If you follow the (-f+2(n-1), (n-1)) sequence to zero, it continues into the positive space with t+=1. x and n stay the same. a, b and d continue only increasing by 1. In the negative space, f grows by 2(n-1) and e grows by 2n. Once it gets into the positive space, t increases by 1 (while x and n stay the same), and e begins growing by 2(n-1) while f begins growing by 2(n-2).
Starting from {18:27:215:84:131:353} t=43 f=-413
{-413:26:216:85:131:353} t=43 e=18
{-361:26:217:85:132:354} t=43 e=72
{-309:26:218:85:133:355} t=43 e=126
{-257:26:219:85:134:356} t=43 e=180
{-205:26:220:85:135:357} t=43 e=234
{-153:26:221:85:136:358} t=43 e=288
{-101:26:222:85:137:359} t=43 e=342
{-49:26:223:85:138:360} t=43 e=396
{3:26:224:85:139:361} t=44 f=-446
{55:26:225:85:140:362} t=44 f=-396
{107:26:226:85:141:363} t=44 f=-346
{159:26:227:85:142:364} t=44 f=-296
{211:26:228:85:143:365} t=44 f=-246
{263:26:229:85:144:366} t=44 f=-196
{315:26:230:85:145:367} t=44 f=-146
{367:26:231:85:146:368} t=44 f=-96
{419:26:232:85:147:369} t=44 f=-46
I'm not sure how much sense this makes, but theoretically, we could create a big chain of connected cells from the (e,1) cell's negative counterpart, transformed into the positive space with -f+2(n-1) to where t+=1, then using the new t value , find another cell at n+=1, transform that into the positive space, etc. That's a completely spitballed idea, but the number of steps this would take would be the number of times you have to add 1 to the d in the (e,1) cell to reach the correct d.