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If the past few parts have given us anything important to discard form the original two equations, it is the plus case for the D-set. All weird exceptions have arisen from this algorithm, so let us, for a second, discard it and look at the implications of only having the two equations: Next, for a kind of common ground to derive some kind of relationship between the two equations, we can let R_x = R_y for the first iteration. I will adopt the convention of the x for the D-set and y for the T-set, with a number following showing how many iterations have taken place. To make it clearer to the reader, for the first iteration, I will make the two R's equal. This is demonstrated here: In fact, we can adopt a new set of equations which will get the same idea across, but be easier to work with: And as long as the original E is still divisible by 4, we should be getting the same results. Like I said, make both initial R_x and R_y equal: This next part will be a bit messy, but I will list a few examples I did and we can find any relationships between T and D:
A second example:
And a final example:
If you play with the relationship between the respective D and T iterations, you will notice that it appears a pattern emerges: This can be proven simply enough: If you take other iterations, it seems like the only things generated are prime numbers, related to their respective R_xn and R_yn numbers. (Look at example two, and apply the same idea with R_x2, R_y2, etc. if you want to see those interesting results.) And now that we can put all of the cases where things get weird in the plus case of the D-set, I can implement the rules found in post 2 and post 6 before. I will also try to set up something like JF's R-circles from post 3. We can actually determine a good amount of information from these two things, with the minus case in the D-set, and where R_x1 and R_y2 are equal. I apologize for the bad quality of drawing, but I have grouped the R_x's in blue, the R_y's in yellow, and I will use a green line to show a possible line of equality. For instance, we know that no R_x1 = R_x2, and no R_y1 = R_2, so they were colored red. The R_x1 = R_y1 is green since we have started with the two values being equal. And here we arrive at one of our first controversies: can an R_xn = R_yn where n is not 1? Some of my progress on this: It is a long way of saying "I do not have an answer yet". The truth is, I believe that, because there is no contradiction in assuming some other R_yn = R_xn, we can probably assume that some exceptions exist. So let us assume out last graph and R_x2 = R_y2, and look at the implications: It is also already implied, but this is a sort of case in point for when more vertices are added. For another example: If R_x3 = R_y3, then it is implied by what we already know from the posts before that R_2y != R_3x and R_2x != R_3y. I will lay this out logically for clarity:
But what about if an R_xn+1 = R_yn, or an R_yn+1 = R_xn? Well, take a look at the graph now: Now, R_x3 and R_y3 cannot be equal, since it would lead to a contradiction (R_y3 would then equal R_y2). Second, you could not have an R_x2 = R_y2, since you would lead to the same contradiction on the other side. Finally, big picture, there would never be a loop back to R_x1, since then you would need to say R_y1 = R_y2, which is also impossible. This leads to a very important theorem when looking at these relationships between R_x's and R_y's, which could be very important to the proof: You could swap x for y, it works vice versa. But if there is a way to show that a loop can never occur, and since the number or R_x's and R_y's must be finite, then there must be a point where you get two primes adding up to 2E.
There is still much work to be done, - AF
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