"number of distinct values taken by i^i^...^i (with n i's and parentheses inserted in all possible ways)" where i = sqrt(-1) and ^ denotes the principal value of the power function [A198683]

--If i were replaced by some generic number, then all the parenthesizations would yield distinct values and the count would be a known Catalan number (which is an upper bound on this i-based count, therefore). Right? WRONG!!! Because (x^x)^(x^x) = (x^(x^x))^x FOR *ANY* x !!! so you can actually get a stronger "generic" upper bound than the Catalan numbers! So the more interesting question seems to be: what is this (generic x) count? However, certain special numbers lower the count. In particular using 1 or -1 as x, there always is only one distinct value. So Reduced-Catalan is the maximum, and 1 is the minimum, count. I'm not sure what is so special about i from this point of view. A more systematic question would be to ask what real (or complex) values replacing "i" yield the minimum counts, in order of increasing count. I note (2^2)^2 = 4^2 = 16 and 2^(2^2) = 2^4 = 16 g^((g^g)^g) = (g^g)^(g^g) = (g^(g^g))^g where g= (1+sqrt(5))/2 = golden ratio = 1.618033988 [also 1/g works instead of g] L^(L^(L^L)) = ((L^L)^L)^L where L = ln(3)/LambertW(ln(3)) = 1.825455023 Q^(Q^(Q^Q)) = (Q^Q)^(Q^Q) = (Q^(Q^Q))^Q where Q=1.776775040 r^((r^r)^r) = ((r^r)^r)^r where r=sqrt(3)=1.732050808 [also -sqrt(3)] -- Warren D. Smith http://RangeVoting.org  <-- add your endorsement (by clicking "endorse" as 1st step)