Gap between $BB(n)$ and "second largest" $BB(n)$ If $HT(n)$ is the set of halting times of $n$-state Turing machines on a binary alphabet with empty initial tape, then $BB(n) = \max HT(n)$. 

What can we say about the second largest number in $HT(n)$?  Call this $BB_2(n)$. 

$BB_2(n)$ is trivially uncomputable, since it lets one compute $BB(n)$: just wait for one more machine to halt.  Naively, I would expect the gap $BB(n) - BB_2(n)$ to be "busy beaver-like", growing faster than any computable function.  Is this provable?
The number of states is just a notion of complexity of the description of computable functions in a model, you can pick any model of computation and any encoding of them as binary strings and then take the length as n and define BB(n) based on that and all the interesting results about BB(n) would still be true, there is boring special about TM model and number of states.
There is nothing that prevents them from picking any modified model of TMs. Generally the questions which are not invariant under such changes of representation of TMs are not about computability or TMs but about the particular representation (like BB(n) mod 2, etc. ) and unless there is some particular reason for them being interesting they don't worth pursuing imho. They are nice puzzles but not of much value. 
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Note that "BB(n) is not computable" is invariant under change of representations of TMs.
So is this question invariant under change of representation of computable functions? The answer I think is no.


i. Consider a representation where we have two special states 0 and 1 and either 0 is initial and just can transition to 1 or 0 is unreachable and 1 is initial. In this encoding the difference is 1. 

ii. Consider another representation where we have a UTM plus a part that writes n bits on tape before transitioning to UTM. So the question becomes max f(x) - 2ndmax f(x) where maxes are over n bits strings and where f is an arbitrary computable function. We only need to find a computable function where this is not computable. I haven't thought about it much but my gut says there is such a computable function.