Have taken this a bit further, and found a way to calculate all triangle portions of the aan(n-1) formula for the initial factor record for any odd n-1 in (0,n).
Pics attached for c21025 (145^2) and c104329 (323^2) show the aan(n-1) derivations for all factor records, and then all square multiples in terms of triangles for the first factor record.
The revised formulas are:
aan(n-1) = aa * 2T(n-1) = aa * 2(T(u) + remainder)
where u = m * x - 1
and remainder = m * x / 2 - m^2
m is just a multiplier, but it turns out to work inversely to each "a" square multiple.
For c21025, as an example, there are 12 possible values that represent the aa square multiple of aan(n-1). 1,2,3,4,6,8,9,12,18,24,36,72.
Using the above formulas, and using m=4 as an example, we can find an "a" factor as:
53742528 = aa * (T(4144 - 1) + (4144/2 - 4^2)) = aa * (T(575) + 272)
aa = 53742528 / 165872
a = 18
And for each possible value of a, a * m will have the same value of 72, hence the inverse relationship between a and m.
Therefore, to find all valid square multiples, we can simply start at the lowest m = 1, find the largest "a" value, and then iterate halfway upwards.
This isn't a solution to our general factoring problem, as it doesn't yet appear to provide a way to jump to any other factor record for our starting c.
It does, however, show that there is a way to calculate both the triangle base and remainders to enable extracting factors from larger products.