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

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  • 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 1 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 2 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 3 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 4 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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