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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 3, 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. 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
      --
      () ASCII ribbon campaign () Hopeless ribbon campaign
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      [stolen with permission from Daniel B. Cristofani]
    • 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 2 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
        --
        () ASCII ribbon campaign () Hopeless ribbon campaign
        /\ against HTML mail /\ against gratuitous bloodshed

        [stolen with permission from Daniel B. Cristofani]
      • 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 3 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 4 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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