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Re: [PrimeNumbers] Union of 2 Sets of Primes in Arithmetical Progression

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  • Jens Kruse Andersen
    ... A (CP-N), N =4, is N consecutive primes with gaps alternating between two different values. ... Let a, b, c be random consecutive primes around N. They
    Message 1 of 4 , Feb 4, 2005
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      Bill Sindelar wrote:

      > (CP-4) means a set of 4 CONSECUTIVE PRIMES NOT IN ARITHMETICAL
      > PROGRESSION such that the smallest prime of the set is equal to the sum
      > of the two next smallest primes minus the largest prime of the subset.

      > (CP-N) means a set of N CONSECUTIVE PRIMES NOT IN ARITHMETICAL
      > PROGRESSION such that EACH POSSIBLE SUBSET of 4 CONSECUTIVE primes is a
      > (CP-4).

      A (CP-N), N>=4, is N consecutive primes with gaps alternating between two
      different values.

      > Question (1); I discovered that the actual number of (CP-4)s that exist
      > in a set of N CONSECUTIVE primes, counting 2 as the first, can be closely
      > approximated by (N/logN).

      Let a, b, c be random consecutive primes around N. They start a (CP-4) iff x =
      c+b-a is prime and there are no primes between c and x.
      x is odd when a>2. If we otherwise assume x is random (maybe not accurate),
      then x has around 2/log N chance of being prime by PNT.
      There are no primes between c and x iff the prime gap following c is not
      smaller than the random gap from a to b. The probability is near 1/2.
      The probability of x completing a (CP-4) should then be
      1/2 * 2/log N = 1/log N.
      It is not surprising if the number of (CP-4) among the first N primes is
      around N/log N.

      Most of the first N primes are above N so the prime chance of x should
      actually be a little below 1/log N. I haven't analyzed the detailed
      heuristics. Maybe they should take into account that x is not a completely
      random odd number.

      > Question (2); How large a (CP-N) can one expect to find? I found (CP-N)s
      > with up to N=7 consecutive primes, but so far no luck with N=8. I can't
      > go much further with my limited resources. They get increasingly harder
      > to find.

      One way to get a potential (CP-N) is to remove every 3rd number in an AP.
      If the prime tuple conjecture (the variant with no other primes inside the
      pattern) is true then (CP-N)s can be arbitrarily long.
      I expect that to hold.

      Sextuplets are (CP-6). The largest known is at
      http://www.ltkz.demon.co.uk/ktuplets.htm#largest6 :
      138765468778 * 850# + 2822707 + 0, 4, 6, 10, 12, 16
      (362 digits, 2004, Norman Luhn)

      --
      Jens Kruse Andersen
    • Jens Kruse Andersen
      A (CP-N), N =4, is N consecutive primes with gaps alternating between two different values. The first (CP-8) has gaps 10 and 8:
      Message 2 of 4 , Feb 4, 2005
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        A (CP-N), N>=4, is N consecutive primes with gaps alternating between two
        different values.

        The first (CP-8) has gaps 10 and 8:
        67944073,67944083,67944091,67944101,67944109,67944119,67944127,67944137

        The first (CP-9) has gaps 2 and 28 (sum 5#):
        1860017189,1860017191,1860017219,1860017221,1860017249,
        1860017251,1860017279,1860017281,1860017309

        The first (CP-10) has gaps 16 and 14 (sum 5#):
        5373097559,5373097573,5373097589,5373097603,5373097619,
        5373097633,5373097649,5373097663,5373097679,5373097693

        --
        Jens Kruse Andersen
      • Jens Kruse Andersen
        Sorry for not making a single post after completing computations. I have stopped now. ... The second (CP-10) is 16 times larger with gaps 28 and 2: 87873432313
        Message 3 of 4 , Feb 4, 2005
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          Sorry for not making a single post after completing computations.
          I have stopped now.

          I wrote:
          > The first (CP-10) has gaps 16 and 14 (sum 5#):
          > 5373097559,5373097573,5373097589,5373097603,5373097619,
          > 5373097633,5373097649,5373097663,5373097679,5373097693

          The second (CP-10) is 16 times larger with gaps 28 and 2:
          87873432313 + 0,28,30,58,60,88,90,118,120,148

          A probably non-minimal (CP-11) with gaps 16 and 14:
          196723765163557 + 0,16,30,46,60,76,90,106,120,136,150

          And a probably non-minimal (CP-12) with gaps 14 and 16:
          438536033046239 + 0,14,30,44,60,74,90,104,120,134,150,164

          A (CP-13) must contain an AP6 inside an AP7. The common difference in an AP7
          not starting at 7 is always a multiple of 7# = 210.
          Inside the AP7 there has to be at least (210-2)*6 = 1248 simultaneous
          composites.
          That makes a (CP-13) computionally infeasible.
          If the numbers are small then 1248 composites is too hard.
          If the numbers are large then 13 primes is too hard.

          --
          Jens Kruse Andersen
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