The x value of the cell in (1,1) where d is the n value from (0,n)'s a=aa b=bb cell is equal to b-a-1 (the b and a from the solution cell, not this cell).
Example:
a=3, b=177, n from (0,n) a=aa b=bb = 15138
(1,1) d=aabbn = (1,1,87) = {1:1:15138:173:14965:15313}, f=-30276, c=229159045, u=87, i=15139, j=174
x from this cell = 173
173=b-a-1=177-3-1
This seems to apply to all cells.
There are a lot of things you can do with the (1,1) cell. VQC said it's the most important cell in the grid at one point. Specific cells in (0,1) and (1,1) with a[t] added to e/2 or (e-1)/2 (depending on parity) give you an and bn, for example.
The t value of the (1,1) d=aabbn cell for a=1 b=c is equal to the d value of the (f,1) cell where (e,1) and (f,1)’s polite x values add together to give 2d+1 (I’m not sure if this works for all parities in the same way but it worked for multiple cases).
e.g. c=1427
The cells in (e,1) and (f,1) where x+x=2d+1=75
(-17,1,19) = {-17:1:713:37:676:752}, f=1408, c=508352, u=19, i=714, j=38
(58,1,20) = {58:1:789:38:751:829}, f=-1521, c=622579, u=19, i=790, j=39
[1, 1427]
(1,1) d=aabbn = (1,1,713) = {1:1:1016738:1425:1015313:1018165}, f=-2033476, c=1033756160645, u=713, i=1016739, j=1426
d=713, t=713
You said factoring d and e allows for the factoring of c. I've put together a program that outputs every known and unknown (unless I missed something) down the factor tree. I've been looking for something in a given d and e's unknowns that somehow shows up in c's unknowns but there are so many possible combinations of numbers. Could you maybe give a hint as to where we should be looking, or even just what concepts we should focus on in this context?
You might be onto something. I've checked with most c values where d=23 and it appears that for most of them the t value for one of the unknown (1,1) cells appears as a d value in either (e,1) or (f,1). I think a few of them didn't but I forgot which, and I also didn't check what happened when you added the x values together. I've look into it properly and report back later.
Turns out the t value of the (1,1) cell where d is the n value from (0,n)'s a=aa b=bb cell was actually j or (x+n) all along. I don't know if anyone figured that out already. (1,1,j)'s d value is the n value from (0,n)'s a=aa b=bb cell. It also appears that j turns up as a d value in (e,1) or (f,1) sometimes. I'm just looking into it some more.
And the j of that (1,1) cell is 2j. So if we're looking for (x+n)(x+n), we'll find 2(x+n)^2 in (1,1) at (1,1,(x+n)).
I looked at all of the odd c values with a d of 23 (529 to 575). I took the t value of the cell in (1,1) where d is equal to the n in (0,n)’s a=aa b=bb cell, and tried to find it as a d value in either (e,1) or (f,1).
Every time you have a=1 b=c, t from (1,1) shows up as d in (e,1) and (f,1).
Example:
c575 = (46,265,12) = {46:265:23:22:1:575}, the relevant (1,1) cell is (1,1,287), and 287 appears as d in (46,1,12) and (-1,1,12). This works for all a=1 b=c.
For solution cells, it gets a bit weird. t for at least one of the relevant (1,1) cells (so solution j or solution (x+n)) appears as a d value in (e,1) for most of them. I didn't notice anything in particular that could be used to calculate this cell, but it's a possibility.
Some of them (c535, c551, c575, c567) have it turn up in (f,1).
Occasionally (c539, c549), the t value of the cell in (0,1) where a is equal to the n value from (0,n)’s a=aa b=bb cell (which is actually the t value we were already looking for plus 1) appears as a d value in (e-1,1).
Some of them (c543, c545, c565, c573) just don’t turn up. Who knows why.
So while j definitely turns up as a d value in (e,1) quite often, it doesn't seem to follow any stable patterns, considering it just kinda doesn't happen sometimes.