@MatthewBolan Wow, well done! I guess I stand corrected. It would be nice if you have the time to write that up as an answer on my relevant post, ideally explaining something about how you found it. I know I'm curious.

This seems Turing-complete to me...? If that's the case, and this will gradually enumerate every computable function, then $B$'s growth could have no computable bound. (I'm guessing I am missing something, or somebody else would have raised this objection.)

These are the same thing. $S(SS)SSSS \to SSS(SS)SS$ after its first S-combinator evaluation.