I have a fairly elementary question but rather than feeding it straight to the wolves I thought I would ask it here. Well, via a more general question:

Are questions of the form "

has $X$ been studied before" on topic?

I guess the nature of $X$ is critical. A question that has prompted the idea of asking my question is this one. I think the $X$ there is quite elementary, as is mine, but the $X$ there almost certainly has better properties/structure.

This isn't my written-out question, just a brief description of my $X$.

My $X$ is the following: where $S:=\operatorname{Seq}_0(\mathbb{N}_0)$ is the set of eventually-zero $\mathbb{N}_0$-valued sequences, using prime decomposition of $n\in\mathbb{N}$, we have a bijection:

$$\pi:\mathbb{N}\to S,$$

e.g. $\pi(1)=(0,0,\dots,)$, $\pi(18)=(1,2,0,0,0,\dots)$, etc.

Where $r_i$ is the $i$th component of $r\in S$, we have operations on $S$:

$$[s\boxed{+}r]_i=s_i+r_i\qquad(r,s\in S),$$ $$[s\boxed{\times}r]_i=s_i\times r_i\qquad (r,s\in S).$$

The bijection allows us to transport addition $+$ and multiplication $\times$ on $\mathbb{N}$ to operations $\boxed{\star}$ and $\boxed{+}$ on $S$, and the addition $\boxed{+}$ and multiplication $\boxed{\times}$ on $S$ to operations $\times$ and $\diamond$ on $\mathbb{N}$.

We have, of course:

$$m\times n=\pi^{-1}(\pi(m)\boxed{+}\pi(n))\qquad(m,n\in\mathbb{N}),$$

We get a strange operation on $\mathbb{N}$:

$$m\diamond n=\pi^{-1}(\pi(m)\boxed{\times}\pi(n))\qquad (m,n\in\mathbb{N}),$$ as well as a strange operation on $S$:

$$r\boxed{\star}s=\pi(\pi^{-1}(r)+\pi^{-1}(s))\qquad (r,s\in S).$$

We have *some* structure therefore on $(\mathbb{N},\times,\diamond)$ and $(S,\boxed{\star},\boxed{+})$.

interesting. The richness of much of mathematics arises from the basic operations interacting through simple laws: in particular, the distribution of multiplication over addition and the associative law for both operations. Do your operations conform to such laws? $\endgroup$`+`

is a binary operation when it is`\boxed`

, and`\mathbin`

reminds it: $r\boxed+s$ vs. $r\mathbin{\boxed+}s$`$r\boxed+s$ vs. $r\mathbin{\boxed+}s$`

. But perhaps just $r \boxplus s$`r \boxplus s`

(although there is only $r \boxtimes s$`r \boxtimes s`

, not a boxed star, as far as I know) will do? $\endgroup$