COMMENTS
Are there any more terms in this sequence?
Evidence that the sequence may be finite, from Rick L. Shepherd, Jun 23 2002:
1) The sequence of last two digits of 2^n, A000855 of period 20, makes clear that 2^n > 4 must have n == 3, 6, 10, 11, or 19 (mod 20) for 2^n to be a member of this sequence. Otherwise, either the tens digit (in 10 cases), as seen directly, or the hundreds digit, in the 5 cases receiving a carry from the previous power’s tens digit >= 5, must be odd.
2) No additional term has been found for n up to 50000.
3) Furthermore, again for each n up to 50000, examining 2^n’s digits leftward from the rightmost but only until an odd digit was found, it was only once necessary to searc
7 Comments
WithinReason
No additional terms up to 2^(10^10). – Michael S. Branicky, Apr 16 2023
How did he do this?
andrewla
This 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.
netsharc
Somehow I missed the title and wondered what the fuck was going on…
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).
hrldcpr
The base 2 list is even shorter.
1970-01-01
0^2 is even
Please fix. You're welcome.
Edit: I see its NOT (x)^2 although its worded as it is
lanna
How many powers of 2 have just a single even digit? 2, 4, 8, 16, 32, 512…
IsTom
It might be finite, but it also has a "fast growing sequence" kind of smell too.