Here is a much simpler version of Taranovsky's question: Let f(0)=1, f(n+1)=2^f(n). Thus f(0)=1, f(1)=2, f(2)=4, f(3)=16, f(4)=65536, etc. What is the "natural" value for f(3.5)? (to the nearest integer if this is easier, but an algorithm for calculating arbitrarily close approximations would be preferable). -- JS