It is unfortunately not that easy.  Consider the following function, if n is even then f(n) = BB(n&#x2F;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) &lt; f(n).<p>However you have encapsulated the reason why I would be confident of this result. :-)