Consider the following five functions defined on positive reals: f1(x) = ((x^x)^x)^x; f2(x) = (x^(x^x))^x; f3(x) = x^((x^x)^x); f4(x) = x^(x^(x^x )); f5(x) = (x^x)^(x^x); Note that f2 = f5. There are really only 4 different functions of this type. We know the Catalan number C(n-1) gives the number of ways to parenthesize a product of n variables. If we use x^x for the binary operation and identify products that give equal functions when x is positive, how many products do we get? If the number of x's is n let e(n) be the number of distinct functions so obtained. I used Maple to simplify the set of C(n-1) expressions and got the following cardinalities for the simplified sets for n from 1 to 11: (*) 1,1,2,4,9,20,48,115,286,719,1842 Except for some of the smaller values, I cannot be sure that Maple has not failed to identify some equal functions. So this may not be the sequence e(n). The sequence (*) coincides with A000081: rooted trees with n nodes --- at least as far as I have computed. Is this known? If not can anyone prove that the sequences coincide? Maple undoubtedly uses standard laws of exponents to simplify these expressions. This in itself raises a number of questions. Edwin Clark