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Re: Proth conjecture

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  • leavemsg1
    ... I m not mad at you Paul; I m pissed at myself for not having a better understanding of my own idea. Yes, n would need to be odd... and Yes, Y should be
    Message 1 of 4 , Jan 18, 2007
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      --- In primenumbers@yahoogroups.com, "Paul Underwood"
      <paulunderwood@...> wrote:
      >
      > --- In primenumbers@yahoogroups.com, Bill Bouris <leavemsg1@> wrote:
      > >
      > > Hello, Group.
      > >
      > > The usual Proth numbers are described as Y= n*2^k +1
      > > where all 'k's are such that n < 2^k +1.

      I'm not mad at you Paul; I'm pissed at myself for not having a better
      understanding of my own idea.

      Yes, 'n' would need to be odd...
      and Yes, 'Y' should be more conventionally restated as Y= k*2^n+1.

      > >
      > >
      > > 2nd-kind Proth's are described (by me) as Z= n*2^k +1
      > > where 'n' is strictly inside the range 2^k+1 < n <
      > > 2*(2^k+1) for some 'k'.
      > >
      > >

      and my "curiosity" is... that... (curiosity killed the cat,...)

      > > My conjecture is... that...
      > >
      > > If 'k' is further restricted to be a prime, then
      > > 2nd-kind Proth's could be detected as prime iff
      > > 2^(Z-1) == 1 (mod Z)... in one single test.
      > >

      I'm trying to discover a large prime number that may later be proven
      prime using a more proper method. (but, satisfaction brought him (the
      cat) back.)

      > >
      > > In other words, when Z is 2-PRP, it would in fact be
      > > prime iff 'k' is prime and n is modified as stated in
      > > the above definition of a 2nd-kind Proth.
      > >
      > >
      > > Others have informed me... that only very few 2-PRP
      > > tested numbers would in fact be composite.

      I don't doubt that this statement is true.

      > >
      > > My

      statement...

      > > again is that "NO" psuedoprimes can be
      > > constructed iff 'n' is inside the range 2^k+1 < n <
      > > 2*(2^k+1) and 'k' is prime.
      > >
      > > Any suggestions... or comments?... besides 'Stop
      > > feed-ing the troll'.
      > >
      >

      Sometimes, my own hunches... make me feel like an ice pick.
      Bill


      > Biting, for a start we usually talk of k*2^n+1 and not "n*2^k+1".
      >
      > I am guilty for the over use of the word "conjecture" -- some people
      > would rather say "a shot in the dark". A conjecture, I suppose, must
      > have some kind of basis, intimating a series of logical/mathematical
      > steps or reasons that lead partially to a fully fledged theorem.
      >
      > The received wisdom is that conjectures such as yours have N^(1-e)
      > counterexamples where "e" can be chosen as small as you like -- N
      just
      > has to very big, more often than not, bigger than we humans can ever
      > calculate with all our computers,
      >
      > ?
      > forprime(n=2,13,for(k=2^n+2,2*(2^n+1)-1,N=k*2^n+1;if(!isprime(N)
      &&Mod(2,N)^(N-1)==1,print(k"
      > "n" "N))))
      > 54 5 1729
      > 4080 11 8355841
      > 4094 11 8384513
      > 12816 13 104988673
      >
      > Paul
      >
      > > Bill Bouris, USA
      > >
      > >
      > >
      > >
      >
      ______________________________________________________________________
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      >
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