In Coq-BB5 machines are not run for 100M steps but directly thrown in the pipeline of deciders. Most halting machines are detected by Loops using low parameters (max 4,100 steps) and only 183 machines are simulated up to 47M steps to deduce halting.
In legacy bbchallenge's seed database, machines were run for exactly 47,176,870 steps, and we were left with about 80M nonhalting candidates. Discrepancy comes from Coq-BB5 not excluding 0RB and other minor factors.
Also, "There are 181 million ish essentially different Turing Machines with 5 states", it is important to note that we end up with this number only after the proof is completed, knowing this number is as hard as solving BB(5).
Why is knowing that there are 181 M 5-state machines difficult? Is it not just (read*write*move*(state+halt))^state? About 48^5, modulo "essentially different".
Number of 5-state TMs is 21^10 = 16,679,880,978,201; coming from (1 + writemovestate)^2*state; difference with your formula is that in our model, halting is encoded using undefined transition.
"essentially different" is not a statically-checked property. It is discovered by the enumeration algorithm (Tree Normal Form, see Section 3 of https://arxiv.org/pdf/2509.12337); in particular, for each machine, the algorithm needs to know it if will ever reach an undefined transition because if it does it will visit the children of that machine (i.e. all ways to set that reached undefined transition).
Knowing if an undefined transition will be ever reached is not computable, hence knowing the number of enumerated machines is not computable in general and is as hard as solving BB(5).
I guess the meaning of "essentially different" essentially depends on the strength of the mathematical theory that you used to classify them!
When I first heard it I thought about using some kind of similar symmetry arguments (e.g. swapping left-move and right-move). Maybe there are also more elaborate symmetry arguments of some kind.
Isn't it fair to say that there is no single objective definition of what differences between machines are "essential" here? If you have a stronger theory and stronger tools, you can draw more distinctions between TMs; with a weaker theory and weaker tools, you can draw fewer distinctions.
By analogy, suppose you were looking at groups. As you get a more sophisticated group theory you can understand more reasons or ways that two groups are "the same". I guess there is a natural limit of group isomorphism, but perhaps there are still other things where group structure or behavior is "the same" in some interesting or important ways even between non-isomorphic groups?
Turing machine program states are conventionally represented with letters: A, B, C, etc. The starting state is A.
Now suppose you are running a Turing machine program from the beginning. The only state it has visited so far is state A. It runs until it reaches a state that has not been visited yet. What state is it? B? C? D? According to "tree normal form", the name for that next state just is the earliest unused state name, which in this case is B.
Visited states so far are A and B. Run until an unvisited state is reached again. What is its name? C? D? E? Tree normal form says that the state will be called C. And the next newly visited state will be D, etc. In general, the canonical form for a Turing machine program will be the one that puts initial state visits in alphabetical order. (This concept also applies to the multi-color case.)
It's not possible to tell of an arbitrary TM program whether or not that program is in tree normal form. But this proof doesn't look at arbitrary programs. It generates candidate programs by tracing out the normal tree starting from the root, thereby bypassing non-normal programs altogether.
That is what "essentially different" means here.
If you count arbitrary transition tables, then you get a count of 63403380965376 [1].
If you count transition tables in which states are reachable and canonically ordered, then you get a count of 632700 * 4^10 = 663434035200 [2]. These machines can behave differently on arbitrary tape contents.
TNF further reduces these counts by examining machine behaviour when starting on an empty tape.
[1] https://oeis.org/A052200
[2] https://oeis.org/A107668
Just to directly address this - the key thing wrong in this formula is that it's ^2*state, not ^state. The state transition table operates both on the current state and the input you read from the tape, so for a binary turing machine with 5 states, you have a 10-entry transition table.
Ah, yes, of course the read bit should move into the exponent, since it's an input, not an output of the function. But the key point I was making is that there exists a formula. (I don't really care what the formula is.) The part I was not understanding was the complexity of "essentially different".
In this context, two syntactically-different TMs are considered "essentially the same" if all reachable states are the same up to reordering their labels (except for the fixed starting label A) and globally swapping the L/R directions.
The problem is knowing how many states are actually reachable, and how many are dead code. This is impossible to decide in general thanks to Rice's theorem and whatnot. In this case, it involves deciding all 4-state machines.
My intuition is that this would include every 4 state (and fewer) machine that has a fifth state that is never used. For every different configuration of that fifth state I would consider the machine to be essentially the same.