Continuing with the aan(n-1) work from the previous bread >>7584, have found an alternate way to calculate the triangle base and remainders in some cases for more than the first factor record.
Drawing from these VQC hints, though not sure if related and/or relevant.
>Think -1.
>The key is how many squares make up the remainder.
All aan(n-1) calculations for (-f,1) and (e,1) will have at least 1 square multiple of aa for the first factor record, and 2 or more for subsequent factor records. Idea being that if we can understand how those triangles and remainders are calculated, we would be able to jump to another record.
Attached pics for c255025 (101^2 * 25^2) show the first 3 factor records where a[t] is 1, 5, and 25.
The "u calc" column is calculated relative to (n-1) as follows:
if a a[t]: (n-1) - (((a - a[t]) * (n-1))/a) - 1
if a < a[t]: (n-1) + (((a[t] - a) * (n-1))/a) + 1
if a = a[t]: always (n-1)
where "a" is a valid square multiple in aan(n-1), and a[t] is the "a" value from the factor record.
For triangle bases u < (n-1), the a a[t] formula seems to be relatively accurate as reflected in the "u diff" column for various test cases in both (0,n) and (1,n).
When u (n-1), however, something else starts to creep into the calculation as can been seen in the a[t]=25 example. Larger a[t] values will have a larger discrepancy, but not yet sure how these are calculated. An "(x)" in the "u calc" column indicates an incorrect value.
The triangle remainders where a[t] is 1 and 5 can also be calculated in terms of (a-a[t]) as shown in the "rm/(a-a[t])" and "sqrt" columns. Again showing the inverse relationship between the "aa" square multiple and the triangle remainder formula. More discrepancies also appear in calculating these remainders as a[t] increases and for u (n-1) as can be seen in the a[t]=25 example.