Ackermann function - how to calculate the number of times it calls itself

The function $RA(m,n)$ (as defined by the computation process described in the question description) satisfies the following conditions (where $A(m,n)$ is the standard Ackermann function): $$ RA(m,n)= \begin{cases} 1 & \text{for } m=0\\ 1 + RA(m-1, 1) & \text{for } m>0 \text{ and } n=0\\ 1 + RA(m, n-1) + RA(m-1, A(m, n-1)) & \text{for } m>0 \text{ and }n>0 \\ \end{cases} $$

First two cases are trivial: $A(0, n)$ is given directly, without recurring any further, and $A(m,0)$ is defined as $A(m-1,1)$, so we just need to perform a single additional recursive call.

The third case is the slightly tricky one: in order to evaluate $A(m-1,A(m,n-1))$, we first need to compute the value of $A(m,n-1)$, which takes $RA(m,n-1)$ steps. After that, we are computing $A(m-1,\cdot)$, where $\cdot$ is the just-computed value of $A(m,n-1)$ (i.e. it doesn't bring in any further recursive calls), which is represented by the last term.

It is now not too difficult to see that $RA(m,n)\geq A(m,n)$ for $m\geq 1$ ($m=0$ is a special case, since it involves no additional recursion).

We can also use the third line repeatedly until we reach $n=0$ and finish by applying the second one to obtain the following equality which expresses $RA(m,n)$ using just $RA(m-1,\cdot)$:

$$RA(m,n)=(n+1) + RA(m-1,1) + \sum_{k=0}^{n-1} RA(m-1,A(m,k))$$

Since $A(3,n)=2^{n+3}-3$ and we already know $RA(2,n)=2n^2+7n+5$, we can evaluate $RA(3,n)$ exactly as: $$RA(3,n) = 128\frac{4^n-1}{3} - 40\left(2^n-1\right) + 3n + 15$$

(the strange-looking terms are results of summing geometric progressions)

This subsequently allows us to reach a little bit further in the computations and find that $RA(4,1)=2862984010$ but beyond this point the numbers grow very fast. Furthermore, unlike the nice polynomials (for $m\leq 2$) and exponentials (for $m=3$), exponential towers and further hyperoperations which are lurking in $A(m,n)$ for $m>3$ do not play well with simple summation, so I doubt there would be any simple exact expression for $RA(m,n)$ with $m>3$.

Of course, this doesn't stop this function from having some explicit (and possibly even tight) bounds in terms of $A(m,n)$ or some similar function; I even have an idea of how such bounds might looks like; I will try to play with and update the answer if it yields anything interesting.

According to the German Wikipedia page about the Ackermann function, what you call $RA(3,n)$ can be expressed as $RA(3,n)=A(3,n)+12^n-2$. Unfortunately, there are no links given.

However, I once read an old paper about this function which gave some good insight into the computations involved to find the number of calls needed to compute $A(m,n)$ (see "The Ackermann function: a theoretical, computational, and formula manipulative study" by Yngve Sundblad). The author does not give a general expression for $RA(m,n)$, as "it is difficult to give [one]". I think this paper can be found online.