Find (if exists) a subset of the non negative integers $X$ such that for every non negative integer $n \in \mathbb{N}\cup\{0\}$ there is exactly one solution of the form $a+2b=n$ with $a,b \in X$
I have "conjectured" those numbers are of the form:
$2 \cdot 3^{m-1} -1$ and $2 \cdot 3^{m-1} -2 $ with $m \in \mathbb{N}$ but I can't prove it is true.
Let $n$ be any nonnegative integer.
Using the base $4$ expansion of $n$, we have
$$n=\sum_{i=0}^k d_i\cdot 4^i,\text{ where each $d_i$ is one of $0,1,2$, or $3$. }$$
Now define $a_i=d_i\% 2$ (here $\%$ is the mod (remainder) operator); and define $b_i=\lfloor\frac{d_i}{2}\rfloor$
Note: (1) For all $i$, $a_i+2b_i=d_i$; and (2) For all $i$, $a_i$ and $b_i$ are each either $0$ or $1$.
Define $a=\displaystyle\sum_{i=0}^k a_i\cdot 4^i$ and $b=\displaystyle\sum_{i=0}^k b_i\cdot 4^i$.
We have that $n=a+2b$.
Thus we may take $X$ to be the set of nonnegative integers that can be written as a sum of distinct powers of $4$. Note: this includes the empty sum, $0$. That is, $0\in X$.
So $X=\{0,1,4,5,16,17,20,21,64,65,68,69,80,\dots\}$
For example, let's look at $n=147$. We have $147=3+0\cdot 4+1\cdot 4^2+2\cdot 4^3$. We take $a=1+4^2=17$ and $b=1+4^3=65$. An, as expected, $n=a+2b$.