For those curious, one relevant field of mathematics that could be used to prove properties of this sequence would be Sieve theory: https://en.m.wikipedia.org/wiki/Sieve_theory
Powers of 2 with all even digits
71–80 of 123 posts
Re: Powers of 2 with all even digits
#72This 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.
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Re: Powers of 2 with all even digits
#73Earlier quoted context omitted.
Looks like that's all of them. The typical number of even digits of n grows like a constant times n, so you need some very large deviations from t I'd conjecture the number of powers of 2 with exactly m even digits is finite for all m.
Just tested up to 2 * 1000000, and it indeed looks like that's all of them.
Asterisk means exponent here?
Re: Powers of 2 with all even digits
#74Definitely not finite in radix-16 (hexadecimal): [2 4 8 10 20 40 80 100 200 400 800 1000 2000 4000 8000 10000 20000 40000 80000 100000 200000 400000 800000 1000000 ...] or radix-8 (octal): [2 4 10 20 40 100 200 400 1000 2000 4000 10000 20000 40000 100000 ...] Interesting puzzle due to radix representation and sequence interactions.
I'm not a number theorist, but I note that 16 is 2^4 and 8 is 2^3 (both powers of 2). Maybe there is a provable statement about whether these lists are finite in bases that are not 2^k, and maybe there is a bound on the length of the list by the value of log_2(base). I'm not going to write it out, there is certainly a proof that the list is infinite in base 2^k (for integer k >= 2). I'm more wondering about how hard…
when dealing with only even and odd they are not finite in base 2^k.
if we marked sequences of integers with 3 options. even, odd, other. then these lists are not finite in bases of 3^k.
for four options. even, odd, other, another. then these lists are not finite in bases of 4^k.
there is an intersection in the infinite lists where the base is equivalent to the power of an earlier base.
so infinite lists for 2^k would overlap a subset of the infinite lists for 2^2^k=4^k
all prime bases, p, p^k would admit infinite lists that cover all the infinite lists for some composite base, c, c^k.
Re: Powers of 2 with all even digits
#75Earlier quoted context omitted.
yeah that's weird - its kind of a pointless comment without an included algorithm or something
Here's a really dumb algorithm: for i in range(1, 10**10): for k in range(1, 5): s = str(pow(2, i, 10**(10**k))) if '1' in s or '3' in s or '5' in s or '7' in s or '9' in s: break else: print(2**i) It's really easily to parallelize, I was able to run it up to 10**8 in about 15min, so you would be able to run it up to 10**10 in a few hours with parallelization.
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Re: Powers of 2 with all even digits
#76Earlier quoted context omitted.
I'm not a number theorist, but I note that 16 is 2^4 and 8 is 2^3 (both powers of 2). Maybe there is a provable statement about whether these lists are finite in bases that are not 2^k, and maybe there is a bound on the length of the list by the value of log_2(base). I'm not going to write it out, there is certainly a proof that the list is infinite in base 2^k (for integer k >= 2). I'm more wondering about how hard…
when dealing with only even and odd they are not finite in base 2^k. if we marked sequences of integers with 3 options. even, odd, other. then these lists are not finite in bases of 3^k. for four options. even, odd, other, another. then these lists are not finite in bases of 4^k. there is an intersection in the infinite lists where the base is equivalent to the power of an earlier base. so infinite lists for 2^k woul…
there is another similar problem about the largest number where all digits are prime numbers. which afaik has only been proven in base 10.
similarly there the largest number with all prime digits actually differs if you ask the question in different bases.
and there is also a pattern that exists to predict what the number will be in a given base.
Re: Powers of 2 with all even digits
#77Earlier quoted context omitted.
> whether 2^k mod 10 is odd 2^k mod 10 is never odd; it's the cycle (2, 4, 8, 6). Related here is the length of the cycles mod 2^k, https://oeis.org/A005054 . Interestingly, the number of all-even-digit elements in those cycles does not appear to be in the oeis, I get 4, 10, 25, 60, 150 as the first five terms. This does appear to get more efficient as k gets higher; for k=11 I get a cycle length of 39,062,500 with a…
So I suppose if you ever find a cycle where the full cycle has no all-even members, you can prove that there are no more all-even numbers to find.
That's not possible, at minimum 2,4,8,64,2048 would all be the the cycle for `k >= 4`.
Re: Powers of 2 with all even digits
#78No additional terms up to 2^(10^10). - Michael S. Branicky, Apr 16 2023 How did he do this?
To prove that there is no value of k between 12 and 10^10 such that 2^k has all even digits, you only have to prove that there is an odd digit among the lowest X decimal digits for all 12 ≤ k ≤ 10^10.
The value of X necessary to prove this grows rather slowly compared to k. For example, the smallest power of 2 that doesn't have an odd digit in its last 16 digits is 2^12106. The smallest power of 2 that doesn't have an odd digit in its last 32 digits is 2^3789535319. So it makes sense to try increasingly large values of X until you are able to rule out all values of 2^k for k up to 10^10.
Here's a C++ program you can run to replicate this proof. It takes around 20 minutes to run, and can probably be optimized further, but it shows the principle: https://pastebin.com/DVK2JKdq
Re: Powers of 2 with all even digits
#79I worked on this once after an argument with my boyfriend.
The original argument was "the ones digit has permanent pattern in 2^n {2,4,8,6,2...}.
We made a system to generate digits for powers of two, although eventually we just made one that can take arbitrary bases, and found that you can decompose digit frequency and find a variety of NMR like resonances that vary based on where you terminate data collection.
It was really fun and this makes me want to get back into this so I could check the properties of those resonances across bases and stopping points for data collection.
Re: Powers of 2 with all even digits
#80How many powers of 2 have just a single even digit? 2, 4, 8, 16, 32, 512...
This is equivalent to asking: how many powers of 2 are there such that all digits except the leading digit are 5 or greater?
The powers of 2 with a single even digit are just those double those numbers (i.e., the next higher power of 2).