I was watching this video on YouTube where it is told (at 6:26) that $2^{16} = 65536$ has no powers of $2$ in it when represented in base-$10$. Then he - I think as a joke - says "Go on, find another power of $2$ that doesn't have a power of $2$ digit within it. I dare you!"
So I did. :) I wrote this little Python program to check for this kind of numbers:
toThePower = 0
possiblyNoPower = True
while True:
number = str(2**toThePower)
for digit in number:
if int(digit) in [1,2,4,8]:
possiblyNoPower = False
print('Not ' + number)
break
if possiblyNoPower:
print(number + ' has no digit that is a power of 2.')
toThePower += 1
possiblyNoPower = True
Sidenote: I could use the programming language Julia instead of Python, which may be much quicker, but I already checked for really big numbers and such a program (and brute-force in general) will never proof that there are no other powers of $2$ having this property. It might disprove it, but I think the chance is really really small.
I checked all the way to $2^{23826}$, which is a 7173 digit number, but no luck. Since the numbers are getting more and more digits with bigger powers of $2$, the chance of a number having no digit that is a power of $2$ becomes smaller and smaller.
I made a plot of $\frac{\text{number of digits being a power of 2}}{\text{total number of digits}}$ versus the $n$th power of $2$ on a logarithmic scale.
This graph is wrong! See the edit.
As I predicted, the graph drops really fast to almost $0$. I think that $\frac{\text{number of digits being a power of 2}}{\text{total number of digits}}\rightarrow 0$ as $n \rightarrow \infty$ and thus $\text{P}(n\text{th power of 2 having no digits of 2 in it}) \rightarrow 0$, but this is just my intuition.
So my question: Is $2^{16} = 65536$ the only power of $2$ that has no digit in it that is a power of $2$ (so no digit $\in \{1,2,4,8\}$) when represented in base-$10$? Is there a proof, a counterexample, or is it an open question? I'm also curious about powers of $2$ having no digits that are a power of $2$ in other bases than $10$.
Edit: As @Aweygan noted, the graph above is wrong. I accidentally divided by the number itself, instead of the amount of digits the number has. Below a good version, on a linear scale.
From this graph it appears that $\frac{\text{number of digits being a power of 2}}{\text{total number of digits}} \rightarrow 0.4 $ as $n \rightarrow \infty$. This seems to make sense, since $\text{P}(\text{digit} \in {1,2,4,8}) = 4/10 = 0.4$ and since the number of digits becomes larger and larger, the law of large numbers becomes "visible".
About the possible duplicate: that question was posted 4.5 years ago. The status of the problem (proven, disproven, open question) might well be changed in the maintime. :)