&gt; In fact I would be willing to bet serious money that BB(n+1)&#x2F;BB(n) is greater than BB(n) if 3 &lt; n.<p>BB(n) grows faster than any computable function. In order for BB(n+1)&#x2F;BB(n) &gt; 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&#x27;s computable, indeed primitive recursive, so BB(n) definitely grows dramatically faster than it.<p><a href="https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;Primitive_recursive_function" rel="nofollow">https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;Primitive_recursive_function</a><p>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!