Earlier quoted context omitted.
0^2 is a power of 0, not a power of 2.
https://en.wikipedia.org/wiki/Power_of_two#First_64_powers_o... ??
That list includes 2^0 = 1, which is odd. It does not include 0^2.
Powers of 2 with all even digits
31–40 of 123 posts
Re: Powers of 2 with all even digits
#32Re: Powers of 2 with all even digits
#33Earlier quoted context omitted.
Proofs of non-existence aren't usually straightforward.
I mean, clearly it isn't in this case. But given that the digits of 2^n are cyclical at each decimal position, it does feel like this should fall out of some sort of chinese remainder theorem manipulation.
True. It might also just be that the question hasn't attracted the attention of number theorists, and finding a proof wouldn't be unreasonably difficult to an expert in the field.
Re: Powers of 2 with all even digits
#34This is remarkable! I always find it fascinating that simple to express properties lack a proof. This is a very simple thing to evaluate and seems like it should be straightforward to establish that 2048 is the highest such power.
Everything about this seems so arbitrary. You look at the powers of an arbitrary number (here, 2), you pick an arbitrary base (here, 10) in which to express those powers, and ask for a random property of its digits (whether they belong to the set {0,2,4,6,8}).
Nothing about this question feels natural. I've noticed that random facts often don't have simple proofs.
Re: Powers of 2 with all even digits
#35Earlier quoted context omitted.
You are thinking of squares, not of powers of 2.
I see the difference in wording now, as its not very clear what they meant
It's actually very clear, and there's nothing wrong with admitting you're unfamiliar with the terminology, we all were at some point.
Re: Powers of 2 with all even digits
#36Earlier quoted context omitted.
You are thinking of squares, not of powers of 2.
I see the difference in wording now, as its not very clear what they meant
Saying “powers of two” is a universal way of denoting 2^n. It’s okay to admit being wrong rather than blaming OEIS for being vague.
Re: Powers of 2 with all even digits
#37[flagged]
Looking at the replies to your comment thread, I feel like you have to be trolling.
Re: Powers of 2 with all even digits
#38No additional terms up to 2^(10^10). - Michael S. Branicky, Apr 16 2023 How did he do this?
Hah, I had Michael Branicky as a professor (for his AI course) at the time (Jan - May 2023). Didn't expect to see his name here. He's a brilliant guy.
Re: Powers of 2 with all even digits
#39Earlier quoted context omitted.
I mean, clearly it isn't in this case. But given that the digits of 2^n are cyclical at each decimal position, it does feel like this should fall out of some sort of chinese remainder theorem manipulation.
True. It might also just be that the question hasn't attracted the attention of number theorists, and finding a proof wouldn't be unreasonably difficult to an expert in the field.
Nope, it's not that easy in this case. E.g., Erdős conjectured in 1979 that every power of 2 greater than 256 has a digit '2' in its ternary expansion [0]. This makes sense heuristically, but no methods since then have come close to proving it.
Digits of numbers are a wild beast, and they're tough to pin down for a specific sequence. At best, we get statistical results like "almost all sequences of this form have this property", without actually being able to prove it for any one of them. (Except sometimes for artificially-constructed examples and counterexamples, or special classes like Pisot numbers.)
Re: Powers of 2 with all even digits
#40This is remarkable! I always find it fascinating that simple to express properties lack a proof. This is a very simple thing to evaluate and seems like it should be straightforward to establish that 2048 is the highest such power.
Base‑10 is just our chosen way of writing numbers, it doesn’t need to have any deep relationship with the arithmetic properties of sequences like the powers of 2. For most series (Fibonacci numbers, factorials etc), the digits for large members will be essentially random, their digits don't obey any pattern - it's just two unconnected things. It seems extremely likely that 2048 is the highest, but there might not be a good reason that could lead to a proof - it's just that larger and larger random numbers have less and less chance of satisfying the condition (with a tiny probability that they do, meaning we can't prove it).
Interestingly, there are results in the other kind of direction. Fields medalist James Maynard had an amazing result that there are infinitely many primes that have no 7s (or any other digit) in their decimal expansion. This actually _exploits_ the fact that there is no strong interaction between digits and primes - to show that they must exist with some density. That kind of approach can't work for finiteness though.