- View Source"David Broadhurst " wrote:
> I noted that Louis Helm gave this advice to his SoB users in

Yes, Mr. Helm has erred in his use of the formulae put forth by

>

> 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.

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 - View Source--- In primenumbers@yahoogroups.com, Jack Brennen <jack@b...> wrote:
> ...

Could someone make such a prediction for the riesel problem?

>

> 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

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.