My "probably works" 100, was quite wrong. (And so was everything afterwards)

@Grimy I think you had the same problem as me. It can be fixed with ^((x*)(\2x)*)(?=xx\1$)((?=\2+(x*))\5(?=(\5xx)*$)(?=((x*)(?=\6(\8*$))x)\7*$)\9(?=\2\3*\6$))*\2\2$|^xx?$ (which can likely be golfed)

oh thanks for noticing that

sub-100 should still be possible

Here's another of the same length (unfortunately has \10 and is much slower): ^(((x*)(\3x)*)(?=((?=\3+(x*))\6(?=(\6xx)*$)(?=((x*)(?=\7(\9*$))x)\8*$)\10(?=\4+\6x$))*\3\3$)x\2|x?)x$

Ah, cool. I couldn't think of overlapping the look-aheads like that

Slow 99: ^(((x*)(\3x)*)(?=((?=\3+(x*))(?=\6(?=(\6xx)*$)((x*)(?=\7\9*$)x)\8*$)\3\4*(?=\7\9$))*\3\3$)x\2|x?)x$

Here's my "counting downwards" 100: ^(((x*)\3*)(?=\3\3\2$)x((?=\3+(x*))(?=\5+$)(?=\5((x*)(?=\7+(?=\5$)xx)x)\6+$)\3*(?=\5\6$)x)*x\3|x?)x$

Here's a counting downwards 95: ^((x+)\2+)(?=\2\2\1$)x((?=\2+(x*))(?=\4(?=\4*$)((x*)(?=\4\6*$)x)\5*$)\2+(?=\4\5$)x)*\2xx$|^xx?$

And of course you can get a slow 94 from it

Nice. I see you got rid of the check that \4 > 2 just by changing some of the quantifiers

(I don't really get it)

The 94 won't be slow. It's slow for the counting upwards one one since it needs more reordering

Cool, of course taking the x to the start of the loop makes no difference

That's not where the golf is, though

^(((x+)\3+)(?=\3\3\2$)x((?=\3+(x*))(?=\5+$)(?=((x+)(?=\5\7*$)x)\6*$)\3+(?=\5\7$)x)*\3x|x?)x$ is an equivalent 92

I know. The golf is that subtracting \5 decreases to the quotient by exactly 1