No, because the halting problem requires implementing a function that can compute whether any other function halts. It has to work for any possible function you give it. You can, of course, compute whether some functions halt. Two easy ways to do so are to observe that it halts or observe an infinite loop.

Isn’t the crux of the issue that you can’t “observe” an infinite loop?

No, the crux of the issue is that you can't observe every infinite loop.

But if you know BB(n) for any n then you can just make run a program of length n for BB(n) steps. If after BB(n) steps it hasn't halted you know that the program will loop forever. Since you can do that for any arbitrary program then you could solve the halting problem.

If you recall from predicate logic, these two statements are not the same:

1) For all Turing Machines A, there exists a Turing Machine M such M can compute whether A halts.

2) There exists a Turing Machine M such that for all Turing Machines A, M can compute whether A halts.

Statement 1 is true, statement 2 is false.

You can apply a similar reasoning to BB(n) or your notion of Omega. For any particular choice of n, there is an algorithm that can compute BB(n) and similarly for any arbitrary precision p, there is an algorithm that can compute Omega up to that precision.

None of this is to be taken that there exists an algorithm that can compute BB(n) for all n, or that there exists an algorithm that can compute Omega up to precision p for all p.

But imagine that we could compute BB(n) for any n, even if we need to use a different algorithm for each n. Then, we could build the mapping n -> BB(n). But the existence of this mapping is forbidden by the unsolvability of the Halting Problem, isn't it?

> But imagine that we could compute BB(n) for any n, even if we need to use a different algorithm for each n.

This is itself uncomputable; therefore

> Then, we could build the mapping n -> BB(n).

This mapping is not computable.

Edit as I'm rate-limited:

What I mean is, informally, the incomputability of the algorithm to select the correct algorithm to compute BB(n) is sufficient to make that mapping uncomputable.

However, that does not mean that an algorithm to compute BB(n) for a specific n does not exist. (Trivially, it does; "output BB(n)".)

yes, this is why I said

> the existence of this mapping is forbidden

:)

No you can't conclude this and in fact the more general principle is that being able to prove P(0), P(1), P(2), etc... does not mean that it's also possible to prove P(n) for all n. The property of a formal system for which this does follow is known as Omega completeness, but Peano arithmetic and ZFC more broadly are both Omega-incomplete systems, hence there are propositions for which P(0), P(1), P(2)... can all individually be proven and yet it is not possible to prove P(n) for all n.

Wikipedia touches on this subject with an article on omega-consistent systems, which is a very closely related property:

https://en.wikipedia.org/wiki/%CE%A9-consistent_theory

And this gives a very brief description of Omega completeness:

https://encyclopediaofmath.org/wiki/Omega-completeness