MathDB
f_0(x) = x, f_{2k+1} (x) = 3^{f_{2k}(x)}, $f_{2k+2} = 2^{f_{2k+1}(x)}

Source: Austrian Polish 1989 APMC

April 30, 2020
recurrence relationfunctionalgebra

Problem Statement

Functions f0,f1,f2,...f_0, f_1,f_2,... are recursively defined by f0(x)=xf_0(x) = x and f2k+1(x)=3f2k(x)f_{2k+1} (x) = 3^{f_{2k}(x)} and f2k+2=2f2k+1(x)f_{2k+2} = 2^{f_{2k+1}(x)}, k=0,1,2,...k = 0,1,2,... for all xRx \in R. Find the greater one of the numbers f10(1)f_{10}(1) and f9(2)f_9(2).