Thinking this through again:
>As stated at the start, the square root of (D)avid and what remains are the key to unsealing the The End, the grid.
I'm thinking it isn't that we treat d as c recursively (even though VQC's reply to my post saying that implied it was), but maybe it's literally as he said, and it the square root of d and the difference between d and its square root.
I tried seeing if this stuff PMA has been posting examples of with the first three times a[t]=c shows any relationship with these root(d) and remainder values but it's quite hard actually finding examples without just making a brute force algorithm. I do have part of another example if anyone's interested.
7*29=203
d=14, root(d)=3, 14-3=11 (before you get any ideas, I realize this is equal to e+n but this is definitely not the case universally)
The true cells for this are {7:4:14:7:7:29:4} -22} and {-22:3:15:8:7:29:4} 7}
-f — a[t] = c
First: {-22:63:363:160:203:649} t=80 e=703
Second: {-22:87:391:188:203:753} t=94 e=759
Third: {-22:117:421:218:203:873} t=109 e=819
Fourth: {-22:149:449:246:203:993} t=123 e=875
e — a[t] = c
First: {7:88:392:189:203:757} t=95 f=-778
Second: {7:116:420:217:203:869} t=109 f= -834
Third: I couldn't find
I have no idea what any of this has to do with 3 or 11. Possibly coincidentally, x[t] = c at {7:64:525:203:322:856} t=102 f=-1044, which is the cell in the sequence that exists in the negative space but was missed in the positive space. I also did this with 13*43=559 and found x[t]=c at (30,1) when the first cell where a[t]=c in the negative space was (-17,4), so I don't know if this is actually relevant or a pattern.