But does "a computable function that grows faster than BB(N)" actually mean that f(N) is always greater than BB(N), or could it be taken to simply mean that f(N

But does "a computable function that grows faster than BB(N)" actually mean that f(N) is always greater than BB(N), or could it be taken to simply mean that f(N) > BB(N) for a proportion of N that trends towards 1? In other words, could you have a function that almost always upper bounds BB(N), except for a tiny (but still technically infinite) proportion of exceptions? Such a function would be larger than BB in a meaningful sense, but if the sequence of exceptions is not computable, it wouldn't be possible to solve the halting problem with it. There may be a problem with this concept, but it doesn't appear obvious to me.