- --- In primenumbers@yahoogroups.com, Jud McCranie <judmccr@b...> 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

I have the same problem. I hope it is fixed soon.

> message is rejected. Are others having the same problem? Anyone know

> what's going on?

--Mark - Jose RamÃ³n 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

having?

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

I cannot access www.primepuzzles.net either. I also get

> message is rejected. Are others having the same problem? Anyone know

> what's going on?

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

www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=002717

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,

1413,1665,1945,2255,2596,2970,3378,3822,4303,4823,5383,5985,

6630,7320,8056,8840,9673,10557,11493,12483,13528,14630,

15790,17010,18291,19635,21043,22517

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

solution.

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:

41

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:

0

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:

3

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

2?

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