My buddy, my friend. You've showed us the way. We're not there yet, it's true, but we still believe in the VQC, in mandelbrot set as a computer. We are here to learn how to create our own quantum machines, we are here for the truth. I hope only the best for you. The past year has been great. The patterns, triangles and the amount of u's we've looked at is astonishing. I've never been as dedicated to a problem I couldn't understand as this one. Maybe you've fucked up, but then again "to err is human". God still loves you.
If it is a constant and let's say it is 8. Then we should be able to group triangles by their group, right?
We know our squares (x + n)^2 - 1 is divisible by 8 (for odd squares). Hence the 8 families?
We know n is even, because that's the pattern we've been working with, odd x and even n. Thus we know (n - 1)(n - 1) - 1 is divisible by 8.
Our entire equation is this:
(n - 1)(n - 1) - 1 + 2n + 2(n-1)d + f - 1 = (x+n)^2 (or have we been working with another one?)
We know that (n - 1)(n - 1) - 1 is divisible by 8.
This leaves 2n, 2(n - 1)d and f - 1. If we could align d and f - 1 so that both of them are divisible by 8 we would be left with 2n as the only unknown with regards to that.
That means we want to figure out how to do this:
d = 8 * y (for some y)
2d + 1 - e - 1 = 8 * x (for some x)
If we could force this then the only element of the triangle that would NOT be guaranteed to fit evenly would be 2n. If we could do this we could start trying to figure out triangular tilings (if there are any). Like assume (f - 1)/8 is on the outside and the inside is filled with 2(n-1)d/8.
For each 2d/8 we add to the triangle we get a clue about n.
Say we add (f - 1)/8 on the outside and we add one 2d. This means n = 2 (since we've added 1 2d). We then attempt to fill the inside with 1 followed by 2*n (4). If this is a complete triangle we try to validate it.
If it is now, we add 2 more 2d's. Giving our triangle the following units added:
(f - 1)/8 + 3*2d. Meaning n = 4 (and the center is (9 - 1)/8.
Essentially, we'll fill in the triangle from the outside and center, moving to meet in the middle.