each frequency precisely. But we know
        that we have about no chance to solve the problem by testing the few next
        ranges because the average value of the frequencies is correctly estimated
        by Jack's probabilities.

        I don't understand Louie's remark: 14 candidates left at 2^20 at 12.8 were
        expected. The result is worst than expected, not better! But hopefully it
        doesn't change the fact that the chance to solve the problem at n=2^43 is
        about 50% :o)

        Yves
      • David Broadhurst <d.broadhurst@open.ac.u
        ... I think this is what Jack meant: w=[.20,.25,.23,.08,.19,.07,.20,.12,.11,.19,.28,.25,.07,.18]; Take Yves weights and assume all 14 tested up to 2^20
        Message 3 of 11 , Dec 9, 2002
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          Jack Brennen wrote:

          > if he would rerun his numbers with Yves' weights but with
          > the "times 2" removed from the exponent, he will see that
          > the analysis of the Sierpinski problem hasn't changed nearly
          > as dramatically as he says

          I think this is what Jack meant:

          w=[.20,.25,.23,.08,.19,.07,.20,.12,.11,.19,.28,.25,.07,.18];
          \\ Take Yves' weights and assume all 14 tested up to 2^20
          for(N=23,30,print("2^"N" : "prod(i=1,14,1-exp(-w[i]*(N-20)))))

          limit : chance

          2^23 : 0.000000822
          2^24 : 0.0000158
          2^25 : 0.000128
          2^26 : 0.000609
          2^27 : 0.00202
          2^28 : 0.00523
          2^29 : 0.0112
          2^30 : 0.0209

          Note that Yves estimated a 5% chance by the time
          one reaches n=2^30. That has gone down to 2%
          because SoB are slightly lagging Yves' mean
          expectation of 12.8 remaining by 2^20.

          However, it seems that k=27653, with w=0.12,
          is tested only up to 2^19. Its Gallot weight is
          (roughly) the chance that it will crack when we
          double the limit on n. Suppose that SoB struck lucky
          and cracked this k before reaching n=2^20.
          (There is a chance of 1-exp(-0.12)=11%.)
          Then we would remove its weight and rerun
          the simple formula to guess

          2^23 : 0.00000272
          2^24 : 0.0000415
          2^25 : 0.000284
          2^26 : 0.00118
          2^27 : 0.00356
          2^28 : 0.00848
          2^29 : 0.0169
          2^30 : 0.0299

          As you can see it's pretty stable:
          the 2% chance by n=2^30 for FoB
          goes up to a 3% chance for ToB.

          Might someone like to estimate the number
          of "P90-years" needed to get up to n=2^30,
          assuming that the candidates peel off roughly
          as Yves expects?

          That might be quite informative for FoB participants...

          David
        • David Broadhurst <d.broadhurst@open.ac.u
          ... The home page http://sb.pns.net/ now says ... so maybe it needs only a proof by your Proth.exe to land them at ToB, uncannily close to your expectation.
          Message 4 of 11 , Dec 10, 2002
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            Yves:
            > 14 candidates left at 2^20 at 12.8 were expected
            The home page
            http://sb.pns.net/
            now says
            > four down, thirteen to go
            so maybe it needs only a proof by your
            Proth.exe to land them at ToB,
            uncannily close to your expectation.
            David
          • Yves Gallot
            ... With a fourth prime, the ToB is definitely convicted that they will solve quickly Sierpinski problem :-D ... but it s just because December is a good month
            Message 5 of 11 , Dec 11, 2002
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              > Yves:
              > 14 candidates left at 2^20 at 12.8 were expected
              > The home page http://sb.pns.net/ now says
              > four down, thirteen to go so maybe it needs only
              > a proof by your Proth.exe to land them at ToB,
              > uncannily close to your expectation.

              With a fourth prime, the ToB is definitely convicted that they will solve
              quickly Sierpinski problem :-D ... but it's just because December is a good
              month for large primes:
              9999 440846^65536+1 369904 GC1 02 Generalized Fermat
              9999 291726^65536+1 358153 GC2 02 Generalized Fermat
              9999 292550^65536+1 358233 GC2 02 Generalized Fermat
              where the 'C' of GC is for http://fatphil.org/maths/GFN/index.html,
              the '1' for Michael Angel and the '2' for Yary Hluchan.

              Many down, infinitely to go!

              Yves
            • Gary Chaffey
              A very small point.... The url below to link to Phil s page doesn t work it should be http://fatphil.org/maths/GFN/index.html (No comma) Gary (ever the
              Message 6 of 11 , Dec 12, 2002
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                A very small point.... The url below to link to Phil's
                page doesn't work it should be
                http://fatphil.org/maths/GFN/index.html
                (No comma)
                Gary
                (ever the perfectionist!)
                . .
                |
                \_/

                --- Yves Gallot <galloty@...> wrote: > > Yves:
                > > 14 candidates left at 2^20 at 12.8 were expected
                > > The home page http://sb.pns.net/ now says
                > > four down, thirteen to go so maybe it needs only
                > > a proof by your Proth.exe to land them at ToB,
                > > uncannily close to your expectation.
                >
                > With a fourth prime, the ToB is definitely convicted
                > that they will solve
                > quickly Sierpinski problem :-D ... but it's just
                > because December is a good
                > month for large primes:
                > 9999 440846^65536+1 369904 GC1 02
                > Generalized Fermat
                > 9999 291726^65536+1 358153 GC2 02
                > Generalized Fermat
                > 9999 292550^65536+1 358233 GC2 02
                > Generalized Fermat
                > where the 'C' of GC is for
                >
                > the '1' for Michael Angel and the '2' for Yary
                > Hluchan.
                >
                > Many down, infinitely to go!
                >
                > Yves
                >
                >

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              • yummie_55555
                ... Could someone make such a prediction for the riesel problem? What does n[i] means in the formula? Is it max n tested? I can t do it, cause that applet
                Message 7 of 11 , Oct 22, 2003
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                  --- In primenumbers@yahoogroups.com, Jack Brennen <jack@b...> wrote:
                  > ...
                  >
                  > When I wrote the formula which Mr. Helm has referenced, I used my
                  > own calibrated Proth weights. Look closely at the formula:
                  >
                  > Product {over all relevant i} of:
                  >
                  > 1-Exp(-ProthWeight(k[i])*Log(N/n[i])/Log(2)*2)
                  >
                  > You see the multiplication of the exponent by 2 -- that "adjusts"
                  > my Proth weight to correspond to Yves' Proth weight.
                  > ...
                  >
                  > N : 10^7 --> chance to solve : 0.00000263, now 0.00000308
                  > N : 10^8 --> chance to solve : 0.000949, now 0.00168
                  > N : 10^9 --> chance to solve : 0.0151, now 0.0245
                  > N : 10^10 --> chance to solve : 0.0681, now 0.0983
                  > N : 10^11 --> chance to solve : 0.166, now 0.218
                  > N : 10^12 --> chance to solve : 0.291, now 0.353
                  > N : 10^13 --> chance to solve : 0.420, now 0.482
                  > N : 10^14 --> chance to solve : 0.538, now 0.593
                  > N : 10^15 --> chance to solve : 0.639, now 0.684
                  > N : 10^16 --> chance to solve : 0.720, now 0.756
                  > N : 10^17 --> chance to solve : 0.785, now 0.812
                  > N : 10^18 --> chance to solve : 0.836, now 0.856
                  > N : 10^19 --> chance to solve : 0.875, now 0.889
                  > N : 10^20 --> chance to solve : 0.905, now 0.915
                  > N : 10^25 --> chance to solve : 0.976, now 0.977
                  > N : 10^30 --> chance to solve : 0.994, now 0.994
                  >
                  > ...
                  >
                  >
                  > Jack Brennen

                  Could someone make such a prediction for the riesel problem?
                  What does n[i] means in the formula? Is it max n tested?

                  I can't do it, cause that applet won't work on my pc.
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