Zeration

[quickfur]

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)

I forgot to note, what I had in mind here is \( (ab)^c = a^{c}b^c \) and \( (a+b)c = ac + bc \). Of course, exponentiation/multiplication has other properties that I didn't take into account, which is not a surprise, since they impose other constraints on the resulting operation and ♦ would not qualify. But I was just using ♦ as an example that depending on what properties are desired, one can derive different operations that qualify under the generic label of "before addition".