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• ... 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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> I noted that Louis Helm gave this advice to his SoB users in
>
>
> > 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
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
• ... 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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> > 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

Yves
• ... 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
• ... 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
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,
David
• ... 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
> 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
• 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
> > 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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• ... 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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