Is there an operation when repeated $n$ times on a number $x$ gives $\max(x, n)$?

Question

It has been asked if there is an operation, that when repeated $n$ times results in addition. The solution given is that one can use the $\max$ function in a clever way:

$$\max(a, b) + 1 + \delta_{ab} = \text{succ}_{ab}(\text{succ}(max(a, b)))$$

Where $\text{succ}_{ab}(x) = \begin{cases} \text{succ}(x) & \text{if } a = b \\ x & \text{otherwise} \end{cases}$.

I use the successor function to emphasize that it actually does not depend on addition. This has also been described as Zeration. I am trying to see is there an operation $@$ such that $\underbrace{x@x@\cdots@x}_{n\text{ times}} = \max(x, n)$ (or if considerably easier: $\underbrace{x@x@\cdots@x}_{n\text{ times}} = \max(x, n) + 1 + \delta_{ab}$) that does not appeal to $+$, $\cdot$, or $\max$. The name "zeration" suggests that this is some how the lowest on the hierarchy and there might not be such an operation, however, I have not see a proof of this which is why I am wondering if there is such an operation.

Answer 1

Such an operation cannot exist, for both formulas.

In the following I'm always assuming that the operators are left-associative, i.e. $x@y@z=(x@y)@z$. The same arguments work similarly with right-associativity or any other form of consistent bracketing.

If $\underbrace{x@x@\ldots@x}_{n\text{ times}}=\max(x,n)$ for all $x,n\in\mathbb N$, then in particular $1@1=\max(1,1)=1$, hence $$2=\max(1,2)=1@1@1=1@1=1.$$

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Similarly, if $x@x@\ldots@x=\max(x,n)+1+\delta_{ab}$, let $\alpha:=3@3=\max(3,2)+1+\delta_{ab}=4+\delta_{ab}$. Then $\alpha@3=3@3@3=\max(3,3)+1+\delta_{ab}=\alpha$, so $$\alpha+1=\max(3,4)+1+\delta_{ab}=3@3@3@3=\alpha@3@3=\alpha@3=\alpha.$$

Answer 2

Let us suppose a function $f^n(x)=$ max$(x,n)$. Consider a number $x>n$. This means max$(x,n)$ will yield x. In this sense we are talking about an identity function that gives us the initial input after applying operation $n$ times. This changes when $x<n$. Now the function must yield output $n$ after applying operation $n$ times.

Thus both examples provide counter answers and thus a contradiction. Hence such a function cannot exist. However if we follow one approach say $x>n$ and $\overrightarrow{x}$ is a vector such that it is the Eigen vector, then $f^n(\overrightarrow{x})$ = $x$ can be thought of power of the $f^n(\overrightarrow{x})$ = $\lambda^n\overrightarrow{x}$ with Eigen value of $1$ the operation is satisfied.