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Re: [PrimeNumbers] PrimePuzzles.net

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  • Jens Kruse Andersen
    ... Sure you did not get your own cache? If you are sure: I know little about the Internet. Could it be an error on a regional domain name server or whatever
    Message 1 of 4 , Feb 1, 2003
      Jose Ramón Brox wrote:
      > I just clicked http://www.primepuzzles.net/ and I didn't get any problems.

      Sure you did not get your own cache?
      If you are sure: I know little about the Internet. Could it be an error on a
      regional domain name server or whatever it's called. I am trying to access
      from Denmark. Can you e-mail Carlos and tell him about the problem others are

      Jud McCranie wrote:
      > I can't get onto PrimePuzzles.net (last night or this morning). It takes
      me to
      > http://www.landois.com/ instead. If I try to email Carlos Rivera, the
      > message is rejected. Are others having the same problem? Anyone know
      > what's going on?

      I cannot access www.primepuzzles.net either. I also get
      http://www.landois.com/ instead.
      I get a second error for www.primepuzzles.net/puzzles (browser switches
      between two messages in status line) and a third for
      www.primepuzzles.net/puzzles/puzz_209.htm (nothing happens).
      If anyone is interested in the solution for puzzle 209 which should have come
      today, my mail to Carlos is below. I don't know what the new puzzle or problem
      today should be.

      --- Begin e-mail to Carlos Rivera ---
      Puzzle 209. Triangles of primes

      1. Can you provide a formula to calculate the quantity of embedded equilateral
      triangles in an K-triangular array?

      This has to be in the EIS. A quick lookup for the first terms 1,5,13,27 finds

      ID Number: A002717 (Formerly M3827 and N1569)
      Sequence: 0,1,5,13,27,48,78,118,170,235,315,411,525,658,812,988,1188,
      Name: Floor(n(n+2)(2n+1)/8).
      Comments: Number of triangles in triangular matchstick arrangement of side n.

      Who needs to think when the EIS is there :-)

      2. Can you find one solution for every 4<K<=10?

      Each solution can be rotated and mirrored in 6 ways. I only count this as one
      There is a single solution for K=3.
      There are 104 solutions for K=4.
      There are 1261 solutions for K=5. This is one of them:
      31 37
      7 29 5
      23 17 13 19
      53 3 11 47 43

      There are no solutions for K>5.

      I first tried my search program for K=6 and found no solutions. I suspected
      there would never be more solutions. This can be proved by considering
      potentiel solutions modulo 3. The only prime which is 0 (modulo 3) is 3. The
      sum of 3 triangle vertexes must not be divisible by 3 if it has to be a prime.
      This means the modulo 3 triples (0,1,2), (1,1,1) and (2,2,2) are impossible.
      I quickly modified my existing search program to show there are no solutions
      for K>5, not even if the 0 (from 3) is omitted. This could also be proved on
      paper relatively easy.
      The above means there are no K>5 solutions for any set of distinct primes.
      There are solutions for K=6 if we allow repetitions of 3. This modulo 3
      template has two 0's, corresponding to two 3's:
      2 2
      2 0 2
      1 2 2 1
      2 1 1 1 2
      1 1 2 2 1 1

      There is only one other modulo 3 template with only two 3's. That is the above
      with all 1's and 2's swapped. This would give one 2 too few to match the
      smallest primes modulo 3.
      If we replace the largest prime 73 with a second 3 then we get 214 solutions,
      all matching the above template. Here is one of them:
      53 71
      11 3 23
      13 29 41 37
      59 31 67 19 47
      61 7 5 17 43 73

      For K>6 there are no template solutions and thus no prime solutions (or even
      integer solutions with prime sums) for any number of 3's allowed.

      3. Do you devise a systematic approach in order to get the solutions asked in

      I used pretty brute force for K<=6. My C program went through the vertices
      with a recursive function. It placed each allowable prime at a vertex before a
      recursive call for the next vertex. It seemed like it would be too slow for
      K=7 and I did not expect solutions. I thought of the modulo 3 simplification
      instead of optimizing the program.
      --- End e-mail to Carlos Rivera ---

      Jens Kruse Andersen
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