- View SourceHow 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 -----

2-xPRP

> 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

> with further divisions than just the .../4) -1)...

includ-> ed.

> formula or up to X divisions of 2's with -1's

> My actual final 2-xPRP test would be the following:

2^[((Z-1)/2-1)/2] == [(Z+1)/2 or (Z-1)/2] mod Z

>

> 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

>

interesting

> 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

> to me. Look below.

whereby

>

> > Hi Bill,

> >

> > this sounds like a "strong Fermat test" [1]

> > 2^(N-1)==1 implies 2^((N-1)/2)==+-1. If it is "+1"

http://primes.utm.edu/glossary/page.php?sort=StrongPRP

> > and N==1 (mod 4) then 2^((N-1)/4)==+-1 etc.

> >

> > [1]

> >

> >

____________________________________________________________________________________

> > 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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