That's a rather narrow definition of comparability. If nothing else said, one would understand the term with regards to the relation of any partially ordered set – not only the one implicitly defined on that page.
I define a non-word "uncomparable" to mean something specific and rather arbitrary, but I do not intend it to be the antonym of the real word "comparable". I'll try to clarify this on the webpage you linked to. - Robert Munafo
I am certain it is greater. BB(n) grows super-exponentially.
In fact I would be willing to bet serious money that BB(n+1)/BB(n) is greater than BB(n) if 3 < n.
(This is, of course, assuming that one assumes that BB(n) is well-defined. That is an interesting point of philosophy given the existence of Turing machines which can't be proven to not halt.)
> In fact I would be willing to bet serious money that BB(n+1)/BB(n) is greater than BB(n) if 3 < n.
BB(n) grows faster than any computable function. In order for BB(n+1)/BB(n) > BB(n) to hold, BB(n) merely has to grow faster than a sequence whose new terms are obtained by repeated squaring (like k^2ⁿ). That's computable, indeed primitive recursive, so BB(n) definitely grows dramatically faster than it.
Edit: another way of looking at this is that the Ackermann function grows unbelievably faster than functions that easily satisfy the property you describe, and the Busy Beaver function grows unbelievably faster than the Ackermann function. Somehow putting it this way feels like an understatement, though!
It is unfortunately not that easy. Consider the following function, if n is even then f(n) = BB(n/2), else f(n) = BB(2n). Then f(n) grows faster than any computable function, but still if n is odd, then f(n+1) < f(n).
However you have encapsulated the reason why I would be confident of this result. :-)
