Problems & Puzzles: Puzzles

Puzzle 209. Triangles of primes

Here you asked to find a K-triangular arrange of the first K*(K+1)/2 odd primes, such that the sum of the three primes in the vertex of every equilateral triangle embedded in the triangular arrange, add up to a prime number.

Example for K=3

05
17-07
11-13-03

05+17+07 = 29
05+11+03 = 19
17+11+13 =
41
17+07+13 = 37
07+13+03 = 23

Example for K=4

19
11-23
13-07-29
31-17-05-03

19+11+23 = 53
19+13+29 = 61
19+31+03 = 53
11+13+07 = 31
11+31+05 = 47
23+07+29 = 59
23+17+03 = 43
13+31+17 = 61
07+17+05 = 29
07+11+23 = 41
29+05+03 = 37
17+13+07 = 37
05+07+29 = 41

Questions:

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

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

3. Do you devise a systematic approach in order to get the solutions asked in 2?

Solution: Q1. Correct solutions were sent by: J. Wharf, J.C. Rosa, J. Arioni, J. L. Pe, J. Heleen, J. vanDelden, Rudolph Knjzek, J. K Andersen & J. Wiesenbauer.

The formulas sent are in a variety of forms (A(k) is the asked quantity of embedded equilateral triangles):

• A(K+1) = A(K) + K(K+1)/2 + floor(K^2/4) ........Wharf
• A(K) = floor((K-1)(K+1)(2K-1)/8)...............Wharf
• A(k) = int[(k(k+2)(2k+1))/8] ............. Pe, Heleen, Andersen
• k=2*m even: sum f(2*m,2*s) s from 1 to m-1:   A(n)=m*(4*m+1)*(m-1)/6
• k=2*m+1 odd: sum(2*m+1,2*s) s from 1 to m:  A(m)=m*(4*m-1)*(m+1)/6.....vanDelden
• K even , K=2*m , N(K)=m*(4*m^2-m-1)/2
K odd , K=2*m-1 , N(K)=m*(m-1)*(4*m-3)/2 ...........J. C. Rosa
• A(K) = 1/4*k^3 - 1/8*K^2 - 1/4*K + x/8 (x=0 when K even x=1 when K odd), J. Arioni
• A(k) := ((k^2 - 1)(2k - 1) - 0^MOD(k, 2))/8 .........Wiesenbauer
Come of them also noticed that this question was also solved as a sequence in the OIS::

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.

or in in the "Book of Numbers" by John H. Conway and Richard K. Guy on p.83.
***

Q2. Solutions for K=5 were sent by Wharf, Rosa, Andersen & Wiesenbauer.

But only Wharf and Andersen discovered the reasons why there are no solutions for K=>6. Both more or less say the same:

"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 potential 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." (Andersen)

Both also found all the solutions for K=4 & K=5

There is a single solution for K=3. There are 104 solutions for K=4. There are 1261 solutions for K=5

***

Q3 becomes irrelevant after Q2.

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