quickfur Wrote:This isn't really related to zeration per se, but some time ago I searched for an operator ♦ over which addition is distributive (i.e., a+(b♦c) = (a+b)♦(a+c)). (I guess my motivation was that just as exponentiation distributes over multiplication, and multiplication distributes over addition, addition should also distribute over an operation that may be regarded to be, in some sense, "before" addition.)This distribution law does not fit into the distribution laws for exponentation and multiplication

a[3](b+c)=(a[3]b)[2](a[3]c)

a[2](b+c)=(a[2]b)[1](a[2]c)

The natural continuation would be:

a[1](b+c)=(a[1]b)[0](a[1]c)

And one can prove that there is only the trivial operation [0]:

Let c=0 then

a+b=(a+b)[0]a

which means that x[0]y=x.

What you are referring to is another kind of hierarchy differently from the hyper operations hierarchy. It is the hierarchy of operation {n} built by the law:

a{n+1}(b{n}c)=(a{n+1}b) {n} (a{n+1}c)

a{1}b=a+b

as opposed to a hierarchy of operations [[n]]:

a[[n+1]](b+c)=(a[[n+1]]b)[[n]](a[[n+1]]c)

a[[1]]=a+b

which is not even satisfyable on the natural numbers (however possible with binary trees).

one can easily verify that one solution to the {n} hierarchy is

which for matches exactly your:

Quote:.

And all these operations {n} are associative and commutative.

However I never heard something interesting about this hierarchy, perhaps because the domain of definition with the operations {n} and {n+1} (any n) is just isomorphic to the reals with + and *.