Let n= 1
f(f( 1))=3
f( 1)< f(2)& lt； F(3) is a positive integer, and f( 1) cannot be 1, otherwise f(f( 1))= 1. There must be.
f( 1)=2
f(2)=3
therefore
f(3)=f(f(2))=6
f(6)=f(f(3))=9
There are only 7 and 8 between the function values of 6 and 9.
Only f(5)=8.