My naive guess was that about sqrt(N) numbers less than N are palindromes to base B; If N = B^K, then there are roughly N^(K/2) palindromes, because once we specify the first half of the digits, the other half are determined. And B^(K/2) = sqrt(B^K). So the probability that a number < N is palindromic in base B is roughly 1/sqrt(N). So if they behave like independent events, the chance that a number < N is palindromic in 2 bases is roughly 1/N, suggesting there will be roughly one such number, regardless of N. So I would also guess that there are finitely many numbers palindromic in any 2 fixed relatively-prime bases, but I have no idea how to prove this. Andy On Thu, Nov 21, 2013 at 8:13 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Thanks for the pointer to OEIS, Hans, and thanks to Neil for enabling it!
OK, my most recent estimate of the number of dual palindromes -- assuming the two number bases are in a sense "independent" as I'd expect 2 and 3 to be -- is that the total number is probably *finite*.
So, QUESTION: Is the number of dual palindromes to bases 2 and 3 finite? ------ What about to any two different prime number bases?
--Dan
On 2013-11-21, at 2:44 PM, Dan Asimov wrote:
Here's a question a friend and I have been looking at very casually, with only the partialest results:
Given two bases B and C, what is the asymptotic fraction of positive integers that are palindromes when expressed to each base?
Perhaps to make things simpler, assume that B and C are relatively prime.
Or are each prime.
In fact, just for definiteness, what is the expected number of integers between 1 and N that are palindromes both to base 2 and to base 3 ???
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