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RE: Proth(last e-mail)

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  • Bill Bouris
    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
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      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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