Re:A cave in Romania that was sealed for 5.5M years
by btilly on Thursday May 12, 2016 @05:34PM
(#11686940)
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. :-)
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.
https://en.wikipedia.org/wiki/Primitive_recursive_function
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!