2 Dec
2017
2 Dec
'17
12:15 p.m.
Suppose ^ means exponentiation and we want to count the *different integers* that 2^2^...^2 is equal to (N 2's in all) using various parenthesizations. E.g. #(2^2) = #{4} = 1 #(2^2^2) = #{16} = 1 #(2^2^2^2) = #{64, 256, 65536} = 3 #(2^2^2^2^2) = ??? Of course, any 3 twos grouped together make 4^2 = 2^4 = 16. Is there an asymptotic formula? Or better, an exact one? —Dan