Left-bracketed Ackermann function also not primitive recursive?

Question

The original Ackermann function $\varphi\colon \mathbb{N}\times\mathbb{N}\times\mathbb{N}_0\to \mathbb{N}$ as defined in [1] was invented to prove that there is a function that is recursive but not primitive recursive. It can be given by the following recursion:

• $\varphi(a,b,0) = a+b$
• $\varphi(a,b,n+1) = (x\mapsto \varphi(a,x,n))^b(\alpha(a,n))$

Where $\alpha(a,0)=0$, $\alpha(a,1)=1$ and $\alpha(a,n)=a$ for $n\ge 2$ are initial values and $(x\mapsto f(x))^k$ is the k-times composition of the function $x\mapsto f(x)$. The function $n\mapsto \varphi(n,n,n)$ is not primitive recursive because - informally speaking - it grows too quickly.

These operations are right-bracketed, this does not matter for $\varphi(a,b,1)=ab$ and $\varphi(a,b,2)=a^b$, but for the next higher rank it is important $\varphi(a,b,3)=\underbrace{a^\land (a^\land (...(a^\land a)))}_{b+1\; \text{occurences of}\; a}$ where $a^\land b:=a^b$.

If we would choose left-bracketing the functions would not grow so quickly. The left-bracketed operations would be defined as:

• $\psi(a,b,0) = a+b$
• $\psi(a,b,n+1) = (x\mapsto \psi(x,a,n))^b(\alpha(a,n))$

Again $\psi(a,b,1)=ab$ and $\psi(a,b,2)=a^b$, but here the forth operation would be $\psi(a,b,3)=a^{a^b}$

My question is now whether the left-bracketed operations still grow fast enough for not being primitive recursive, i.e. is $n\mapsto \psi(n,n,n)$ still not primitive recursive?

[1] Ackermann, W. (1928 ). Zum Hilbertschen Aufbau der reellen Zahlen. Math. Ann., 99, 118–133.

Answer 1

Sorry for reviving this ancient post, but it seemed to me that this question shouldn’t be left unanswered.

It is easy to establish by induction that for $a,b,n\ge2$, the function $\psi(a,b,n)$ is strictly increasing in all three coordinates, and $$\psi(a,b,n)>a,b,n.$$ Then we can show $$\tag{$*$}\psi(a,b,n+1)\ge\psi(b,a,n)$$ for $a,b,n\ge2$ by induction on $b$: clearly, $\psi(a,2,n+1)=\psi(\psi(a,a,n),a,n)\ge\psi(2,a,n)$, and going from $b$ to $b+1$, we have $$\psi(a,b+1,n+1)=\psi(\psi(a,b,n+1),a,n)\ge\psi(\psi(b,a,n),a,n)\ge\psi(b+1,a,n)$$ using monotonicity and $\psi(b,a,n)\ge b+1$.

Thus, by $(*)$ and monotonicity, we get $$\psi(a,b,n+2)\ge\bigl(x\mapsto\psi(a,x,n)\bigr)^b(a),$$ which implies $$\psi(a,b,2n-2)\ge\varphi(a,b,n)$$ for all $a,b,n\ge2$ by induction on $n$. Consequently, $\psi$ (or its diagonal, for that matter) is not bounded by any primitive recursive function.