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Pritchard 1981, A sublinear additive sieve .... Comm ACM 24:18-23

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  • Phil Carmody
    I have issues with something in Crandall & Pomerance, their algorithm and their claimed big-Oh don t match (they are optimistic , heheh). However, they
    Message 1 of 22 , Apr 25, 2002
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      I have 'issues' with something in Crandall & Pomerance, their
      algorithm and their claimed big-Oh don't match (they are
      'optimistic', heheh). However, they refer to the above paper, which
      might provide an explanation. I believe I've 'fixed' their algorithm
      (it requires splitting stages into different parts, and changing the
      order of things, so it's almost a different algorithm), but if
      Pritchard already documents the alterations I propose, then there's
      less point in me investigating further.

      Does anyone have access to that paper so that I could briefly take a
      peek?

      I've picked up later papers by Pritchard, Sorensen and Dunten (3
      different papers that is) off Citeseer, which all refer to it;
      there's a chance that my fix may be in those but from the abstracts I
      don't think that's the case, and if it were to be then C&P need to
      fix their references.

      Cheers,
      Phil

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    • Paul Jobling
      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
      Message 2 of 22 , May 1, 2002
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        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.


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      • rlberry2002
        Paul, Just a couple of observations regarding primes of the form 2^p+3. First, there should be infinitely many primes of this form. When p is even, the result
        Message 3 of 22 , May 1, 2002
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          Paul,

          Just a couple of observations regarding primes of the form 2^p+3.
          First, there should be infinitely many primes of this form. When p
          is even, the result is a number congruent 1 Mod 6; when p is odd, the
          result is a number congruent 5 Mod 6. Since all primes other than 2
          or 3 are congruent 1 Mod 6 or 5 Mod 6 and since there are infinitely
          many primes contained in either of the two congruences, it follows
          that there should be infinitely many primes of the form 2^p+3. I
          haven't a clue as to the largest prime to date of this form, but
          certainly it should be (or could be, with a little focused attention)
          very large.

          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.

          Robert

          --- In primenumbers@y..., "Paul Jobling" <Paul.Jobling@W...> wrote:
          > 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.
          >
          >
          > __________________________________________________
          > Virus checked by MessageLabs Virus Control Centre.
        • Paul Jobling
          Hi Robert, ... I believe that we are only interested in p prime here. ... Careful... that sort of reasoning is wrong - you are saying that given an infinite
          Message 4 of 22 , May 1, 2002
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            Hi Robert,

            > Just a couple of observations regarding primes of the form 2^p+3.
            > First, there should be infinitely many primes of this form. When p
            > is even, the result is a number congruent 1 Mod 6; when p is odd, the
            > result is a number congruent 5 Mod 6.

            I believe that we are only interested in p prime here.

            > Since all primes other than 2
            > or 3 are congruent 1 Mod 6 or 5 Mod 6 and since there are infinitely
            > many primes contained in either of the two congruences, it follows
            > that there should be infinitely many primes of the form 2^p+3.

            Careful... that sort of reasoning is wrong - you are saying that given an
            infinite set N, where an infinite number of its member have some property A,
            then any infinite subset M of N must contain an infinite number of members
            with the proper A as well (consider N=the integers; A=prime; M=the composite
            numbers).

            > I
            > haven't a clue as to the largest prime to date of this form, but
            > certainly it should be (or could be, with a little focused attention)
            > very large.
            >
            > 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.

            But they do not seem to. The earliest Mersenne primes are quite small, whereas
            I am not sure that even one example of a pseudoprime of this form has been
            found.

            Regards,

            Paul.


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          • Paul Jobling
            ... Let me correct that to say for reasonably large p . __________________________________________________ Virus checked by MessageLabs Virus Control Centre.
            Message 5 of 22 , May 1, 2002
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              > But they do not seem to. The earliest Mersenne primes are
              > quite small, whereas
              > I am not sure that even one example of a pseudoprime of this
              > form has been found.

              Let me correct that to say "for reasonably large p".

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            • paulunderwooduk
              ... reached, and ... being done a ... results were ... http://groups.yahoo.com/group/primenumbers/message/1023 might contain some results for 2^p+3 Paul U.
              Message 6 of 22 , May 1, 2002
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                --- In primenumbers@y..., "Paul Jobling" <Paul.Jobling@W...> wrote:
                > 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?

                http://groups.yahoo.com/group/primenumbers/message/1023

                might contain some results for 2^p+3

                Paul U.
              • Paul Jobling
                Well, the OLEIS (A057736) only has 2, 3, 7, 67. But looking at http://groups.yahoo.com/group/primeform/message/1218 it appears that some searching was going
                Message 7 of 22 , May 1, 2002
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                  Well, the OLEIS (A057736) only has 2, 3, 7, 67. But looking at
                  http://groups.yahoo.com/group/primeform/message/1218
                  it appears that some searching was going on. So now the question is what
                  search limits were reached - Christ; Chris?

                  Paul.


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                • rlberry2002
                  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. I stand by my earlier statements (let s
                  Message 8 of 22 , May 1, 2002
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                    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.

                    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.

                    I enjoy your comments and perspective...

                    Robert

                    --- In primenumbers@y..., "Paul Jobling" <Paul.Jobling@W...> wrote:
                    > Hi Robert,
                    >
                    > > Just a couple of observations regarding primes of the form 2^p+3.
                    > > First, there should be infinitely many primes of this form. When
                    p
                    > > is even, the result is a number congruent 1 Mod 6; when p is odd,
                    the
                    > > result is a number congruent 5 Mod 6.
                    >
                    > I believe that we are only interested in p prime here.
                    >
                    > > Since all primes other than 2
                    > > or 3 are congruent 1 Mod 6 or 5 Mod 6 and since there are
                    infinitely
                    > > many primes contained in either of the two congruences, it follows
                    > > that there should be infinitely many primes of the form 2^p+3.
                    >
                    > Careful... that sort of reasoning is wrong - you are saying that
                    given an
                    > infinite set N, where an infinite number of its member have some
                    property A,
                    > then any infinite subset M of N must contain an infinite number of
                    members
                    > with the proper A as well (consider N=the integers; A=prime; M=the
                    composite
                    > numbers).
                    >
                    > > I
                    > > haven't a clue as to the largest prime to date of this form, but
                    > > certainly it should be (or could be, with a little focused
                    attention)
                    > > very large.
                    > >
                    > > 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.
                    >
                    > But they do not seem to. The earliest Mersenne primes are quite
                    small, whereas
                    > I am not sure that even one example of a pseudoprime of this form
                    has been
                    > found.
                    >
                    > Regards,
                    >
                    > Paul.
                    >
                    >
                    > __________________________________________________
                    > Virus checked by MessageLabs Virus Control Centre.
                  • 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 9 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 10 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 11 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 12 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 13 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 14 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 15 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 16 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 17 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 18 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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                                      • 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 19 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 20 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 21 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
                                              hard disk." - Prof. Stuart E Madnick, MIT.
                                              So called "expert" witness for Microsoft. 2002/04/02

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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 22 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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