Numbers of the form $\sum_{n=0}^{N} 2^{a_n} 3^n$

Question

Just out of curiosity. Does the numbers of the form \begin{align} \sum_{n=0}^{N} 2^{a_n} 3^n \text{,} \end{align} with $a_n \ge 0$ and $N \ge 0$, cover all integers not divisible by 3?

These numbers can never be divisible by three. It is also easy to discover integers with multiple representations (e.g., $4 = 2^2 3^0 = 2^0 3^0 + 2^0 3^1$). I am however stuck to prove that they cover all integers not divisible by 3.

Second question: The numbers of the form \begin{align} \sum_{n=0}^{N} 5^{a_n} 3^n \text{,} \end{align} with $a_n \ge 0$ and $N \ge 0$, are numbers congruent to $\{1, 4, 5, 8\} \pmod{12}$. These are the numbers not divisible by 3 and not $\{ 2, 3 \} \pmod{4}$. Many other such forms may be considered. What is the name of such numbers (numbers of such a form)?

Answer 1

If $x \le 8$, then all numbers not divisible by $3$ have the desired form as one can check by hand (e.g. $1=1$, $2=2$, $4=4$, $5=2+3$, $7=1+2 \cdot 3$, $8 = 8$).

If $x > 8$, and $x \equiv 1,2,4,5,7,8 \bmod 9$, then let $m = 4,8,1,2,4,2$ respectively, so $m$ is a power of $2$ and $x \equiv m \bmod 3$ but $x \not\equiv m \bmod 9$. Now write

$$x = m + 3y$$

Now, by construction, $y<x$ is not divisible by $3$. By induction, it has the required form, which implies that $x$ also has the required form. Indeed from this proof you see that one can take the power of $2$ to be in the set $\{1,2,4,8\}$.

Answer 2

Well as far as the first question every integer$M$ can be written $\sum_{n=0} a_i3^n$ where $a_i =0,1=2^0$, or $2^1=2$ i.e., $M$'s tertiary expansion. So yes to the first question.

Answer 3

Erdos asked a similar question some time ago (about “3-smooth representations” of the natural numbers). See here for a discussion on both of your questions.