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Re: goldbachian prime progression puzzle

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  • Mark Underwood
    Yes, different rules. I like Ratwain s idea of using adjacent primes, whether increasing or decreasing. It extends the path and makes it more interesting, yet
    Message 1 of 17 , Mar 31 7:36 AM
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      Yes, different rules. I like Ratwain's idea of using adjacent primes,
      whether increasing or decreasing. It extends the path and makes it
      more interesting, yet still forms a continuity. Ratwain, is 5449 the
      last prime in the path, or did you just choose to stop there?

      If the path does end there, we could (probably) extend the path again
      by loosing the rules so that even the main prime p(n) is not forced
      to increment to the next prime p(n+1), but can stay the same or even
      decrease to p(n-1). Still continuous. In this case we would consider
      the *quickest* (shortest) path to get to a destination prime.

      It would be very cool if any prime could be reached by such a
      continuous path from the original 7 = 2+2+3.

      Mark








      --- In primenumbers@yahoogroups.com, "jbrennen" <jb@b...> wrote:
      >
      > --- In primenumbers@yahoogroups.com, "ratwain" <ratwain@y...> wrote:
      > >
      > > I think I can get to 10 different triplets that sum to 5449.
      > > The widest fanout in the process was 1301 ways to represent 1307.
      > >
      > > Since it seems I've come up with a different answer from the great
      > > Jack Brennen, I'd definitely like someone to either verify my
      > > results or show me where I've gone wrong.
      >
      > > The rules I was sticking to are:
      > > If { p_{i}, p_{j}, p_{k} } such that p_{i}+p_{j}+p_{k} = p_{x} is
      > > in the tree, then { p_{i'}, p_{j'}, p_{k'} } such that
      > > max(|i-i'|,|j-j'|,|k-k'|)<=1 and
      > > p_{i'}+p_{j'}+p_{k'} = p_{x+1} is also in the tree
      >
      > It seems that we used different rules. Your rules seem to permit
      > "back-tracking" of one or two of the primes, while my rules do not.
      > Two different problems, with two different answers...
      >
      > By your rules, (5,5,19) = 29 could lead to (7,7,17) = 31.
      > By my rules, such a step would not be allowed, because the 19
      > would not be permitted to step backward to 17.
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