[Solved]: A variant of the busy beaver function

I can’t believe I didn’t see this before — but yes, with an oracle for $L$ you can solve the halting problem. Obviously an oracle for $L$ gives us ‘recursively’ all non-Busy Beaver halting machines, so the question is ‘can we figure out recursively in $L$ what the busy beavers are?’. Define $Sigma_2(n)$ as the count function of the ‘second busiest beaver’; that is, the second-highest attainable score among all halting two-symbol $n$-state TMs. The trick here is that there’s a recursive function $f()$ such that $Sigma(n)leqSigma_2(f(n))$ (it’s almost certain that $f(n)=n+1$ will do the trick, in fact, but that requires knowing that the BB function is strictly increasing) : given a machine $M$ of size $n$ that prints $Sigma(M)$ 1s on its tape and then halts, there’s some $cgt 1$ and two machines each of size $leq cn$ that print exactly $Sigma(M)$ 1s and exactly $Sigma(M)+1$ 1s, respectively, on their tapes — and this holds true for a ‘busy beaver’ machine $M$ even though we don’t know $M$ explicitly. This means that having a bound on the ‘second busy beaver’ function for $f(n)$ gives a bound for the busy beaver function on $n$; but then having this, it’s easy to solve the halting problem for a TM $M$ of size $n$ – if $langle Mranglein L$ then say that $M$ halts; otherwise, find the longest-running machine of size $f(n)$ in $L$ (which can be done recursively since there are only finitely many machines of size $leq f(n)$) and simulate $M$ for as many steps as that machine takes to halt. If $M$ doesn’t halt within that time then $M$ can’t possibly halt.