Generating numbers by repeated doubling and digit reversal Let $S$ be the smallest set of positive integers satisfying the following conditions:


*$1 \in S$,

*If $n \in S$ then $2n \in S$,

*If $n \in S$ then the digit reversal of $n$ is also in $S$.
We assume that any leading zeros are dropped after digit reversal. For example, the digit reversal of $12300$ is $321.$
Is it true that $S$ contains all positive integers, except those divisible by $3$ or $11$?
EDIT: I have verified the conjecture up to $n = 10\,000$.

This is to show one can arrive at $5$ by several steps. I did this "backward" by continually starting with a goal number and allowing the moves: multiply by $5$, reversal, divide an even number by $2$ To explain the multiply by $5$ step, suppose the goal number was $17.$ Then $17 \cdot 5=85,$ and if we got to $85$ somehow, then $2 \cdot 85=170,$ reversed is $71,$ and reversed again is our goal number $17.$ So this is a kind of attempt at backtracking from some goal number.
For the goal of $5$ I multiplied by $5$ to get $25,$ reversed to get $52,$ then divided by $2$ twice to get $13.$ Thus if $13$ can be obtained, so can $5.$
From here on I'll just present the links, each step being doubling or reversal, without showing the backtracking method above outlined.
From 13 to 5: $13,26,52,25,50,5.$
From 7 to 13: $7,14,28,56,65,130,31,13.$
From 19 to 7: $19,91,182,281,562,265,530,35,70,7$
From 64 to 19: 
$$64,46,92,184,368,736,1472,2741,5482,2845,5690,965,\\ 
1930,391,193,386,772,277,554,455,910,19.$$
Then since powers of $2$ such as $64$ are obtained from $1$ by doubling, we can put the above together and get a (long) derivation of the goal number $5.$
There is a lot of choice to be made while "backtracking" but a subgoal is to backtrack to a reasonably small number, and hope that from there more backtracking will eventually lead to something known to be attainable already (such as a power of 2). As in the above example for obtaining 5, the numbers arrived at in between may become large, and still later small again. Not a sure-fire method, for example a failure to backtrack in one (or several) way(s)doesn't rule a number out.
