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Re: [PrimeNumbers] New Type of Prime Arithmetic Progression?

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  • Jens Kruse Andersen
    ... This follows from the widely believed conjecture that all admissible prime constellations have infinitely many occurrences. If the conjecture is true then
    Message 1 of 12 , Sep 24, 2007
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      Bill Sindelar wrote:
      > For any integer k=3 or greater, and any integer n=0 or greater, one can
      > find an arithmetic progression of prime numbers (PAP) of length k, such
      > that the number n of consecutive primes between adjacent terms of the PAP
      > is constant.

      This follows from the widely believed conjecture that all
      admissible prime constellations have infinitely many occurrences.
      If the conjecture is true then one can first choose specific admissible
      positions for all primes, and then choose additional prime positions
      outside the whole PAP interval in a way which fixes chosen primes
      modulo the interval start such that every other number in the interval
      gets a known prime factor.

      > Consecutive primes in arithmetic progression can be considered as a
      > (PAP-k, n) since there are n=0 primes between adjacent terms. I would
      > interpret Green and Tao as covering this type.

      No. Green and Tao does not say that the primes in AP are
      consecutive or have a constant number of primes between them.

      > I would very much appreciate the advice of those that have worked on
      > (PAP-k, 0) as to the computational complexity of finding (PAP-k, n>3).

      (PAP-8, 5): 1376308269430349 + 210*i is prime for i = 0 to 7,
      and there are 5 primes between in each case.
      It's probably not the smallest (PAP-8, n).

      (PAP-9, n>3) and (PAP-10, n>3) look easier than n=0 which has known
      cases (a single case for PAP-10) at http://hjem.get2net.dk/jka/math/cpap.htm

      (PAP-11, n>0) looks very hard, but not as hard as for n=0 which
      could take trillions of GHz years.
      A PAP-11 has a common difference which is a multiple of 11# = 2310,
      and 10 intervals of that size would very rarely have the same number
      of primes - unless the numbers are so large that the intervals have few
      or no primes, but then the 11 primes in the PAP-11 would be large
      and extremely hard to find.

      --
      Jens Kruse Andersen
    • w_sindelar@juno.com
      Thank you once again, Jens. I value your responses. I would like to add a few comments in reply. ... I m lost here. Seems like a convoluted approach. Here s
      Message 2 of 12 , Sep 26, 2007
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        Thank you once again, Jens. I value your responses. I would like to add a
        few comments in reply.

        > Bill Sindelar wrote:
        > > For any integer k=3 or greater, and any integer n=0 or greater,
        > one can
        > > find an arithmetic progression of prime numbers (PAP) of length k,
        > such
        > > that the number n of consecutive primes between adjacent terms of
        > the PAP
        > > is constant.

        Jens K Andersen wrote:
        > This follows from the widely believed conjecture that all
        > admissible prime constellations have infinitely many occurrences.
        > If the conjecture is true then one can first choose specific
        > admissible
        > positions for all primes, and then choose additional prime
        > positions
        > outside the whole PAP interval in a way which fixes chosen primes
        > modulo the interval start such that every other number in the
        > interval
        > gets a known prime factor.

        I'm lost here. Seems like a convoluted approach. Here's how I came to
        making the above statement. A few weeks ago I realized that, in all the
        literature I read on PAP's, including Green and Tao, no reference to the
        numbers enclosed between adjacent primes of the PAP was made. The sole
        requirement was a constant difference between adjacent primes. I decided
        to add the requirement that the number of consecutive primes between
        adjacent primes also be constant, and see if I could find any PAPs of
        this type.

        The approach I used required making the following assumption, which is
        equivalent to the above statement; Let S(p, n) represent an infinite
        subset of the universal set of all consecutive odd primes, where p is the
        first prime of the subset, and n (including 0) represents the number of
        consecutive primes from the universal set that have been omitted between
        adjacent primes of the subset.Then any S(p, n) contains a set of any
        number k of primes in arithmetic progression. The program I wrote is
        based on this.

        Sindelar wrote:
        > Consecutive primes in arithmetic progression can be considered as
        > a
        > (PAP-k, n) since there are n=0 primes between adjacent terms. I
        > would
        > interpret Green and Tao as covering this type.

        Andersen wrote:
        > No. Green and Tao does not say that the primes in AP are
        > consecutive or have a constant number of primes between them.

        As I said in my first post, I do not have the ability to understand the
        Green-Tao proof to see if specific exceptions are included. Tao in a
        presentation slide showed the theorem worded as "The primes contain
        arbitrarily long arithmetic progressions". In another slide he says "In
        particular, for any given k, the primes contain infinitely many
        arithmetic progressions of length k". To me, this is a very broad claim
        covering any type of (PAP-k, n). Jens can you explain your answer a bit
        more?

        Bill Sindelar

        Bill Sindelar
      • Jens Kruse Andersen
        ... You asked for comments on your statement (which is an unproven guess). I briefly showed that it would follow from a well-known and trusted conjecture,
        Message 3 of 12 , Sep 26, 2007
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          Bill Sindelar wrote:
          >> Bill Sindelar wrote:
          >>> For any integer k=3 or greater, and any integer n=0 or greater,
          >>> one can find an arithmetic progression of prime numbers (PAP)
          >>> of length k, such that the number n of consecutive primes
          >>> between adjacent terms of the PAP is constant.
          >
          > Jens K Andersen wrote:
          >> This follows from the widely believed conjecture that all
          >> admissible prime constellations have infinitely many occurrences.
          >> If the conjecture is true then one can first choose specific
          >> admissible positions for all primes, and then choose additional
          >> prime positions outside the whole PAP interval in a way which
          >> fixes chosen primes modulo the interval start such that every
          >> other number in the interval gets a known prime factor.
          >
          > I'm lost here. Seems like a convoluted approach.

          You asked for comments on your statement (which is an unproven
          guess). I briefly showed that it would follow from a well-known
          and trusted conjecture, sometimes called the prime k-tuple conjecture
          (that name is sometimes restricted to special cases). Of course it's
          less "convoluted" to not relate your guess to anything else like
          previosly studied things. If you want something unconvoluted
          (but not very useful by itself) then here it is: I guess your guess is
          right.

          > A few weeks ago I realized that, in all the
          > literature I read on PAP's, including Green and Tao, no reference to the
          > numbers enclosed between adjacent primes of the PAP was made. The sole
          > requirement was a constant difference between adjacent primes. I decided
          > to add the requirement that the number of consecutive primes between
          > adjacent primes also be constant

          There is plenty of literature about consecutive primes in arithmetic
          progression which is what you call (PAP-k, 0).
          I have not seen mention of your (PAP-k, n>0) before.

          > Sindelar wrote:
          >>> Consecutive primes in arithmetic progression can be considered as
          >>> a (PAP-k, n) since there are n=0 primes between adjacent terms. I
          >>> would interpret Green and Tao as covering this type.
          >
          > Andersen wrote:
          >> No. Green and Tao does not say that the primes in AP are
          >> consecutive or have a constant number of primes between them.
          >
          > As I said in my first post, I do not have the ability to understand the
          > Green-Tao proof to see if specific exceptions are included. Tao in a
          > presentation slide showed the theorem worded as "The primes contain
          > arbitrarily long arithmetic progressions". In another slide he says "In
          > particular, for any given k, the primes contain infinitely many
          > arithmetic progressions of length k". To me, this is a very broad claim
          > covering any type of (PAP-k, n). Jens can you explain your answer a bit
          > more?

          Just to be clear: My "No" was only to your second sentence:
          "I would interpret Green and Tao as covering this type."

          A PAP-k (often just called AP-k) is k primes in arithmetic progression.
          A PAP-k has no condition on how many other primes are between them.

          Tao's two formulations are trivially equivalent.
          1) "The primes contain arbitrarily long arithmetic progressions".

          By definition of "arbitrarily long" this means that:
          For any given natural number n, there is a PAP-k with k>=n.

          2) "In particular, for any given k, the primes contain infinitely many
          arithmetic progressions of length k".

          By definition of "infinitely many" this means that:
          For any given m and k, there are more than m PAP-k.

          2) obviously implies 1) by setting m=1.

          1) implies 2). Proof: If 1) is true then for any m and k, there exists
          a PAP-(m+k), and that contains m+1 overlapping PAP-k.


          You defined (PAP-k, n) as a PAP-k with n primes between each of
          the k-1 pairs of successive primes in the AP. Tao and Green don't
          mention this concept of equal prime counts and their theorem says
          nothing about your (PAP-k, n) for n=0 or any other n value.
          I don't know what else you want me to explain.
          All I can say is that the theorem simply doesn't say it.

          --
          Jens Kruse Andersen
        • elevensmooth
          ... I ll try. Are you aware of the concept of admissible constellations? For example, twin primes, x and x+2, are believed to occur infinitely often. A
          Message 4 of 12 , Sep 26, 2007
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            --- In primenumbers@yahoogroups.com, w_sindelar@... wrote:

            > Jens can you explain your answer a bit more?

            I'll try.

            Are you aware of the concept of "admissible constellations?" For
            example, twin primes, x and x+2, are believed to occur infinitely
            often. A triplet of the form x, x+2, x+4 is not admissible because
            one of these numbers always divisible by 3. Hence (3, 5, 7) is the
            only example - a finite number. However x, x+2, x+6 is admissible and
            is believed to be prime infinitely often.

            Jens point is that you can construct an admissible constellation that
            has all kn-1 primes in fixed locations so the the arithmetic
            progression and primes between are honored. This is more restrictive
            than your rules require, but would be an example as long as none of
            the the other values within the constellation's range are also prime.
            Jen's further point is that you could further restrict the
            constellation so that the intermediate values were divisible by
            selected primes, and hence composite. Again more restrictive than you
            rules, but it would qualify as an example. Finally, from the
            constellation conjecture there would be an infinite number of these
            constellation, each and every one an example.

            For example, x, x+2, x+6, x+8, x+12 is an example of admissible
            constellation that has 3 primes in arithmetic progression with(al
            least) one prime between them. In this case there is exactly one
            prime because x+4 and x+10 must be divisible by 3. In general, we
            would need to jiggle the start point as x=ay+b for selected fixed
            values if a and b to ensure the omitted points are composite.

            Finally, every example that you find can could be reverse-engineered
            to such an admissible constellation, so this search would be a subset
            of a search for examples of admissible constellations.

            William Lipp
            Poohbah of OddPerfect.org
          • w_sindelar@juno.com
            ... Jens, I think I may have offended you by writing I m lost here. Seems like a convoluted approach. Looking back at this, I can see that it can be taken as
            Message 5 of 12 , Sep 29, 2007
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              Sindelar wrote:
              > > I'm lost here. Seems like a convoluted approach.

              Andersen wrote:
              > You asked for comments on your statement (which is an unproven
              > guess). I briefly showed that it would follow from a well-known
              > and trusted conjecture, sometimes called the prime k-tuple
              > conjecture
              > (that name is sometimes restricted to special cases). Of course
              > it's
              > less "convoluted" to not relate your guess to anything else like
              > previously studied things. If you want something unconvoluted
              > (but not very useful by itself) then here it is: I guess your guess
              > is
              > right.

              Jens, I think I may have offended you by writing "I'm lost here. Seems
              like a convoluted approach." Looking back at this, I can see that it can
              be taken as arrogant criticism. Gad, that is not what I meant it to be. I
              should have written "I am unable to follow your explanation. It seems
              complicated to me because I know nothing about admissible prime
              constellations, but I accept your opinion". I regret my choice of words
              and hope you accept my sincere apology.

              Sindelar wrote in regard to Green and Tao:
              >>>To me, this is a very broad claim covering any type of (PAP-k, n).
              Jens can you explain your answer a bit more?

              Andersen wrote:
              > Just to be clear: My "No" was only to your second sentence:
              > "I would interpret Green and Tao as covering this type. (meaning
              (PAP-k, n=0))"
              >
              > You defined (PAP-k, n) as a PAP-k with n primes between each of
              > the k-1 pairs of successive primes in the AP. Tao and Green don't
              > mention this concept of equal prime counts and their theorem says
              > nothing about your (PAP-k, n) for n=0 or any other n value.
              > I don't know what else you want me to explain.
              > All I can say is that the theorem simply doesn't say it.

              Jens thank you. Nothing more to explain. You made it clear that the Green
              and Tao theorem does not apply to type (PAP-k, n=0 or greater). And thank
              you for an example of a (PAP-8, 5). Don't know how you calculated that so
              quickly. I was beginning to think there might be a limit on k.

              Bill Sindelar

              [Non-text portions of this message have been removed]
            • w_sindelar@juno.com
              ... Jens, I think I may have offended you by writing I m lost here. Seems like a convoluted approach. Looking back at this, I can see that it can be taken as
              Message 6 of 12 , Sep 29, 2007
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                Sindelar wrote:
                > > I'm lost here. Seems like a convoluted approach.

                Andersen wrote:
                > You asked for comments on your statement (which is an unproven
                > guess). I briefly showed that it would follow from a well-known
                > and trusted conjecture, sometimes called the prime k-tuple
                > conjecture
                > (that name is sometimes restricted to special cases). Of course
                > it's
                > less "convoluted" to not relate your guess to anything else like
                > previously studied things. If you want something unconvoluted
                > (but not very useful by itself) then here it is: I guess your guess
                > is
                > right.

                Jens, I think I may have offended you by writing "I'm lost here. Seems
                like a convoluted approach." Looking back at this, I can see that it can
                be taken as arrogant criticism. Gad, that is not what I meant it to be. I
                should have written "I am unable to follow your explanation. It seems
                complicated to me because I know nothing about admissible prime
                constellations, but I accept your opinion". I regret my choice of words
                and hope you accept my sincere apology.

                Sindelar wrote in regard to Green and Tao:
                >>>To me, this is a very broad claim covering any type of (PAP-k, n).
                Jens can you explain your answer a bit more?

                Andersen wrote:
                > Just to be clear: My "No" was only to your second sentence:
                > "I would interpret Green and Tao as covering this type. (meaning
                (PAP-k, n=0))"
                >
                > You defined (PAP-k, n) as a PAP-k with n primes between each of
                > the k-1 pairs of successive primes in the AP. Tao and Green don't
                > mention this concept of equal prime counts and their theorem says
                > nothing about your (PAP-k, n) for n=0 or any other n value.
                > I don't know what else you want me to explain.
                > All I can say is that the theorem simply doesn't say it.

                Jens thank you. Nothing more to explain. You made it clear that the Green
                and Tao theorem does not apply to type (PAP-k, n=0 or greater). And thank
                you for an example of a (PAP-8, 5). Don't know how you calculated that so
                quickly. I was beginning to think there might be a limit on k.

                Bill Sindelar

                [Non-text portions of this message have been removed]
              • w_sindelar@juno.com
                On Thu, 27 Sep 2007 01:55:00 -0000 elevensmooth ... I m glad you did, and I thank you. You must be a mind reader. You somehow sensed why I got lost trying to
                Message 7 of 12 , Sep 29, 2007
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                  On Thu, 27 Sep 2007 01:55:00 -0000 "elevensmooth"
                  <elevensmooth@...> writes:
                  > --- In primenumbers@yahoogroups.com, w_sindelar@... wrote:
                  >
                  > > Jens can you explain your answer a bit more?

                  William Lipp wrote:
                  > I'll try.

                  I'm glad you did, and I thank you. You must be a mind reader. You somehow
                  sensed why I got lost trying to follow Jens reasoning. Right off the bat
                  I'm confronted with "admissible prime constellations" and right there I'm
                  lost.

                  Your neat little introductory on this greatly helped me understand what
                  Jens meant. I'm going to study this concept in more detail. Regards with
                  appreciation.

                  Bill Sindelar
                • w_sindelar@juno.com
                  On Thu, 27 Sep 2007 01:55:00 -0000 elevensmooth ... I m glad you did, and I thank you. You must be a mind reader. You somehow sensed why I got lost trying to
                  Message 8 of 12 , Sep 29, 2007
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                    On Thu, 27 Sep 2007 01:55:00 -0000 "elevensmooth"
                    <elevensmooth@...> writes:
                    > --- In primenumbers@yahoogroups.com, w_sindelar@... wrote:
                    >
                    > > Jens can you explain your answer a bit more?

                    William Lipp wrote:
                    > I'll try.

                    I'm glad you did, and I thank you. You must be a mind reader. You somehow
                    sensed why I got lost trying to follow Jens reasoning. Right off the bat
                    I'm confronted with "admissible prime constellations" and right there I'm
                    lost.

                    Your neat little introductory on this greatly helped me understand what
                    Jens meant. I'm going to study this concept in more detail. Regards with
                    appreciation.

                    Bill Sindelar
                  • Jens Kruse Andersen
                    ... No problem. You can search more information about admissible constellations with a search engine. If a prime p
                    Message 9 of 12 , Sep 29, 2007
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                      Bill Sindelar wrote:
                      > Jens, I think I may have offended you by writing "I'm lost here. Seems
                      > like a convoluted approach."

                      No problem. You can search more information about admissible constellations
                      with a search engine.

                      If a prime p <= k does not divide the common difference in an AP-k then
                      p will divide at least one of the terms in the AP. In order to be
                      admissible, a PAP-k must therefore have a common difference which is
                      a multiple of k# (k primorial).
                      I guess a PAP-k with small difference (and therefore relatively few primes
                      between the terms) will have a better chance of being a (PAP-k, n),
                      because the number of primes can vary between fewer values.
                      A PAP-11 has minimal difference 11# = 2310, so 10 intervals of 2309
                      numbers must have the same prime count to produce a (PAP-11, n)
                      with minimal difference. That appears computationally too hard for me.

                      PAP-7 to PAP-10 all have minimal difference 10# = 7# = 210.
                      I used my old tuplet finder to systematically search a lot of PAP-10 with
                      difference 210 and count whether there happened to be an equal number
                      of primes between the terms. There were other things to use my only
                      computer for so the search stopped when only (PAP-8, n) had been found.
                      Hans Rosenthal is more patient and has found many (PAP-9, 0), also called
                      CPAP-9, with a version of the same program. (PAP-9, n>3) looks easier.
                      In 2004 he found the smallest known CPAP-8 = (PAP-8, 0) with
                      another version. I just tested the other PAP-8 from the search and found
                      a (PAP-8, 1) with difference 210:
                      64881326075217862991473794035228920286672784697 +
                      0,36,210,264,420,564,630,784,840,942,1050,1086,1260,1360,1470

                      --
                      Jens Kruse Andersen
                    • w_sindelar@juno.com
                      ... I m relieved. I was trying to get up some nerve to ask you what sort of ... I just tested the other PAP-8 from the search and ... equivalent to the above
                      Message 10 of 12 , Oct 2, 2007
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                        Jens K. Anderson wrote:
                        > No problem. You can...
                        >>

                        I'm relieved. I was trying to get up some nerve to ask you what sort of
                        approach you used on (PAP-8, 5) when your mail arrived with the answer:

                        > I used my old tuplet finder to systematically search a lot of PAP-10
                        > with
                        > difference 210 and count whether there happened to be an equal
                        > number
                        > of primes between the terms. There were other things to use my only
                        > computer for so the search stopped when only (PAP-8, n) had been
                        > found.
                        I just tested the other PAP-8 from the search and
                        > found
                        > a (PAP-8, 1) with difference 210:
                        > 64881326075217862991473794035228920286672784697 +
                        > 0,36,210,264,420,564,630,784,840,942,1050,1086,1260,1360,1470

                        Sindelar wrote (Yahoo #19096):

                        >>The approach I used required making the following assumption, which is
                        equivalent to the above statement; Let S(p, n) represent an infinite
                        subset of the universal set of all consecutive odd primes, where p is the
                        first prime of the subset, and n (including 0) represents the number of
                        consecutive primes from the universal set that have been omitted between
                        adjacent primes of the subset.Then any S(p, n) contains a set of any
                        number k of primes in arithmetic progression. The program I wrote is
                        based on this.>

                        I used Pari-gp for this. For every set of k consecutive primes, which has
                        n skipped consecutive primes between its adjacent terms, after an
                        inputted integer, it checks if the terms of that set are in arithmetic
                        progression. Jens, is this slower than your approach with your tuplet
                        finder? If one could prove the above assumption, would that also prove
                        that all admissible prime constellations have infinitely many occurrences
                        as you put it, or only those that have a (PAP-k, n) subset?

                        Sindelar wrote (Yahoo #19093):

                        >>Obviously, the ordinal numbers of the primes in such a PAP are also in
                        arithmetic progression (AP) with a constant difference of (n+1).>

                        This suggested trying this assumption which is just a fancy way of
                        defining a (PAP-k, n): In any infinite arithmetic progression of positive
                        integers with a common difference d, there exists a subset of k
                        consecutive integers, so that if each integer in that subset is
                        considered to represent the ordinal number of a prime, the associated
                        primes will be in arithmetic progression of length k with (d-1)
                        consecutive primes between adjacent terms of that arithmetic progression.
                        (Ordinal number of a prime means its position in the numerically ordered
                        set of all primes, with prime 2 being number 1). It works but is more
                        computationally complicated. What do you think?

                        Bill Sindelar

                        Bill Sindelar
                      • Jens Kruse Andersen
                        ... I would expect your method to be much slower based on how randomly consecutive prime gaps appear to be distributed. ... No, and also no to the only-part.
                        Message 11 of 12 , Oct 2, 2007
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                          Bill Sindelar wrote:
                          > I used Pari-gp for this. For every set of k consecutive primes, which has
                          > n skipped consecutive primes between its adjacent terms, after an
                          > inputted integer, it checks if the terms of that set are in arithmetic
                          > progression. Jens, is this slower than your approach with your tuplet
                          > finder?

                          I would expect your method to be much slower based on how
                          "randomly" consecutive prime gaps appear to be distributed.

                          > If one could prove the above assumption, would that also prove
                          > that all admissible prime constellations have infinitely many occurrences
                          > as you put it, or only those that have a (PAP-k, n) subset?

                          No, and also no to the only-part. Your assumption says nothing
                          about the existence of specific differences between primes,
                          so it says nothing about any admissible constellation.

                          > This suggested trying this assumption which is just a fancy way of
                          > defining a (PAP-k, n):
                          .....
                          > It works but is more
                          > computationally complicated. What do you think?

                          "computationally complicated" refers to something computational,
                          for example the time to compute something with a given algorithm.
                          You have made another formulation of your conjecture but not
                          described an algorithm so "computationally complicated" is a
                          concept which does not apply.

                          I don't have time to discuss more.

                          --
                          Jens Kruse Andersen
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