## Monday, 16 July 2012

### Plane substitutions.

OK, to help explain the logic of the attempted twin prime proof, here is an example of how the plane substitutions work. Here's the starting point of the 4n x 9n grid (in other words how it looks in the prime number hypercuboid we started from).
 4n 4n+1 4n+2 4n+3 9n 36 9 18 27 9n+1 28 1 10 19 9n+2 20 29 2 11 9n+3 12 21 30 3 9n+4 4 13 22 31 9n+5 32 5 14 23 9n+6 24 33 6 15 9n+7 16 25 34 7 9n+8 8 17 26 35

Now I am going to duplicate the 9n +1 and 9n+7 plane. The 9n+2 and 9n+8 plane are coloured black (as well as the 4n+2 column). The 9n+5 is coloured yellow.
 4n 4n+1 4n+2 4n+3 9n 36 9 18 27 9n+1 28 1 10 19 9n+1 28 1 10 19 9n+2 20 29 2 11 9n+3 12 21 30 3 9n+4 4 13 22 31 9n+5 32 5 14 23 9n+6 24 33 6 15 9n+7 16 25 34 7 9n+7 16 25 34 7 9n+8 8 17 26 35

If we omit the black planes at 9n+2 and 9n+8 we have two horizontal black planes as in the original diagram (in the proof I talk about merging planes but I think it makes more sense to simply discard them.) The grid contains some repetitions and some numbers in the number chain are missing (but we know they aren't twin primes).

 9n 36 9 18 27 9n+1 28 1 10 19 9n+1 28 1 10 19 9n+3 12 21 30 3 9n+4 4 13 22 31 9n+5 32 5 14 23 9n+6 24 33 6 15 9n+7 16 25 34 7 9n+7 16 25 34 7

Next we need to extend this up to the 25n plane. The first few slices look like this (before they are sorted by their residue mod 25):

 36 9 18 27 28 1 10 19 28 1 10 19 12 21 30 3 4 13 22 31 32 5 14 23 24 33 6 15 16 25 34 7 16 25 34 7

 72 45 54 63 64 37 46 55 64 37 46 55 48 57 66 39 40 49 58 67 68 41 50 59 60 69 42 51 52 61 70 43 52 61 70 43

 108 81 90 99 100 73 82 91 100 73 82 91 84 93 102 75 76 85 94 103 104 77 86 95 96 105 78 87 88 97 106 79 88 97 106 79

To make it easier to see what the next stage looks like, I'm going to leave out the black planes and the red plane (in which there are no primes other than 3, which is an exception - it isn't a twin because it is only on one red plane. )

 4n+1 4n+3 9n+1 1 19 9n+1 1 19 9n+4 13 31 9n+4 13 31 9n+5 5 23 9n+7 25 7 9n+7 25 7

Here, the numbers in the green and yellow planes up to 900 have been rearranged by their residue mod 25 (picture this as 25 slices of a cuboid):

25n+2, 25n+12, 25n+17 and 25n+22 are black planes as they can't contain the higher of two twin primes. 25n+5 is red, 25n+7 is yellow.

Next, we are going to substitute:

25n+3 for 25n+2
25n+13 for 25n+12
25n+18 for 25n+17
25n+ 23 for 25n+22

I've highlighted 5 - this is the only number on this grid that won't appear on any more red or yellow planes. There are two instances of 7 (one of which is highlighted) - these will both be on the 49+7 red plane in the next dimension and then no other red planes.

The first prime that is a non-twin is 11 - this has already been removed from the pattern when we coloured the 9n+2 plane black - we chose to repeat the 9n+1 plane, so you could see the second instance of 19 as a replacement for 11.

Every number that is only in one red plane when this process is reiterated in infinite dimensions will also be on only one yellow plane, so we can treat the yellow plane as being the same as a green plane.*

Given this, the intersection of green, black and red planes in the grid above is the same topological pattern as we would find if we were looking only for primes, but we have multiple repetitions of individual numbers within the pattern.

Also, I talk about rearranging the numbers according to modulo residue above - it would maybe be better to visualise this as a hypercuboid that from the start contains duplications of some planes that aren't either pn or pn+2  and doesn't include any pn+2 planes other than the (p^2)n+p+2 planes.

Hope that all makes sense?

In theory we need to make these plane substitutions in every dimension - I do have one or two thoughts on why that might be problematic, I'll come back to talk about that soon.

*See the algebra in part 7 of the original proof.