Can any integer, not a multiple of three, be represented as $n = \sum_{i=0}^{a-1} 3^i \times 2^{b_i}$?

Question

This question has some relevance to the Collatz conjecture. It was originally based on trying to represent a number like this: Finding whether $\dfrac{2^k - (2\cdot3^{n-1} + 2^{t_0}3^{n-2} + 2^{t_0+t_1}3^{n-3} .... + 2^{t_0+t_1+...+t_{n-1}})}{3^n}$ can describe all integers

However, I generalised that and tried to ask a simpler question instead, now this is simply out of curiosity rather than anything useful, as in often the case with the Collatz conjecture.

Is it possible to represent a number that is not divisible by three as: $$n = \sum_{i=0}^{a-1} 3^i \times 2^{b_i}$$ $$n, a, b_i \in \Bbb{Z^+_0}$$ Where $b_i$ is some arbitrary integer such that $b_{i+1} \leq b_i$

For example, $13=1+3+9$.

EDIT: My original question asked what happened with these conditions: $$n, a, b_i \in \Bbb{Z^+}$$ $$b_{i+1} < b_i$$ (Notice the subtle differences in specifications.) Under these above conditions, some numbers cannot be represented, like $13$. What are some characteristics of numbers that cannot be represented like this?

Any help will be much appreciated. Thanks.

Answer 1

I deduced the same thing when study the Collatz conjecture, here is the proof without the restrictions and some stuff related.

Let $G_n = \{m \in \mathbb{N} \,|\,\gcd(m,n) = 1\}$

We can do this for all $G_n$, for example $n = 2$, but we only interested in $n = 3$.

Lemma: For all $n \in G_3$, exists $a \in \mathbb{Z} : 0 \le a$ such that:

1. $n = 2^a \hspace{5pt} (\text{mod } 3)$

2. $n \neq 2^a \hspace{5pt} (\text{mod } 9)$

Proof: Without loss of generality, we use this table:

      +---+---+---+---+----+----+
      | 1 | 2 | 4 | 8 | 16 | 32 |
+-----+---+---+---+---+----+----+
|Mod 3| 1 | 2 | 1 | 2 | 1  | 2  |
+-----+---+---+---+---+----+----+
|Mod 9| 1 | 2 | 4 | 8 | 7  | 5  |
+-----+---+---+---+---+----+----+

From the table:

1. For all $n \in G_3$, exists $a$ such that $n = 2^a\ (\text{mod }3)$.

2. For all $n \in G_3$, exists $a$ such that $n = 2^a\ (\text{mod }9)$.

3. By contradiction, suppose that for all $a$ we have $n = 2^a\ (\text{mod }3)$ and $n = 2^a\ (\text{mod }9)$.

   By the table, exists $a'$ such that $2^a = 2^{a'}\ (\text{mod }3)$ and $2^{a} \neq 2^{a'}\ (\text{mod }9)$.

   Then, $n = 2^{a'}\ (\text{mod }3)$ and $n \neq 2^{a'}\ (\text{mod }9)$.

Finally, by contradiction, exists $a$.

Corollary: For all $n \in G_3$, exists $a \in \mathbb{Z} : 0 \le a$ and $k \in G_3$ such that $n = 2^a + 3k$.

Notation: If a number $n$ can be represented in this form, for some $l \in \mathbb{Z} : 0 \le l$: $$ n = 2^{\alpha_0} + 2^{\alpha_1}\,3^{1} + 2^{\alpha_2} + ... + 2^{\alpha_{l}}\,3^{l} $$ such that $\alpha_i \in \mathbb{Z} : \alpha_i \ge 0$ for $i = 1, ..., l$.

Then, we said that n can be represented in a $l$-canonical form and if exists $l \in \mathbb{Z} : 0 \le l$ such that n is a $l$-canonical form then we said n can be represented in the canonical form.

Theorem: Every number in $G_3$ can be represented in the canonical form.

Proof: We prove by induction that $n$ can be represented in that form, for all $n \in G_3$.

Base case:

You can find a expression for all $n \in G_3 : n \le 9$.

Inductive case:

Suppose this is holds for all $k \in G_3 : k < n$:

By the corollary, exists $a$ and $k \in G_3$ such that $n = 2^a + 3 k$. Because $k < n$, $k = 2^{\alpha_0} + 2^{\alpha_1}\,3 + ... + 2^{\alpha_i}\,3^i$.

Finally, because $n = 2^{a} + 3\,(2^{\alpha_0} + 2^{\alpha_1}\,3 + ... + 2^{\alpha_i}\,3^i)$, every number in $G_3$ can be represented in the canonical form.

Notation: If $n \in G_3$ can be represented in a $0$-canonical form, we said $n$ can be represented in the canonical principal form.

Notation: If $m$ can be represented in a $l$-canonical form and $n = m + 3^{l + 1}\,t$ then we said $t$ is in the tail of $n$, with $n,m,t \in G_3$.

The Collatz conjecture implies that every number in $G_3$ is in the tail of some canonical principal form.

Conjecture: If $n \in G_3$ can be represented in a $a$-canonical form and $b$-canonical form, for $a \le b$, then $n$ can be represented in a $l$-canonical form, for all $l \in \mathbb{Z} : a \le l \le b$.

Answer 2

This is only a comment, but too long for the comment box

Since your $b$'s are organized as equal or decreasing one can define "layers" of equal $b$ and sum the terms with equal $2^b$ and decreasing powers of $3$ up. That would give something like $$ n = 2^{b_1}\left( {3^{A_1} -1 \over 3-1 }\right) + 2^{b_2}\left( {3^{A_2} -1 \over 3-1 } \right) + ... + 2^{b_m}\left( {3^{A_m} -1 \over 3-1 } \right)$$ and can even be expressed more compact like this form $$ n = 2^{b_1}\left( {3^{A_1} -1 \over 3-1 }+ 2^{b_2}\left( {3^{A_2} -1 \over 3-1 } + ... + 2^{b_m}\left( {3^{A_m} -1 \over 3-1 } \right)\right)\right) $$ (the indexes $A_k$ and $b_k$ are used only as models, not actual indices here).
From this the question can then also be written in base-3 representation $$ n=2^{b_1} \cdot 111111...1_3 + 2^{b_2} \cdot 111...1_3 + ...$$ or so. Perhaps looking at such representations would easier give an idea whether any number $n$ can be expressed in such a form.