by Hbruz0 on 3/20/25, 11:55 AM with 116 comments
by nneonneo on 3/20/25, 9:15 PM
Context: I wrote a search program that is substantially faster - it takes just a few minutes to get up to 2^(10^13), although my laptop's limited memory is starting to be a problem (my intermediate result file is already nearly 1GB in size). Unfortunately, it seems there are no results up to 2^15258789062500, which is a 4.5-trillion digit number.
by bluewin on 3/20/25, 6:16 PM
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.
by WithinReason on 3/20/25, 12:43 PM
How did he do this?
by waffletower on 3/20/25, 4:25 PM
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.
by hrldcpr on 3/20/25, 1:10 PM
by andrewla on 3/20/25, 12:54 PM
by IsTom on 3/20/25, 2:23 PM
by Mae_soph on 3/21/25, 12:32 AM
Assume this list is complete up to 10^n. We find the biggest l such that 2^(5^(l - 1)*4) < 10^n. Let us consider the 10^(n+1) > 2^k > 10^n such that 2^k has all even digits. By cyclicity of powers of 2 mod 10^l (that's why we chose this l), this means that 2^(k - 1) = a*10^l + b, where a is some integer and b is 1,2,4,32 or 1024 (because those are the only options with digits less than 5 mod 10^l). If l > 10,that means that we can divide by b to get 2^(k-1)/b = c*10^d + 1 where c and d are nonzero integers. But this is a contradiction.
Now we only need to show up to 2^(5^10 * 4) to allow l > 10, which has already been done by other comments.
by ted_dunning on 3/26/25, 6:51 PM
https://github.com/tdunning/EvenDigits
This uses much higher order sieves so that it runs about 32000 times faster than the naive algorithm and was able to search to this point on a single core. It is also possible to thread this algorithm relatively easily.
by Aardwolf on 3/20/25, 9:07 PM
by openasocket on 3/20/25, 5:25 PM
by lanna on 3/20/25, 1:34 PM
by vanderZwan on 3/21/25, 8:27 AM
1: it's been too long since we've had a Neil Sloane episode, which are always a highlight
2: it sounds like the kind of thing where just a little bit more attention from maths enthusiasts will result in a proof of the sequence being finite (or not) very quickly
by bitwize on 3/20/25, 4:20 PM
by jmount on 3/20/25, 9:41 PM
by chasing on 3/20/25, 3:31 PM
by kristopolous on 3/21/25, 5:22 AM
As in, could you somehow solve it without knocking the other two down in a major way?
by froh on 3/20/25, 9:01 PM
https://en.wikipedia.org/wiki/Double_dabble
for 2^n only zeroes are shifted in, to all eternity. thus the lowest digits go through a fixed cycle.
as the top but is shifted to the left in each shift+add-threes-where-needed cycle, and leaves "it's" bcd digit after four such cycles, I intuit the next bcd byte will also switch to some cycle, as it's 'input' is boringly deterministic: all zeroes for the lowest digit, leading to 1, 2, 4, 8 (1)6, (1)2, 4, 8, (1)6, (1)2, ... so 0000(1100)* is shifted in to the tens digit.
that gives 0,0,0,0, 0+1, 2+1, 6, (1)2, 4+1, (1)0+1, 2, 4, 8+1, (1)8+1, (1)8, (1)6, (1)2+1, 6+1, (1)4, 8, (1)6+1, (1)4+1, (1)0, 0, 0+1, 2+1, ... for the tens digit. which has a period of 20 ... with a shift to hundreds pattern of 0000(00010100011110101110)* and an odd odd even even rhythm on the tens digit.
noice.
some number nerds will for sure figure or know ways to spin this on for the hundreds digit. and determine the periodicity of having all the lowest n digits even. or the loss of that periodicity... because maybe just maybe this spins into some wheel where one of the digits foo to bar always is odd. and then you can stop searching...
but what do I know.
I just Dunning-Kruger an intuition that the "double dabble" bin2bcd _may_ be useful in this :-D
by millipede on 3/21/25, 12:54 AM
by darepublic on 3/22/25, 2:18 PM
by kazinator on 3/22/25, 8:37 AM
by netsharc on 3/20/25, 1:00 PM
2, 4, 8, 64, 2048 are powers of 2 (i.e. 2^n), and they don't contain odd numbers (e.g. 16, 128, 1024 contain 1 so are not in this list, same with 4096 containing 9).