What numbers are in the set $S$? Here is a problem that I came up with that has been bothering me for a while, and I haven't been able to solve it.
Let $S$ be the set defined by the following rules:


*

*$2\in S$.

*Define the functions $\alpha$ and $\beta$ as
$$\alpha(n)=n^2$$
$$\beta(n)=\lfloor n/3\rfloor$$
If $n\in S$, then $\alpha(n)\in S$ and $\beta(n)\in S$.

*If the previous two rules do not imply that a number $n$ is in $S$, then $n\notin S$.


QUESTION: What numbers are in $S$?
So far, I've verified that every positive integer from $1$ to $14$ is in $S$, since I've found "composition sequences" of $\alpha$ and $\beta$ that, when applied to $2$, yield each of those numbers. For example,
$$3=(\beta^{\circ 4}\circ\alpha^{\circ 3})(2)$$
Things get a little crazy when I get to $n=6$:
$$6=(\beta^{\circ 10}\circ\alpha^{\circ 2}\circ \beta\circ\alpha^{\circ 2})(2)$$
...and even crazier at $n=14$:
$$14=(\beta^{\circ 3}\circ \alpha\circ \beta^{\circ 10}\circ \alpha^{\circ 2}\circ \beta\circ \alpha\circ \beta^{\circ 10}\circ \alpha^{\circ 2}\circ\beta^{\circ 2}\circ \alpha^{\circ 3})(2)$$
Any ideas how to prove (or disprove) that any positive integer can be reached?
NOTE: It is true that all positive integers are in $S$ if (not iff) the following is true:

Blockquote
  For any positive integer $N$, there exist positive integers $a,b$ such that
  $$N=\bigg\lfloor \frac{2^{2^a}}{3^b}\bigg\rfloor$$

If someone figures out how to prove this, then we're done... if it's false, then there remains some work to be done.
 A: I don't know how to answer this, but let me elaborate on my comments about your proposed proof strategy.  The punchline is that it is very likely that $S$ does contain all positive integers via your strategy, but it is an open problem to prove this.
As you say, a sufficient condition for $S$ to contain a positive integer $N$ is that $N$ can be written in the form $$N=\bigg\lfloor \frac{2^{2^a}}{3^b}\bigg\rfloor.$$  More generally, if you modify your definition of $S$ to start with $x\in S$ and modify your functions to $\alpha(n)=n^y$ and $\beta(n)=\lfloor n/z\rfloor$ for some positive integers $x,y,$ and $z$, then a sufficient condition for $S$ to contain $N$ is that $N$ can be written in the form $$N=\bigg\lfloor \frac{x^{y^a}}{z^b}\bigg\rfloor.$$
I now claim that this is possible for all positive integers $N$ iff every finite sequence of digits appears in the base $y$ expansion of $\log_z x$.  So, in your case, that would be the binary expansion of $\log_3 2$.  The proof is basically "take $\log_z$ of everything"; I've written up the details below.
Now, the bad news is that questions like this are typically very hard to answer: if you pick some specific nicely defined irrational number, it is usually very difficult to say much about its base expansions.  In particular, I'm pretty sure that for all positive integers $x,y,$ and $z$ for which $\log_z x$ is irrational, it is an open problem whether the base $y$ expansion of $\log_z x$ contains all finite sequences.
On the other hand, if you pick a random number (uniformly from some interval), then with probability $1$ it will contain every finite sequence in every base.  So it is considered extremely likely that $\log_z x$ contains every finite sequence in base $y$; it would be an amazing discovery if it didn't.  In particular, this means it is extremely likely that your set $S$ contains all positive integers.
Of course, since this condition is only sufficient and not necessary for $S$ to contain all positive integers, it is possible that there is an easier proof which does not require solving this open problem.  I haven't been able to find any promising approach that doesn't end up reducing to a question of this sort, though.

OK, now here's the proof of the claimed equivalence.  Let's write $C=\log_z x$. 
First, suppose every finite sequence of digits appears in the base $y$ expansion of $C$, and let $N$ be any positive integer.  Choose integers $m$ and $k$ such that $$\log_z N<\frac{m}{y^k}<\frac{m+1}{y^k}<\log_z(N+1).$$  Let the string $s$ be the base $y$ expansion of $m$, with enough initial zeroes added (if necessary) so that $s$ has length at least $k$.  This string $s$ appears infinitely many times in the base $y$ expansion of $C$ (since every finite string containing it appears somewhere), and in particular it appears somewhere starting after the decimal point.  This means that there exist positive integers $a$ and $b$ such that the base $y$ expansion of $y^aC-b$ is identical to that of $\frac{m}{y^k}$ up to the $k$th digit after the decimal point.  That is, $$\frac{m}{y^k}\leq y^aC-b<\frac{m+1}{y^k}.$$  By our choice of $m$ and $k$, this implies that $\lfloor z^{y^aC-b}\rfloor=N$.  But $$z^{y^aC-b}=\frac{z^{y^a\log_z x}}{z^b}=\frac{x^{y^a}}{z^b},$$ so this is exactly what we wanted.
Conversely, suppose every $N$ can be written in this form, and let $s$ be a finite string of base $y$ digits.  Note that there exists a positive integer $N$ such that the base $y$ expansions of $\log_z N$ and $\log_z(N+1)$ after the decimal point both start with $s$.  (This is true because the difference between $\log_z N$ and $\log_z(N+1)$ goes to $0$ as $N$ gets large.)  By hypothesis, there exist positive integers $a$ and $b$ such that $$N=\bigg\lfloor \frac{x^{y^a}}{z^b}\bigg\rfloor.$$  This means that $$\log_z \frac{x^{y^a}}{z^b}=y^aC-b$$ is between $\log_z N$ and $\log_z(N+1)$, and thus its base $y$ expansion starts with $s$ after the decimal point.  This implies that the string $s$ appears in the base $y$ expansion of $C$ (starting $a$ digits after the decimal point), as desired.
