[math-fun] dual palindromes

The Latto-Asimov finiteness argument leaves me unimpressed. Essentially same argument instead predicts infinitude: The number of N-digit binary numbers that are palindromic is 2^(floor(N/2)-1+parity(N)) which is bounded between two positive constants times sqrt(2^N). Assuming base-2 and base-3 palindromitude are "independent" then the expected number of N-bit numbers that work in both radices is bounded between two positive constants. Now summing over N, we find the total number of dual palindromes is infinite and the predicted number below B is of order logB. In fact under the probabilistic independence assumption I can prove an infinity exist with probability=1.