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.

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!

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. :-)

Google didn't help me. What's BB(#)?

Busy Beaver

Link: https://en.wikipedia.org/wiki/Busy_beaver#The_busy_beaver_fu...

I think people are downvoting you because they're assuming you're not serious. :(

Fun discussion to be found at: http://www.scottaaronson.com/writings/bignumbers.html