Every cell in (0,n) produces a square c, since e=0. That means when a is a square b also has to be a square, so that c becomes (ab)(ab), rather than (ab)(asomethingelse).

In (0,n) where a and b are both squares, n is equal to twice the sequence of squares from 1^2 up. 2 (12), 8 (42), 18 (92), 32 (162), 50 (252), 72 (362), 98 (49*2), etc. This pattern is meant to be related to (e,1) somehow.

c^2 appears in (0,n) where a and b are equal to a and abb, aa and bb, ab and ab, b and abb, and 1 and cc.

In (0,n), where x=2c and n=2a, a=b*c.

In (0,n), if you set x=2c (and therefore with semiprimes t=c+1), each valid n value is divisible by either a or b (or both).

Where a=cc in (0,n):

>n follows that sequence of numbers (twice the squares). For simplicity's sake in the rest of the variables, I'll make a new variable r which is equal to sqrt(n/2). This is just going to be 1, 2, 3, 4, 5, 6, 7 etc throughout the sequence.

>b is equal to (c+2r)(c+2r)

>t is equal to (originalc*r)+1

>d is equal to c*(c+2r)

>x is equal to originalc*2r

>a is obviously going to be originalc*originalc the whole time not changing

The square at (0,1) is the basis of the patterns of all cells on row one.

D[t] - d = a(n-1)

D[t-1] - d = b(n-1)

Long story short with the algebra, and only for odd e (there’ll be a slightly different formula somewhere for even e), D[t] - d = a(n-1) could also be expressed as (4t^2 - f)/2 = a(n-1).

The values for D[t]-d are equal to 4 times the triangular numbers.

If we go to the record where (D[t]-d) is NOT divisible by (n-1) but where A[t] IS divisible by n, then the GCD of D[t]-d and A[t] is equal to a. Also, the X value for that record is equal to the correct D value for the record in this column that has the correct X value for the final record.