[waffletower]

Definitely 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.

[parsimo2010]

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 it is to prove that the list is finite in a different base.

[ethanwillis]

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.

[ethanwillis]

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.

[RheingoldRiver]

> the largest number where all digits are prime numbers

do you mean the largest prime number where all digits are prime numbers?