This question asks about the reachability of some specific numbers via a procedure that starts from the number 1 and where a valid step is to either double the current number to yield a new number, or to arbitrarily rearrange the digits of the current number to yield a new number (subject to the restriction that zeroes cannot be rearranged to the front). For instance, the two-digit numbers that can be obtained are $16$ (from $1\mapsto2\mapsto4\mapsto8\mapsto16$; the one-digit members of the sequence are, of course, immutable), $32$ (from $16$), $64$ (from $32$), $61$ (from $16$), $23$ (from $32$), $46$ (from either $23$ or $64$), $92$ (from $46$), $29$ (from $92$), $58$ (from $29$), and $85$; in total, there are $4$ reachable 1-digit numbers and $10$ reachable 2-digit numbers.
My question is about the asymptotic order of growth of this set: (approximately) how many $\leq n$-digit numbers are reachable? My gut instinct would be that this number is $\Theta(C^n)$ for some $C\lt 10$, but I don't even have a great heuristic argument (though one could possibly be constructed from an analysis of multiset permutations) and I wouldn't be shocked to learn that it's actually $\frac2310^n-o(10^n)$, or even that every possible sufficiently large number is reachable. (The $\frac23$ factor here comes from the fact that no multiple of $3$ is reachable — rearrangement can't change a number's value $\mod 3$ and doubling can never make a number that wasn't already a multiple of $3$ into one. No number $\gt 1$ with all of its digits odd is reachable either, because such a number can't have any valid ancestor not of this form, but there are only approximately $\frac54\cdot5^n$ such numbers). Problems of this sort tend to be notoriously hard, but not necessarily completely inaccessible; has anyone studied this?
