Problems & Puzzles:
Collection 20th
Coll.20th-007.
Symmetrical (Dihedral) prime tiling of the plane.
On March 14 2018, Jan
van Delden wrote:
Consider the following hexagon:

In the center-hexagon (cyan) the consecutive primes 7 ..
23 are given such that all sums of 3 neighbouring tiles add up to a
prime number.
Here these sums have the additional property that they form 6
consecutive prime numbers, from 31=7+11+13 until 59=23+17+19.
In new triangles primes may be added if they form prime
numbers, when added to existing triangles using similar constraints.
In the large hexahedron (yellow-blue-cyan) strips of the
following form can be found (and their rotations):

In the example above only type A is used.
Let’s extend this procedure to the star (blue-cyan).
For new primes in the blue triangles we use all strips op
type A and type B.
In order to compute p the following constraints are
imposed (using type A):
·
p+11+7 is prime
·
p+11+13 is prime
But p and q are situated in a strip of type B, so the
next constraint is also imposed:
·
p+11+13+23+q is prime
Similarly for the other blue-triangles (and all new
strips).
So a total of 6x2 constraints of type A and 6 constraints
of type B are added.
The result is that for the star-shape (cyan-blue) all constraints of
type A and B are imposed.
We call a solution minimal if the sum of the primes
involved is minimal. Please note that in these solutions might not be
unique.
Q1: The given solution is not minimal. Find the
minimal solution.
Q2: Find a minimal solution for the star (blue-cyan).
Is there a solution using consecutive primes?
Q3: Find a minimal solution for the large hexagon
(yellow-blue-cyan).
Is there a solution using consecutive primes?
Try to impose type A,B and C constraints if possible.
Q4: Is it possible to find solutions to Q2/Q3
where all resulting prime sums are consecutive?
Contributions came from Emmanuel Vantieghem,
***
Emmanuel wrote on July 13, 2018:
Here are my results about the puzzle Coll.20th-007.
Q1. This is a minimal solution

Q2. This is a minimal solution : all primes from 5 to
43 are in use

Q3. This might be minimal, but I have
serious doubt.
It would take too much time to work
through all possibilities.

***
On July 17, 2008, Jan van Delden wrote:
Emmanuel’s solutions to Q1 en Q2 are fine.
Q2:
A solution to Q2 with 24 distinct prime sums (sum=348 ,
minimal given 24 distinct prime sums).

Q3:
Emmanuel’s solution is fine, however it is not minimal.
I found (sum = 1198):

It
uses a subset of 26 consecutive primes starting
with 5, 71 and 97 are not present.
Prime
number 2 and 3 can’t be part of the solution
(simple modular proof).
The
sum of the first 24 primes starting with 5
equals 1156.
So there is little slack to improve upon this
solution.
The sum of the first 24 primes starting with 7
equals 1254, so 5 should be present in the
minimal solution.
If one starts with 5 and omits 7, the sum of the
first 24 primes equals: 1252.
If we repeat this analysis one can show that the
minimal solution should at least contain the
primes 5..61.
In the
displayed figures the colors indicate the way
the algorithm works.
***
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