Something I didn't see the anon who wrote that second post mention is that, of course, BigN-n appears in (f-2(n-1),1) too. The a[t] values there represent the BigN-n values for (f,n-1). If we take t for BigN-n in (e-2n,1) and (f-2(n-1),1) as a center point, pic related shows the differences between the a[t] values as you move up or down from that point (for all c as far as I can tell). You could directly calculate if you knew these n values but, I mean, you'd already know the n values.
>The smooth value for a=8, b=38 is equal to 7*37/2 (129). This also implies that at (e, 1, d/2) (even) and (e, 1, (d+1)/2) (odd) will be the BigN * c transformation.
I wonder if there's anything to find taking this concept and working backwards from the (e,1) a[t] = c * shadowN cell or any of the other cells in (e,1) whose a[t] values we've looked at.