A Diophantine property of these numbers: ((a(n+1) - a(n-1))/2)^2 - 255*a(n)^2 = 1.
More generally, for t(m) = m + sqrt(m^2-1) and u(n) = (t(m)^(n+1) - 1/t(m)^(n+1))/(t(m) - 1/t(m)), we can verify that ((u(n+1) - u(n-1))/2)^2 - (m^2-1)*u(n)^2 = 1. (End)
a(n) equals the number of 01-avoiding words of length n on alphabet {0,1,...,31}. - Milan Janjic, Jan 26 2015