Re: Proth conjecture
- --- In firstname.lastname@example.org, "Paul Underwood"
>I'm not mad at you Paul; I'm pissed at myself for not having a better
> --- In email@example.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.
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.
> >and my "curiosity" is... that... (curiosity killed the cat,...)
> > 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'.
> > My conjecture is... that...I'm trying to discover a large prime number that may later be proven
> > 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.
prime using a more proper method. (but, satisfaction brought him (the
> >I don't doubt that this statement is true.
> > 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.
> > My
> > again is that "NO" psuedoprimes can beSometimes, my own hunches... make me feel like an ice pick.
> > 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'.
> Biting, for a start we usually talk of k*2^n+1 and not "n*2^k+1".just
> 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
> has to very big, more often than not, bigger than we humans can ever&&Mod(2,N)^(N-1)==1,print(k"
> calculate with all our computers,
> "n" "N))))______________________________________________________________________
> 54 5 1729
> 4080 11 8355841
> 4094 11 8384513
> 12816 13 104988673
> > Bill Bouris, USA
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