First, I ran some bruteforce myself (I don't have access to Mathematica at the moment), and I'm fairly sure the mysterious numbers obtained in OP ($1, 2, 2, 2, 4\ldots$) are the *minimal possible numbers that are present in any $S_m(2n)$*. I'm going to explain the pattern in this assumption.

Long story short, the minimal possible number in $S_m(n)$ is $2^{\left\lfloor \frac{\log n}{\log 3} \right\rfloor}$.

As follows from the OP's explanation, for any odd number $x \leq n$ there is *exactly one* representative of form $x 2^k$ in any $S_m(n)$. Let us call the exponent $k = d(x)$ the *degree* of $x$ in a particular $S_m(n)$. Obviously, for non-equal $x \vert y$ we must $d(x) > d(y)$.

One can see that the "critical" values where the OP's minimal numbers change are exactly $\frac{3^k + 1}{2}$, that is, the minimum changes whenever a new power of 3 arrives in the set. One can indeed see that in any $S_m(n)$ we have $d(1) \geq \left\lfloor \frac{\log n}{\log 3} \right\rfloor = k_0$ by simply applying the inequality above for a divisor chain $1 \vert 3 \vert \ldots \vert 3^{k_0}$. The numbers $2^{k_0}$ follow the OP's pattern exactly.

But why any number that is not a binary power cannot be less than $2^{k_0}$ in an $S_m(n)$? Let $1 < x \leq n$ be an odd number. Then for any $S_m(n)$ the numbers $\{2^{d(x)}, 3\cdot 2^{d(3x)}, \ldots\}$ must be an $S_m(\left\lfloor \frac{n}{x} \right\rfloor)$. As we have shown above, that implies $d(x) \geq \left\lfloor \frac{\log \lfloor n/x \rfloor}{\log 3} \right \rfloor$. It suffices to show $2^{\left\lfloor \frac{\log n}{\log 3} \right\rfloor} < x \cdot 2^{\left\lfloor \frac{\log \lfloor n/x \rfloor}{\log 3} \right\rfloor}$. But $2^{\left\lfloor \frac{\log n}{\log 3} \right\rfloor - \left\lfloor \frac{\log \lfloor n/x \rfloor}{\log 3} \right\rfloor} \leq 2^{\frac{\log x}{\log 3} + 1} = 2 \cdot x^{\frac{\log 2}{\log 3}} < x$ for all $x > 5$ (for $x = 3, 5$ we may still have the claim by handling the rounding error more carefully).

To finish the proof, notice that the set $\{x \cdot 2^{\left\lfloor \frac{\log \lfloor n/x \rfloor}{\log 3} \right\rfloor}\}$ with $x$ ranging over odd numbers at most $n$ is an $S_m(n)$. Note that this set is minimal with respect to any measure, e.g. no proper divisor of any of its elements can be in an $S_m(n)$, hence it's also lexicographically minimal when sorted.

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