22 Nov
2013
22 Nov
'13
2:09 p.m.
Poor wording on my part. There are sqrt(N) base-2 palindromes and the probability any one of them is a base-3 palindrome is 1/sqrt(N), so the expected number of dual palindromes is order 1. But that applies for every N. -Veit On Nov 22, 2013, at 2:30 PM, Andy Latto <andy.latto@pobox.com> wrote:
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.
No, the probability is approximate 1/(sqrt N) (from my estimates) or 2/(sqrt N) (Dan's more careful estimates) that a number between 2^N and 2^N+1 - 1 is a base-3 palindrome, and this value is not N-independent.
Andy