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Re: 2^p+3 and more

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  • paulunderwooduk
    Paul, btw this ABC2 2^($a)-2^($a/2+1)+1 a: from 2 to 3000 produced: 2^3-2^(3/2+1)+1 2^7-2^(7/2+1)+1 2^47-2^(47/2+1)+1 2^73-2^(73/2+1)+1 2^79-2^(79/2+1)+1
    Message 1 of 22 , May 1, 2002
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      Paul,

      btw this

      ABC2 2^($a)-2^($a/2+1)+1
      a: from 2 to 3000

      produced:

      2^3-2^(3/2+1)+1
      2^7-2^(7/2+1)+1
      2^47-2^(47/2+1)+1
      2^73-2^(73/2+1)+1
      2^79-2^(79/2+1)+1
      2^113-2^(113/2+1)+1
      2^151-2^(151/2+1)+1
      2^167-2^(167/2+1)+1
      2^239-2^(239/2+1)+1
      2^241-2^(241/2+1)+1
      2^353-2^(353/2+1)+1
      2^367-2^(367/2+1)+1
      2^457-2^(457/2+1)+1
      2^1367-2^(1367/2+1)+1

      Note the first exponents are all prime. Are there anymore of these?

      Paul U.
    • Phil Carmody
      ... [SNIP - see Paul J s reply] ... I wouldn t expect them to have the same distribution. 2^p+1, being a cyclotomic form, has restrictions on what its factors
      Message 2 of 22 , May 1, 2002
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        --- rlberry2002 <rlberry2002@...> wrote:
        > Just a couple of observations regarding primes of the form 2^p+3.

        [SNIP - see Paul J's reply]

        > 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.

        Phil

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      • Chris Caldwell
        ... Such as: 2^364289-2^182145+1 Sure. These are norms of the Gaussian Mersenne primes. Look on the prime list.
        Message 3 of 22 , May 1, 2002
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          At 04:02 PM 5/1/02 +0000, paulunderwooduk wrote:
          >2^167-2^(167/2+1)+1
          >2^239-2^(239/2+1)+1
          >2^241-2^(241/2+1)+1
          >2^353-2^(353/2+1)+1
          >2^367-2^(367/2+1)+1
          >2^457-2^(457/2+1)+1
          >2^1367-2^(1367/2+1)+1
          >
          >Note the first exponents are all prime. Are there anymore of these?

          Such as:

          2^364289-2^182145+1

          Sure. These are norms of the Gaussian Mersenne primes. Look on the
          prime list.
        • jbrennen
          ... No, you got it right :-) ... See Sierpinski numbers for a well-known counterexample: 78557*2^n+1 contains an infinite number of elements which are (5 mod
          Message 4 of 22 , May 1, 2002
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            --- 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
            ... 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 5 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 6 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 7 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 8 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 9 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 10 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
                        >
                        > __________________________________________________
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                        > 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 11 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 12 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 13 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 14 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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