Bit of a dumb miss here but c^2 actually shows up five times in (0,n). The fifth cell is where a=c and b=c, which will appear in (0,0).
From someone's Twitter DMs with you know who:
"Root of d = {0, 2xd, 3xd, 2xd, d, 9xd}. All a and b of ( 0, 2xd, t) are multiple of d for all t.
Yes and that pattern can be used elsewhere."
For the initial 1,c records, some records have t as valid factors of the d[t] diff or the a[t] diff. When these factors exist, the pattern appears to be:
For even n:
d[t] diff / t are even (2,4,6,etc)
a[t] diff / t are sequential (1,2,3,4,etc)
For odd n:
d[t] diff / t are sequential (1,2,3,4,etc)
a[t] diff / t are even (2,4,6,etc)
Since d[t]-d values have the opposite parity (odd or even) of the a[t] values, then that means that for any e, we already know the parity of the d[t] - d values.
if (e is even) (d + n)^2 = (2t)^2 + c
if (e is odd) (d + n)^2 = (2t - 1)^2 + c
if (e is even) (x + n)^2 = (2t)^2
if (e is odd) (x + n)^2 = (2t - 1)^2
if (e is even) n = (floor_sqrt(c + (2 * t)^2)) - (d)
if (e is odd) n = (floor_sqrt(c + (2 * t - 1)^2)) - (d - 1)
Every square is the sum of odd numbers.
4 = 1 + 3
9 = 1 + 3 + 5
16 = 1 + 3 + 5 + 7
25 = 1 + 3 + 5 + 7 + 9
36 = 1 + 3 + 5 + 7 + 9 + 11
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
There is a cell in (0,n) (following this >>7395 sequence of ns) where a=(x+n)(x+n) and b=(d+n)(d+n).
a*b can be represented as the sum of consecutive odd numbers where a is the number of terms and b is the midpoint:
5*29 = 145 = 25 + 27 + 29 + 31 + 33 = 145.
13*43 = 559 = 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 + 51 + 53 + 55