There are many, many crumbs that all seem like they refer to entirely different concepts, but they're meant to be used together to find the solution. I thought it would be useful to get some of these crumbs together and figure out some way in which all of them can coexist.
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It doesn't seem as though anyone has been working on that seemingly quite major crumb about the root of (D)avid and what remains >>6629 here. When I suggested it meant using d as a new c value, VQC responded >>6652 here, implying it was important.
If we know there's a pattern in (e,1) and (-f,1) within which n or n-1 is a factor of a or b, and we also know that a prime number only has 1 and itself as factors, if we apply this b(n-1) etc pattern to a prime number c, in the a(n-1) pattern, we'll be able to directly calculate n-1 (because we know a and b). If we know n-1, we know 2d(n-1) and nn. Another useful thing to note is that if e=0 when we find the square root of c, we can definitely very easily calculate everything, since d=a=b for at least one set of the factors of a square. VQC has said in the past that the e=0 cells 'rule' the others (or something like that).
VQC said here >>6185
>The two sets of equations take the problem and simplify it. Together they handle the "lock and key" nature of the problem/solution, particularly when c is divisible by 1, c, and two other prime numbers, such as in RSA.
This obviously implies that we can do a calculation on the (1,c) cell (the key) that "unlocks" or calculates a and b (the lock).
Also, according to >>5887 this, we're meant to use (f-1)/8 to find the base of the triangles in an (x+n) square larger than (n-1)/8 but smaller than (x+n)/8. There are choices (referring to the fact that we're picking a number in the gap between x+n and n, meaning the number of "choices" is controlled by how many possible x values could potentially apply to a given c) and these choices reduce 'exponentially'.
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So, putting these four points together, I'm thinking there's something about the b(n-1) pattern that applies to all of the (1,c) cells for every c value in the order of c=c, c=d, c=d's d, c=d's d's d, etc that allows us to calculate (n-1). This aligns with these other crumbs. We're finding the root of d and what remains. Doing so reduces things exponentially, since we're constantly taking the square root of a number (exponential including the maximum value of x - I had a method of calculating the range of possible x values for a given c somewhere and if anyone ends up reading and responding to this post and wants me to find it I can, although you'll see it in the examples I'll post below as the x values in each (1,c), which, of course, decrease exponentially). If you go through this pattern far enough, you will definitely eventually end up on a cell that allows you to very easily calculate endxab (e=0, or if c is prime). Once we find one of those cells, we can find the b(n-1) pattern, factor out b since b=c if c is prime, and use that to calculate our (x+n) square for that specific cell. I'm thinking because this is meant to be recursive, there's something about one solved cell that allows us to calculate the (n-1) of the solved cell that we found this cell from. That's the point of recursive algorithms, obviously: you have a thing you can't solve until you solve the problem with a smaller value, which also needs the same problem solved on a smaller value to be solved, etc, until you find a number small enough that you can solve it. So all that's missing is a link between these c=d cells. I haven't really looked at them yet. All I've really done so far is try to navigate this giant dumping of seemingly unrelated information. I'm trying to figure out how these crumbs can all be true at the same time, and this seems to be the only way I can figure out. I'll leave some example cells in the next post.
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I wish I could follow this lead more right now but yet again I find myself only having the time to work on this when it's far too late at night to maintain attention.