So I suppose this is why e and n are the axes, rather than any of the other two variables. They obviously weren't picked arbitrarily.
At risk of pointing out things we already know again, I'll begin posting any diagonal patterns I notice while the rest of you are still asleep (or until I go to sleep).
First one is the relationship between d and f when moving diagonally. When you increase e and n by 1, at the same t, the increase in f is equal to 2*(the increase in d)+1.
Examples:
(19,14,41) = {19:14:316:81:235:425} f=-614
(20,15,41) = {20:15:294:80:214:404} f=-569
d decreases by 22, f increases by 45 (22*2 + 1).
(19,14,6) = {19:14:16:11:5:55} f = -14
(20,15,6) = {20:15:14:10:4:54} f = -9
d decreases by 2, f increases by 5 (2*2 + 1).
(60,40,56) = {60:40:262:110:152:452} f = -465
(61,41,56) = {61:41:262:111:151:455} f = -464
d stays the same, f increases by 1 (0*2 + 1).
When you decrease e by 1 but increase n by 1, the increase in f is equal to 2d-1.
Examples:
(83,42,43) = {83:42:172:85:87:341} f = -262
(82,43,43) = {82:43:167:84:83:337} f = -253
d decreases by 5, f increases by 9 (2*5 – 1).
(75,34,93) = {76:34:683:184:499:935} f = -1291
(75,35,93) = {75:35:675:185:490:930} f = -1276
d decreases by 8, f increases by 15 (2*8 – 1).
(61,23,40) = {61:23:216:79:137:341} f = -372
(60,24,40) = {60:24:206:78:128:332} f = -353
d decreases by 10, f increases by 19 (2*10 – 1).
Hang on a minute. The x and t values in the an and a(n-1) cells are the same as the x and t values in the prime solution. VQC has said a few times that we're meant to find the x or t value of the an cell, which implies that we'd then factor an to find the solution. If we're meant to find that cell based on its x or t value, we can completely bypass it and use the x or t value to find the prime solution without needing to factor an. That renders row 1 meaningless, if x or t are the values we're looking for. So that can't be right.