Generalized arithmetic operator

[hixidom]

These are not the "natural" (left-associative) hyperoperations family, and seems that are different from lower hyperoperations too.

Why you chose this kind of bracketing?

I'm not familiar with the difference between left- and right-associative hyperoperators. I would say that the unary version is definitely righ-associative, as
\( [x][y][z]a=[x]([y]([z]a)) \)

but perhaps that is not what you are referring to.

If I had to write a recursion relation along the lines of what has been posted so far, I would say it is:
\( a[x]b=(a[x]\frac{b}{2})[x-1](a[x]\frac{b}{2}) \)

which is neither left- nor right-associative, if I understand the meaning of that phrase.
The ones that have been posted so far are
\( a [x] b=a[x-1](a [x] (b-1)) \),
and
\( a [x] b=(a [x] (b-1))[x-1]a \).

which are very similar, with the only difference being that the [x-1] operation on the right side is flipped.

But neither of these definitions result in the same [4] operator that I have.
\( \begin{align*}a[4]2&=[3]a&=a^a\\a[4]4&=[3]^2a&=(a^a)^{(a^a)}\\a[4]8&=[3]^3a&=\left ((a^a)^{(a^a)}\right )^{\left ((a^a)^{(a^a)}\right )}\end{align*} \)
etc.