Maybe I'm missing something, but your argument shows that given a random base-2 palindrome between 2^N and 2^(N+1)-1 there is some N-independent probability (in the limit of large N) that this number is also a base-3 palindrome. In other words, the number of dual palindromes is not finite, but has uniform density on a logarithmic scale. -Veit On Nov 22, 2013, at 11:05 AM, Dan Asimov <dasimov@earthlink.net> wrote:
Yep, that's roughly the same argument I used, but with a bit more care, to get a fraction of ~ 4/N numbers that are palindromes to both bases B and C.
Which is why I thought the total number of dual palindromes may be finite, especially if gcd(B,C) = 1.
--Dan
On 2013-11-21, at 11:25 PM, Andy Latto wrote:
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
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