We have of course:
$2\Bbb Z-1$, or $2\Bbb N+1$
These I think of as the integers reduced by the congruence $x\cong x+1$ where in one case $x$ is even and in the other $x$ is odd (it makes no difference really).
But I want to think specifically of reducing integers by the congruence $x\cong 2x$ perhaps more clearly the transitive closure of the equivalence relation $x\sim 2x$.
My random stab in the dark is to write it $\Bbb Z/\langle 2\rangle$
Expressed like this, I guess the set includes $0$.
Since this is a multiplicative group modulo a prime, perhaps it's better to exclude $0$ something like: $\Bbb Z^\times/\langle2\rangle$?
What's acceptable / usual here?
If I understand the question right, we are looking at $\mathbb N$ and the relation $x \sim y \iff $ there exists $i \in \mathbb Z$ such that $x = y \cdot 2^i$. This is an equivalence relation, and - apart from the equivalence class {0} - each equivalence class contains a unique odd integer.
$\mathbb N$ with multiplication is a monoid (like a group, but without the requirement for inverses to exist), and it's easy to show that for any monoid $A$, if you have an equivalence relation $\sim$ such that $a \sim b$ and $x \sim y$ implies $ax \sim by$ - which holds in the case we are considering - then there is a quotient monoid structure on $A\ / \sim$, and it is this quotient monoid structure you are thinking of.
I don't know of any particular standard notation for the equivalence relation $\sim$ or for $\Bbb N \ / \sim$.