Determination of the fifth Busy Beaver value

Not quite, I think this is the relevant part of the paper:

> Structure of the proof. The proof of our main result, Theorem 1.1, is given in Section 6. The structure of the proof is as follows: machines are enumerated arborescently in Tree Normal Form (TNF) [9] – which drastically reduces the search space’s size: from 16,679,880,978,201 5-state machines to “only” 181,385,789; see Section 3. Each enumerated machine is fed through a pipeline of proof techniques, mostly consisting of deciders, which are algorithms trying to decide whether the machine halts or not. Because of the uncomputability of the halting problem, there is no universal decider and all the craft resides in creating deciders able to decide large families of machines in reasonable time. Almost all of our deciders are instances of an abstract interpretation framework that we call Closed Tape Language (CTL), which consists in approximating the set of configurations visited by a Turing machine with a more convenient superset, one that contains no halting configurations and is closed under Turing machine transitions (see Section 4.2). The S(5) pipeline is given in Table 3 – see Table 4 for S(2,4). All the deciders in this work were crafted by The bbchallenge Collaboration; see Section 4. In the case of 5-state machines, 13 Sporadic Machines were not solved by deciders and required individual proofs of nonhalting, see Section 5.

So, they figured out how to massively reduce the search space, wrote some generic deciders that were able to prove whether large amounts of the remaining search spaces would halt or not, and then had to manually solve the remaining 13 machines that the generic deciders couldn't reason about.

You can not rely on brute force alone to compute these numbers. They are uncomputable.

I think you have to exhaustively check each 5-state TM, but then for each one brute force will only help a bit - brute force can't tell you that a TM will run forever without stopping?