Comment Comment by gizmo686 Given a constanct C, you can construct a constant sequence f(n) = C.Since f(n) is trivially computable, we know that BB(n) grows faster than f(n). It follows that there must be some n such that BB(n) > C. parent gizmo686 18 days ago
Since f(n) is trivially computable, we know that BB(n) grows faster than f(n). It follows that there must be some n such that BB(n) > C.