I found an interesting property of the row 1 cells. Each one will have its own e and f values unique from the e,f pair used to find them. The difference between these e and f values is equal to either a or b in my test cases.

c559, 13*43, (30,5,6) = {30:5:23:10:13:43}

(30:1:6) = {30:1:75:10:65:87} f=-121, c=5655

(-17:1:6) = {-17:1:63:11:52:76} e=108, c=9607

e-|f| = 13, which is a

c203, 7*29, (7,4,4) = (7,4,14,7,7,29)

(-22:1:5) = {-22:1:29:8:21:39} e=35, c=819

(7:1:4) = {7:1:35:7:28:44} f=-64, c=1232

|f|-e = 29, which is b

>>7480

Oh hey, another property of this unique pair of e and f values. You can represent a given c with its e and f values, right? Well, this unique pair of e and f values in row 1 for the an and a(n-1) cells also represent valid cs themselves. These new c values are actually divisible by a or b (depending on which the e-f/f-e gap is divisible by) and they're divisible by BigN-n. Now that is interesting.

c559, 13*43, (30,5,6) = {30:5:23:10:13:43}

(30:1:6) = {30:1:75:10:65:87} f=-121, c=5655

(-17:1:6) = {-17:1:63:11:52:76} e=108, c=9607

e-|f| = 13, which is a

an is divisible by e’-|f’’|

Do these e and f values produce another c?

They do: 13104, which is a(BigN-n)2*2

c203, 7*29, (7,4,4) = (7,4,14,7,7,29)

(-22:1:5) = {-22:1:29:8:21:39} e=35, c=819

(7:1:4) = {7:1:35:7:28:44} f=-64, c=1232

|f|-e = 29, which is b

bn is divisible by |f’|-e’’

Do these e and f values produce another c?

Yes, 2436, which is equal to b*(BigN-n)