Show that this fractal contains every odd, positive integer Let $f(x)=2x+\frac{1}{3}$
Let $g(x)=\frac{2x-1}{3}$
$f^n$ represents composition of $f$
Let $G_0=\{1\}$
Let $F_{m}=\{f^n(x):x\in G_{m}, n\in\mathbb{N_{\geq0}}\}$
Let $G_{m+1}=\{g(x):x\in F_m\}$

Show that for any given odd, positive integer $p$ there is some sufficiently high $m$ for which $p$ is in $F_m$

For what it's worth (and this may well not be helpful at all) but I've got some insight into why it's true but I can't translate that into a solution.
I can see that the denominator of $F_m$'s elements is a power of $3$ increasing with each successive $m$.  I can see that these functions $f$ and $g$ (and restricted to the positive integers) form a Euclidean function where the orbits of $f$ and $g$ are in some sense orthogonal away from the vicinity of $1$.
I think this fact, combined with the fact that $F_m\cap F_{m-1}$ is somehow dense in the odd integers within the range from $G_m$ to $G_{m+1}$, proves the statement.  But I have no idea by what metric $F_m\cap F_{m-1}$ are between $G_m$ and $G_{m+1}$.
I'm pretty sure it's a 2- and 3- adic number problem.
It is also a fact that this is a fractal. Each application of $F$ adds, for every element, a new set of numbers spaced exponentially, just above or below those already included. 

This answer is a work in progress...
We need to show that the orbits of $f$ and $g$ are in some sense orthogonal in the positive, odd integers greater than $3$, or that they are in some sense Euclidean under composition in the same range.
The closed form for $f^n(x)$ is given by:
$$f^n(x)=2^nx+\frac{2^n-1}{3}$$
The closed form for $g^m(x)$ is given by:
$$g^m(x)=\left(\frac{2}{3}\right)^mx-\frac{1}{3}\sum_{k=1}^m\left(\frac{2}{3}\right)^{k-1}\\=\left(\frac{2}{3}\right)^m\cdot x+\left(\frac{2}{3}\right)^m-1\\=\left(\frac{2}{3}\right)^m(x+1)-1$$
If $f^n(x)=2^nx+\frac{2^n-1}{3}$ is an action of $\mathbb{N}$
Let $F_x=\{f^n(x)\mid n\in\mathbb{N}\}$ be the orbit of the point $x$ through $\mathbb{Q_{>2}}$ under this action.
And $g^m(x)=\left(\frac{2}{3}\right)^m(x+1)-1$ is also an action of $\mathbb{N}$
Let $G_x=\{g^m(x)\mid n\in\mathbb{N}\}$ be the orbit of the point $x$ through $\mathbb{Q}$ under this action.

To show that there is some metric space in which $F_{g^{-1}(x)}< F_x<F_{g(x)}$ for all $x\in F_x$ for all $F$ will prove the conjecture.


Update: Ok I've worked out that this metric space is a Euclidean combination of; the inequality:
$\log_2{x}-\lfloor\log_2{x}\rfloor>1-\log_2 3$
And the $2-$adic norm of $x$
I'm sorry if that's a bit vague; it needs more thought.  Truth or otherwise of the inequality determines whether the highest power of $2$ which is less than the odd factors of any successor of any $F_{g^{-1}(x)}$, will be $4$ times that of its predecessor in $F_{g(x)}$, or only $2$ times.
The $2-$adic norm tells us whether the binary string of any $F_{g^{-1}(x)}$'s odd factors shrinks from the right by $1$ digit, or more.
