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Re: [PrimeNumbers] Re: 2^p+3

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  • Phil Carmody
    ... They of course have their own divisibility criterea. But one unrelated to (a^n+b^n) forms. Their criteria are more similar to those behind the fixed k
    Message 1 of 22 , May 1, 2002
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      --- Phil Carmody <thefatphil@...> wrote:
      > > The prime number form you propose are very similar to Mersenne
      > > primes
      > > and I would expect prime number results for 2^p+3 to rival those
      > of
      > >
      > > Mersenne primes in size and distribution too.
      >
      > I wouldn't expect them to have the same distribution.
      > 2^p+1, being a cyclotomic form, has restrictions on what its
      > factors
      > can be. 2^p+3 has no such divisibility criterea.

      They of course have their own divisibility criterea. But one
      unrelated to (a^n+b^n) forms. Their criteria are more similar to
      those behind the 'fixed k Proth' problems.

      Phil

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    • jim_fougeron
      ... For an infinity of easy to show counter examples, look at this form: k#+p If p is a fixed prime 3, then the expression will be +-1 mod(6), however, the
      Message 2 of 22 , May 1, 2002
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        --- In primenumbers@y..., "jbrennen" <jack@b...> wrote:
        >--- In primenumbers@y..., "rlberry2002" <rlberry2002@y...> wrote:
        >> Paul,
        >>
        >> I obviously misunderstood your equation; I read it as (2^p) + 3
        >> where p=2,result 7; p=3, result 11, and so on.
        >
        >No, you got it right :-)
        >
        >> I stand by my earlier statements (let's term it as Robert's
        >> Conjecture) though: A one variable equation which, (a) does not
        >> reduce and (b) contains infinitely many 1 Mod 6 and/or 5 Mod 6
        >> numbers, will contain infinitely many prime numbers.
        >
        > See Sierpinski numbers for a well-known counterexample:
        > 78557*2^n+1 contains an infinite number of elements which are
        > (5 mod 6), but has no prime numbers for any integer n.

        For an infinity of easy to show counter examples, look at this form:

        k#+p

        If p is a "fixed" prime > 3, then the expression will be +-1 mod(6),
        however, the expression (for a variable k and fixed p) will produce
        ONLY a finite (if any) amount of primes. Primes can only be generated
        by the above form while k < p. Once k reaches the size of p, then
        p will always be a factor of the expression.

        Take for example k#+13.
        This is prime for k=3, 5, 7 and NO others, since 2#+13=3*5,
        11#+13= 23*101 and when k>=13 then k#+13 is always has a factor of 13.
        However k#+13 == 1mod(6) (except for 2#+13==3mod(6)).

        Jim.
      • mikeoakes2@aol.com
        AFAIK the largest PRP of form 2^n+3 is still the one found by me in July 2001:- 2^122550+3 It is the 19th largest known PRP, according to Henri Lifchitz s
        Message 3 of 22 , May 1, 2002
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          AFAIK the largest PRP of form 2^n+3 is still the one found by me in July
          2001:-
          2^122550+3

          It is the 19th largest known PRP, according to Henri Lifchitz's database at
          http://ourworld.compuserve.com/homepages/hlifchitz/

          The sequence for lower values is Sloane's A057732. I had searched up to
          n=127677, and proved primality for n <=2370 using Titanix, finishing this
          project on 15 Aug 2001.

          (See also my post to the primenumbers group dated 8 Jul 2002.)

          Mike Oakes


          In a message dated 01/05/02 14:41:02 GMT Daylight Time,
          Paul.Jobling@... writes:

          > Hi,
          >
          > I was just wondering what the state of play was with looking for a
          > (pseudo-)prime of the form 2^p+3 - what search limits have been reached,
          > and
          > have any PRP's been found? I recall that there was some searching being
          > done a
          > couple of years ago (I think), but I do not know what (if any) results were
          > found?
          >
          > Regards,
          >
          > Paul.
          >




          [Non-text portions of this message have been removed]
        • rlberry2002
          Please pardon my ignorance; I am here to learn to grow in my knowledge of number theory. However, I will try to be more careful in responding to posts before
          Message 4 of 22 , May 1, 2002
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            Please pardon my ignorance; I am here to learn to grow in my
            knowledge of number theory. However, I will try to be more careful
            in responding to posts before I have fully thought out my response -
            you could extend me the same courtesy.

            Your example of N=2^p + 12213 violates one of the two qualifications
            that I laid down - "the equation cannot be reducible". I typically
            would use this tenet for an equation like 4n + 1 (which does have
            infinitely many primes in it)where n is odd. The tenet holds for
            numbers of the form 2^p + n also, it just is usually harder to find
            the pattern that this type of equation reduces to. In your example,
            you provide the pattern which violates my first condition: that is
            for 2, 3, 1 Mod 12, 5 Mod 12, 7 Mod 12, & 11 Mod 12; there a fixed
            set of possible outcomes each of which will have fixed factors
            depending upon which of the outcomes it fall under.

            For instance, the equation 30Y + 35 = N generates infinitely many 5
            Mod 6 numbers none of which are prime. This equation is easy since it
            reduces to 5*(6Y + 7) = N; numbers of the form 2^p + N require a
            deeper analysis as long as N itself does not have a factor of 2^p.

            Just a few thoughts

            Robert

            --- In primenumbers@y..., "jbrennen" <jack@b...> wrote:
            > --- In primenumbers@y..., "rlberry2002" <rlberry2002@y...> wrote:
            > > Paul,
            > >
            > > I obviously misunderstood your equation; I read it as (2^p) + 3
            > > where p=2,result 7; p=3, result 11, and so on.
            >
            > No, you got it right :-)
            >
            > > I stand by my earlier statements (let's term it as Robert's
            > > Conjecture) though: A one variable equation which, (a) does not
            > > reduce and (b) contains infinitely many 1 Mod 6 and/or 5 Mod 6
            > > numbers, will contain infinitely many prime numbers.
            >
            > See Sierpinski numbers for a well-known counterexample:
            > 78557*2^n+1 contains an infinite number of elements which are
            > (5 mod 6), but has no prime numbers for any integer n.
            >
            > Back to the original question... There are easily found concrete
            > examples of the form 2^p+n which do not have an infinite number of
            > prime values (with p prime) despite having an infinite number of
            > (1 mod 6) and (5 mod 6) numbers:
            >
            > N=2^p+12213 (with p prime) has no prime values whatsoever.
            >
            > This is because:
            >
            > If p == 2, N is divisible by 19
            > If p == 3, N is divisible by 11
            > If p == 1 (mod 12), N is divisible by 5
            > If p == 5 (mod 12), N is divisible by 5
            > If p == 7 (mod 12), N is divisible by 7
            > If p == 11 (mod 12), N is divisible by 13
            >
            > Every prime p meets one of the six cases above, so N is never
            > prime when p is prime.
          • Phil Carmody
            ... I would trust that such courtesy is demonstrated on the list. (Yeah, flame me off list, you won t be the first... (or the second) ... Strictly, it is
            Message 5 of 22 , May 1, 2002
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              --- rlberry2002 <rlberry2002@...> wrote:
              > Please pardon my ignorance; I am here to learn to grow in my
              > knowledge of number theory. However, I will try to be more careful
              > in responding to posts before I have fully thought out my response
              > - you could extend me the same courtesy.

              I would trust that such courtesy is demonstrated on the list.
              (Yeah, flame me off list, you won't be the first... (or the second)
              :-) )

              > Your example of N=2^p + 12213 violates one of the two
              > qualifications
              > that I laid down - "the equation cannot be reducible".

              Strictly, it is irreducible. The letter of the law is obeyed.

              > I typically
              > would use this tenet for an equation like 4n + 1 (which does have
              > infinitely many primes in it)where n is odd. The tenet holds for
              > numbers of the form 2^p + n also, it just is usually harder to find
              >
              > the pattern that this type of equation reduces to. In your
              > example,
              > you provide the pattern which violates my first condition: that is
              >
              > for 2, 3, 1 Mod 12, 5 Mod 12, 7 Mod 12, & 11 Mod 12; there a fixed
              > set of possible outcomes each of which will have fixed factors
              > depending upon which of the outcomes it fall under.

              Sure, but that's not redicibility. What Jack has highlighted is an
              intrinsically interesting property about that sequence of numbers.
              This property will ba shared by an infinite number of other
              sequences, not just the ...+12213. The property isn't reducibility,
              and that term was used, it's a very precisely defined term, so Jack
              and others can't be faulted for taking the term at face falue.
              Perhaps the term "with no intrinsic predicatble factorisations" or
              similar could be used to cover such concepts.

              Such forms are certainly vastly interesting, the Sierpinski/Riesel
              problems (that Jack mentioned, IIRC) are all about whether there are
              primes with forms that have no intricsic predictable factorisations.

              (hmmm, reminder to self or others - is there a link waiting to be
              added to the yahoogroup regarding the Sierpinski/Riesel problems?)

              Phil

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            • rlberry2002
              Phil, As always, your comments and insights are welcome. I had to do a little refresher myself on Sierpinski numbers: a positive, odd integer k for which
              Message 6 of 22 , May 1, 2002
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                Phil,

                As always, your comments and insights are welcome. I had to do a
                little refresher myself on Sierpinski numbers: a positive, odd
                integer k for which integers of the form k*2^p + 1 are all composite.
                I would suggest that there is little here which serves as an adequate
                counter example to the conjecture that I previously made.

                First, for all k less than 78557, at least 1 prime has been found to
                be generated by k*2^p + 1 with only 19 exceptions (and I feel it is
                just that the first prime solution for these 19 k's has not yet been
                found). It is only conjectured that k=78557 is a Sierpinski number.
                It is likely that the smallest Sierpinski number (if indeed one does
                exist) is so large that direct factorization, indirect factorization,
                etc. will not be feasible. One final point concerning k=78557 will
                show the difficulty in analyzing these numbers: For k=78557, p=300
                you get a 95-digit result. In other words, the 300th example for
                k=78557 is already a number of such magnitude that only 1 number in
                300 will be prime. Guess what the odds look like for the next 300
                values of p.

                Indeed, any series of numbers of the form k*N^p +/- c are very
                difficult to analysis except with indirect methods.

                Regards,

                Robert

                --- In primenumbers@y..., Phil Carmody <thefatphil@y...> wrote:
                > --- rlberry2002 <rlberry2002@y...> wrote:
                > > Please pardon my ignorance; I am here to learn to grow in my
                > > knowledge of number theory. However, I will try to be more
                careful
                > > in responding to posts before I have fully thought out my response
                > > - you could extend me the same courtesy.
                >
                > I would trust that such courtesy is demonstrated on the list.
                > (Yeah, flame me off list, you won't be the first... (or the second)
                > :-) )
                >
                > > Your example of N=2^p + 12213 violates one of the two
                > > qualifications
                > > that I laid down - "the equation cannot be reducible".
                >
                > Strictly, it is irreducible. The letter of the law is obeyed.
                >
                > > I typically
                > > would use this tenet for an equation like 4n + 1 (which does have
                > > infinitely many primes in it)where n is odd. The tenet holds for
                > > numbers of the form 2^p + n also, it just is usually harder to
                find
                > >
                > > the pattern that this type of equation reduces to. In your
                > > example,
                > > you provide the pattern which violates my first condition: that
                is
                > >
                > > for 2, 3, 1 Mod 12, 5 Mod 12, 7 Mod 12, & 11 Mod 12; there a
                fixed
                > > set of possible outcomes each of which will have fixed factors
                > > depending upon which of the outcomes it fall under.
                >
                > Sure, but that's not redicibility. What Jack has highlighted is an
                > intrinsically interesting property about that sequence of numbers.
                > This property will ba shared by an infinite number of other
                > sequences, not just the ...+12213. The property isn't reducibility,
                > and that term was used, it's a very precisely defined term, so Jack
                > and others can't be faulted for taking the term at face falue.
                > Perhaps the term "with no intrinsic predicatble factorisations" or
                > similar could be used to cover such concepts.
                >
                > Such forms are certainly vastly interesting, the Sierpinski/Riesel
                > problems (that Jack mentioned, IIRC) are all about whether there are
                > primes with forms that have no intricsic predictable factorisations.
                >
                > (hmmm, reminder to self or others - is there a link waiting to be
                > added to the yahoogroup regarding the Sierpinski/Riesel problems?)
                >
                > Phil
                >
                > __________________________________________________
                > Do You Yahoo!?
                > Yahoo! Health - your guide to health and wellness
                > http://health.yahoo.com
              • Jack Brennen
                ... We re glad you did a little research on Sierpinski numbers. However, you must have missed something. It is PROVEN, and can be shown using nothing more
                Message 7 of 22 , May 1, 2002
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                  rlberry2002 wrote:
                  > As always, your comments and insights are welcome. I had to do a
                  > little refresher myself on Sierpinski numbers: a positive, odd
                  > integer k for which integers of the form k*2^p + 1 are all composite.
                  > I would suggest that there is little here which serves as an adequate
                  > counter example to the conjecture that I previously made.
                  >
                  > First, for all k less than 78557, at least 1 prime has been found to
                  > be generated by k*2^p + 1 with only 19 exceptions (and I feel it is
                  > just that the first prime solution for these 19 k's has not yet been
                  > found). It is only conjectured that k=78557 is a Sierpinski number.

                  We're glad you did a little research on Sierpinski numbers.

                  However, you must have missed something. It is PROVEN, and can be shown
                  using nothing more than very simple arithmetic, that k=78557 is
                  a Sierpinski number. The proof, in condensed form:

                  If n == 0 (mod 2), 78557*2^n+1 is divisible by 3
                  If n == 1 (mod 4), 78557*2^n+1 is divisible by 5
                  If n == 3 (mod 36), 78557*2^n+1 is divisible by 73
                  If n == 15 (mod 36), 78557*2^n+1 is divisible by 19
                  If n == 27 (mod 36), 78557*2^n+1 is divisible by 37
                  If n == 7 (mod 12), 78557*2^n+1 is divisible by 7
                  If n == 11 (mod 12), 78557*2^n+1 is divisible by 13

                  Every integer n satisfies one of these seven congruences.

                  The unproven conjecture is that k=78557 is the
                  *smallest* Sierpinski number.
                • Gary Chaffey
                  Does the idea of 2^p+3 extend to 2^p-3? I have found that 2^233-3 is PRP now p is of the form 60k-7. Is this just a coincedence or is there some sort of
                  Message 8 of 22 , May 3, 2002
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                    Does the idea of 2^p+3 extend to 2^p-3? I have found
                    that 2^233-3 is PRP now p is of the form 60k-7. Is
                    this just a coincedence or is there some sort of
                    pattern.
                    I haven't checked
                    2^233= a mod 233
                    2^(2^233)= 2^n mod a
                    Like Norman did for 2^p+3 with p=7 and 67
                    Gary

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                  • Phil Carmody
                    ... This can be checked as follows. If 5 | 2^x-3 then 2^x==3 (5) then x==3 (4) If 7 | 2^x-3 then 2^x==3 (7) no solution If 11 | 2^x-3 then 2^x==3 (11) then
                    Message 9 of 22 , May 3, 2002
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                      --- Gary Chaffey <garychaffey@...> wrote:
                      > Does the idea of 2^p+3 extend to 2^p-3? I have found
                      > that 2^233-3 is PRP now p is of the form 60k-7. Is
                      > this just a coincedence or is there some sort of
                      > pattern.

                      This can be checked as follows.

                      If 5 | 2^x-3 then 2^x==3 (5) then x==3 (4)
                      If 7 | 2^x-3 then 2^x==3 (7) no solution
                      If 11 | 2^x-3 then 2^x==3 (11) then x==8 (10)
                      If 13 | 2^x-3 then 2^x==3 (13) then x==4 (12)
                      ...

                      These remove
                      3,7,11,15,19,23,27,31,35,39,43,47,51,55,59 (mod 60)
                      8 18 28 38 48 58 (mod 60)
                      4 16 28 40 52 (mod 60)
                      ...

                      There are other primes that add to this mod 60 period, obviously.

                      Not every residue is removed, which leads me to suspect that 53 isn't
                      the only residue primes will be found along.

                      Phil

                      =====
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                      "One cannot delete the Web browser from KDE without
                      losing the ability to manage files on the user's own
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                    • djbroadhurst
                      A little reminder: if you find a PRP of the form 2^n-3 or 2^n+3 (for any n, not just a prime) you may have prospects of a BLS (which failing a KP) proof by
                      Message 10 of 22 , May 3, 2002
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                        A little reminder: if you find a PRP of the form
                        2^n-3 or 2^n+3 (for any n, not just a prime) you may
                        have prospects of a BLS (which failing a KP) proof
                        by looking at work on factorization of Phi(2,k),
                        since, in *either* case, *both* N-1 and N+1 are
                        algebraically factorizable into these intensively
                        studied base-2 cyclotomic numbers:
                        http://www.cerias.purdue.edu/homes/ssw/cun/index.html
                        Apologies to those for whom this is blindingly obvious.
                        David Broadhurst
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