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

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  • 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 1 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 2 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.
        >
        >
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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 3 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 4 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 5 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 6 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 7 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 8 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 9 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 10 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 11 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 12 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 13 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 14 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 15 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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                                • 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 16 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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