TITLE: what operation repeated $n$ times results in the addition operator?
QUESTION [24 upvotes]: I had a difficult time in phrasing my question. But I was wondering if there is an operation that, when repeated n times, results in the addition operator. Same way as repeating addition n times results in the multiplication operator, and repeating multiplication n times results in the exponentiation operator etc.
So $n$ times addition of a number $x$ results in $x\times n$.
And $n$ times multiplication of a number $x$ results in $x^n$
Then my question is $n$ times ...what... results into the number $x+n$.
Let's call this operator: $@$.
For example, the following would then hold:
$$a\times a=a^2$$
$$a+a=a\times 2$$
$$a@a=a+2$$
My question is, does it make any sense thinking of such an operator, is there anything known about it, can it be followed through even further like $a\sim a = a@2$?

REPLY [2 votes]: There are already amazing answers!
But I'll try to give some additional info since this question was asked a lot of times in the past (and it will be asked again in the future right?) and I gave some answers too.

As Hagen von Eitzen and Brevan Ellefsen said $$a\circ b:= \max (a,b)+1+\delta_{ab}$$ is one of the solutions of the equation $a*(a+n)=a+n+1$ , where $a$ and $n$ range over some restricted domains.
This particular solution was called Zeration by Rubtsov and Romerio: you can easily find the details of the story in the Zeration thread that was linked by Brevan Ellefsen.
Another solution is $\max(a,b)+1$ which is commutative and is not discontinuous, but in the linked thread, a user, Tetration Forum's founder, also noticed that it is possible to find a non-commutative solution too.
You can find an updated discussion of the topic in this thread:
Zeration Update @Tetration Forum
Also here at MSE, similar questions were answered: for example, this one is almost identical
Does anything precede incrementation in the operator “hierarchy?”
and also remark 1 of this answer I wrote 2 years ago:
More info on Zeration - Go to remark 1