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Re: [PrimeNumbers] three-prime sum chains

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  • Phil Carmody
    ... I specifically wanted to offer the choice. I wanted the chains to be longer, and also less simple to find. (By which I mean that you can t just do a simple
    Message 1 of 7 , Jun 4, 2013
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      --- On Sun, 6/2/13, Jens Kruse Andersen <jens.k.a@...> wrote:
      > Phil Carmody wrote:
      > > Find sets of 3 consecutive primes whose sum is itself prime,
      > > and associate that set and its members with that sum.
      > >
      > > What's the longest chain of associated primes you can find?
      > > Heuristically - would you expect there to be longer ones
      > > than the longest you can find?
      >
      > http://www.primepuzzles.net/puzzles/puzz_421.htm requires that the
      > sum is always the first of the 3 consecutive primes in the next sum.

      I specifically wanted to offer the choice. I wanted the chains to be longer, and also less simple to find. (By which I mean that you can't just do a simple O(length)-time depth-first search from each root prime, as you now have a tree to probe, not a straight line, and might want to consider how much effort is wasted doing duplicated work as neigbouring roots share common sub-trees.)

      > http://primes.utm.edu/curios/page.php/507995698619.html
      > says for that:
      >
      > "Start with the prime 507995698619. Add it with the two next primes
      > to get a sum which is prime. Repeat. The first 10 sums are prime.
      > This is the first case with more than 8.

      Good work (as per usual!). I presume you just ran a prime sieve and then probed for each one with no deduplication? Given how likely failure is, probing a tree shouldn't be much more than a small factor smaller than probing a straight line.

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

      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.

      Phil
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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 2 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 3 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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