[UncombedCoconut]

Mostly no: we did find some non-halting TMs that required new proofs, but none of those had the flavor of new math, per se. Indeed, we found that all but 30 of them could be proved by finite automata methods, meaning the TM's state/tape at any step could be reduced to one of finitely many states and we'd still know all we needed to know about future steps. I would argue that such a non-halting proof can't have much mathematical content. (Maybe a bit, in about the same way that an integer equation is sometimes proved unsolvable by considering it modulo n and checking every case.) Also, I learned some math I wasn't personally familiar with from the analysis of a particular machine: https://www.sligocki.com/2023/03/14/skelet-10.html (Zeckendorf's Theorem).

[Sniffnoy]

Tangential, but fun fact about Zeckendorf: In addition to Zeckendorf representation, there's also dual Zeckendorf (sometimes also called lazy Fibonacci), where instead of requiring no two consecutive ones, you require no two consecutive zeroes. (Not counting the implicit zeroes at the big end, of course.) It was surprising to me that this also works, but it does!

Actually, just as you can do bijective base-b [0], you can also do bijective Zeckendorf (using 1 and 2 with no two consecutive 1s). Although, as happens with bijective binary, bijective Zeckendorf is closely tied to ordinary Zeckendorf, so it doesn't offer much new. But bijective dual Zeckendorf doesn't work -- lots of numbers can't be represented!

One more fun fact about Zeckendorf and dual Zeckendorf: Write n>0 in Zeckendorf, and count how many zeroes it ends in. This will be even if the dual Zeckendorf representation of n ends in a 1, and odd if it ends in a 0. Similarly, if you write n in dual Zeckendorf and count how many 1s it ends in, this will be even if the (ordinary) Zeckendorf representation ends in a 0 and odd if it ends in a 1.

[0] https://en.wikipedia.org/wiki/Bijective_numeration