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

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  • 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 1 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 2 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 3 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 4 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 5 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 6 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 7 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 8 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 9 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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