Is the inverse ackermann function the slowest growing function that goes to infinity?

Question

Actually, this is not precisely my question. If $a(x)$ is the inverse ackermann function, then obviously $a(a(x))$ grows slower than $a(x)$, as does $\log(a(x))$, and so on. But is there a function f that goes to infinity and cannot be asymptotically bounded below (when x goes to infinity) from any finite composition of logarithms, inverse Ackermann functions, and arithmetic operations?

(Equivalently, is there a "well-behaved" function that cannot be bounded above by a finite composition of Ackermann functions, exponentials, arithmetic operations, etc?)

Edit: I mean functions that are defined for all (positive) reals, not just integers.

Edit 2: Okay, infinitely differentiable functions.

Answer 1

Let $F(n) = A^n(n)$ for any natural $n$

Then $F(n) \in ω(A^k(n))$ as natural $n \to \infty$ for any natural $k$

Let $f(n) = \min( \{ k : k \in \mathbb{N} \wedge F(k) \ge n \} )$

Then $f(n) \in o(a^k(n))$ as natural $n \to \infty$ for any natural $k$

Answer 2

Check out the Busy Beaver function.

Answer 3

In the fast growing hierarchy, we have

$$A(n,n)\approx f_\omega(n)$$

So for any $\alpha>\omega$, $A(n,n)\ll f_\alpha(n)$, and by taking the inverse (with a rounding just like how one takes the inverse of the Ackermann function) you can easily make functions that grow slower than $A^{-1}(n)$.

See Create the slowest growing function you can in under 100 bytes for some more examples.