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Re: [PrimeNumbers] three-prime sum chains

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  • Jens Kruse Andersen
    ... With the choice we quickly get a chain with 12 sums: 281 + 283 + 293 = 857 857 + 859 + 863 = 2579 2551 + 2557 + 2579 = 7687 7681 + 7687 + 7691 = 23059
    Message 1 of 7 , Jun 4, 2013
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      Phil Carmody wrote:
      > I specifically wanted to offer the choice. I wanted the chains
      > to be longer, and also less simple to find.

      With the choice we quickly get a chain with 12 sums:
      281 + 283 + 293 = 857
      857 + 859 + 863 = 2579
      2551 + 2557 + 2579 = 7687
      7681 + 7687 + 7691 = 23059
      23059 + 23063 + 23071 = 69193
      69193 + 69197 + 69203 = 207593
      207569 + 207589 + 207593 = 622751
      622751 + 622777 + 622781 = 1868309
      1868287 + 1868291 + 1868309 = 5604887
      5604881 + 5604887 + 5604901 = 16814669
      16814669 + 16814671 + 16814701 = 50444041
      50444027 + 50444041 + 50444059 = 151332127

      The longest chain with initial primes below 10^11 is probably 14 sums:
      2878090951 + 2878090961 + 2878090967 = 8634272879
      8634272839 + 8634272879 + 8634272891 = 25902818609
      25902818609 + 25902818629 + 25902818663 = 77708455901
      77708455819 + 77708455823 + 77708455901 = 233125367543
      233125367543 + 233125367599 + 233125367609 = 699376102751
      699376102751 + 699376102757 + 699376102783 = 2098128308291
      2098128308239 + 2098128308273 + 2098128308291 = 6294384924803
      6294384924787 + 6294384924803 + 6294384924823 = 18883154774413
      18883154774413 + 18883154774441 + 18883154774447 = 56649464323301
      56649464323301 + 56649464323303 + 56649464323307 = 169948392969911
      169948392969877 + 169948392969881 + 169948392969911 = 509845178909669
      509845178909611 + 509845178909669 + 509845178909687 = 1529535536728967
      1529535536728967 + 1529535536728987 + 1529535536728999 = 4588606610186953
      4588606610186927 + 4588606610186947 + 4588606610186953 = 13765819830560827

      There are also 14 starting with
      44453980303 + 44453980309 + 44453980379 = 133361940991

      > I presume you just ran a prime sieve and then probed for each one
      > with no deduplication?

      I didn't bother with deduplication but I ran two prime sieves in
      parallel, around p and 3p. This meant the first sum never had to be prp
      tested, and its surrounding primes were also generated without prp tests.
      It could be extended to a third sieve around 9p and so on, but each
      extra sieve has to sieve a larger interval and avoids fewer prp tests.

      > I hadn't considered Dickson. It's the big gun, certainly.
      > I wondered initially if it would fail on guaranteeing that the
      > positioned primes are consecutive, but if you ask for a triplet
      > {p,p+3+/-1,p+6}, then that comes for free.

      Dickson's conjecture can be manipulated to ensure consecutive primes
      in this problem for any admissible pattern. We can just demand extra
      primes chosen so specific prime factors are forced to divide certain
      numbers in a gap.
      For example, we can ensure consecutive {p, p+6} by demanding four primes
      {p, p+6, p+8, p+14}. This forces 3 to divide p+4, and 5 to divide p+2.

      --
      Jens Kruse Andersen
    • djbroadhurst
      ... which is a special case of Bateman-Horn: http://tech.groups.yahoo.com/group/primenumbers/message/25133 David
      Message 2 of 7 , Jun 4, 2013
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        --- In primenumbers@yahoogroups.com, "Jens Kruse Andersen" <jens.k.a@...> wrote:

        > Arbitrarily long chains are certainly expected.
        > It would for example follow from Dickson's conjecture:
        > http://primes.utm.edu/glossary/xpage/DicksonsConjecture.html

        which is a special case of Bateman-Horn:
        http://tech.groups.yahoo.com/group/primenumbers/message/25133

        David
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