Problems & Puzzles:
Collection 20th
Coll.20th007.
Symmetrical (Dihedral) prime tiling of the plane.
On March 14 2018, Jan
van Delden wrote:
Consider the following hexagon:
In the centerhexagon (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 (yellowbluecyan) 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 (bluecyan).
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 bluetriangles (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 starshape (cyanblue) 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 (bluecyan).
Is there a solution using consecutive primes?
Q3: Find a minimal solution for the large hexagon
(yellowbluecyan).
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.20th007.
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.
***
