## RE: Proth(last e-mail)

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• How do I implement this idea using pfgw ??? anyone? ... I know that it s not much of an improvement, but look below... ... 2-xPRP ... includ- ed. ...
Message 1 of 1 , Feb 14, 2007
How do I implement this idea using 'pfgw'??? anyone?

--- Bill Bouris <leavemsg1@...> wrote:

I know that it's not much of an improvement, but
look below...

--- Paul Underwood <paulunderwood@...>
wrote:

> ----- Original Message -----
> From: "Bill Bouris" <leavemsg1@...>
> To: paulunderwood@...
> Subject: RE: Proth
> Date: Tue, 13 Feb 2007 13:22:06 -0800 (PST)
>
>
> Hello, Paul.
>
> I have used 'pfgw' to successfully find a large
> amount of different prime numbers wrt ABC2 files.
>
> I know that you fully understand it, but here's
> my official definition of a 2nd-kind Proth:
>
> Let L= (small prime), M= 2^L, and N= (an odd number)
> such that 1 <= N <= M +1, Y= M*(M +N), and Z= Y +1.
>
> I searched for 2-PRP's throughout the entire range
> of N for a given L, and I noticed that my home-grown
> PHP program... also spotted all 2-PRP Z's using
> my 2-xPRP OR 2^(((Z -1) /4) -1) == [(Z-1)/2 or
> (Z+1)/2] (mod Z), again, throughout the entire range
> of N for a given L.
>
> I do appreciate that a handful of numbers that test
> 2-PRP may still be composite, but I'm confident that
> some numbers may even be detected earlier than
2-xPRP
> with further divisions than just the .../4) -1)...
> formula or up to X divisions of 2's with -1's
includ-> ed.

> My actual final 2-xPRP test would be the following:
>
> Number of divis'n by 2 is k=sqrt(2*L*ln(2)/ln(10))+1
> (not sure of the exact formula, yet.)
>
> A small subset of numbers in a list would yield a
> 2-xPRP, like... again...
>
> L=5; k= 1.xyz,... +1= 2; Z= 1249,... OR
2^[((Z-1)/2-1)/2] == [(Z+1)/2 or (Z-1)/2] mod Z

>
> Example 1249 is prime; 1248/2 OR 1250/2 shows up
> as a residue earlier than [(2^1248) mod Z] ...OR
> at 2^311 == (624 or 625) mod 1249
>
> I just want to be able to detect a smaller
> quantity of primes more quickly than 2-PRP and then
> prove them out using the correct N+/-1 'pfgw' test.
>
> Bill, beating a dead horse
>
> not much of a savings at this level, just
interesting
> to me. Look below.
>
> > Hi Bill,
> >
> > this sounds like a "strong Fermat test" [1]
whereby
> > 2^(N-1)==1 implies 2^((N-1)/2)==+-1. If it is "+1"
> > and N==1 (mod 4) then 2^((N-1)/4)==+-1 etc.
> >
> > [1]
> >
http://primes.utm.edu/glossary/page.php?sort=StrongPRP
> >
> > Beating a Fermat 2-PRP test is going to be hard to
> > beat. Even if you can then reducing it to 1 less
> > operation is not going to help much.
> >
> > Keep flogging,
>
> done flogging, thanks for listening.
>
> >
> > Paul
> >

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