TITLE: Which negative integer powers of 2 belong to the Cantor Set?
QUESTION [7 upvotes]: Consider the Cantor set $C$, and negative integer powers $2^{-k}$. 
Clearly, for $k=1$,   $2^{-1} \notin C$ since $1/2 \in (1/3, 2/3)$, the first deleted open interval.
It is known that $1/4 = 2^{-2} \in C$, since it alternates between the lower and upper thirds of each non-deleted interval in every iteration, i.e., it lies in $L_1 = [0, 1/3], U_2 = [2/9, 1/3] ...$ and so on. 
Since $2^{-3} \in (1/9, 2/9)$, a deleted interval, $2^{-3} \notin C$.
A slightly more tedious procedure shows that since $2^{-4} \in (1/27, 2/27)$, $2^{-4} \notin C$.
My question is as follows:
Which negative powers of 2 belong to the Cantor set?  Does it contain any powers other than $1/4$?
It seems easy enough to determine this question for special cases (as shown above), but I wonder about the general case. 
PS: The alternating interval proof for $1/4$ is from the text by Kolmogorov and Fomin.

REPLY [7 votes]: In this paper by Charles R. Wall, it is shown that there are only fourteen numbers with terminating decimal expansion in the Cantor set (sixteen if you include $0$ and $1$), they are:
$$\frac{1}{4}, \frac{3}{4}, \frac{1}{10}, \frac{3}{10}, \frac{7}{10}, \frac{9}{10}, \frac{1}{40}, \frac{3}{40}, \frac{9}{40}, \frac{13}{40}, \frac{27}{40}, \frac{31}{40}, \frac{37}{40}, \frac{39}{40}.$$
As every number of the form $2^{-k}$ has finite decimal expansion, any such number in the Cantor set must appear in the above list. As such, the only number of the form $2^{-k}$ which is an element of the Cantor set is $\frac{1}{4}$.