Proof of Confinement

Let n be alone in the field of triangles and let it be neither divisible by 2 or 3 (such as 5) and let the even number preceding n also not divisible by 3 (such as 4). n may or may not be prime.

Consider n in figure below. As n has questionable primality, we consider left and right spin as we move to the next cell. To the left and right of n’s cell we now have n+1. Given n, and the prior number, are not divisible by 3, n+1 must be both even and divisible by three, so spin remains constant. n+2, is odd and not divisible by three, so possibly prime, so again we must assess both two paths, for n+3. n+3, which must be even, keeps spin, as does n+4, which must be divisible by 3. n+5 is even, so same spin to n+6. From the above we know n+6 is odd (because one greater than the even n+5), is not divisible by 3 (because two greater than n+4, divisible by 4), and may be prime, that is, n+6 meets the definition of n. Figure 4(b) shows the hexagon with all possible positions within the t-hexagon have been considered and three facts become apparent:

  1. Only “n” and “n+2” may host prime numbers, because all other cells are even or three divisible or both.
  2. No n or n+2 cell borders the edge of the hexagon, so spirals near the edge continue past and return to the center of the hexagon.
  3. No escape is possible

QED

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