Why not just write it out?
Put $x = (x_{-K}, \dots, x_K)$, where
\begin{align*}
x_p = \begin{cases}
i & \text{if $p$ is odd}; \\
0 & \text{if $p$ is even}.
\end{cases}
\end{align*}
Your way to denote $\mathbf{M}_r^{/0}$ seems accurate. You are then free to substitute the $r$ in this notation by any number you like, so e.g. $\mathbf M_1^{/0}$ to denote the neighborhood of range one. No additional definition is necessary.
If the exact order of the element you want to index is not of importance, you can just write that you assume that
$$\mathbf M_r^{/0}=\{\mathbf m_{r,1},...,\mathbf m_{r,k(r)}\},$$
where $k(r)$ denotes the size of a Moore-neighborhood of range $r$. Now you are free to write $\mathbf m_{r,i}\in\mathbf M_r^{/0}$ for any element in the neighborhood. Just do not assume that the index $i$ stands for anything else than a distinguishing feature from the other elements.
Of course, if you need a specific order of the neighbors, you can always say that $\mathbf M_r^{/0}=\{\mathbf m_{r,1},...,\mathbf m_{r,k(r)}\}$, where the indices are chosen so that $i<j$ implies $\mathbf m_{r,i}\prec \mathbf m_{r,j}$. Here $\prec$ denotes a specifically defined order of the grid points, e.g. lexicografic $(1,2)\prec(1,3)$ and $(2,3)\prec(3,1)$.
Best Answer
The default notation is that $[1,n]$ is an interval in the set of real numbers: $$[1,n] = \{x \in \mathbb R \mid 1 \le x \le n\} $$ and this includes non-integer numbers such as $x = 1.5$ (assuming $n \ge 2$). So no, that's not correct (unless you explained very carefully and very clearly, in what you are writing, what your intention is for the notation $[1,n]$).
What would be correct is to use the intersection operator $\cap$ and to write $x \in [1,n] \cap \mathbb N$, using the standard notation $\mathbb N$ for the set of natural numbers (a.k.a. the positive integers).
However, it is common notation to literally use the ellipsis in this situation, $$\{1,...,n\} = [1,n] \cap \mathbb N $$ If you used that in your mathematical writing, you would be almost universally understood.