令f(1)=a,若a=1,f(f(1))=f(a)=f(1)=1≠3矛盾,∴a≠1
若a=3,f(f(1))=f(3)=3,与f(x)在(0,正无穷)单调递增矛盾
又f(n)属于N+,得a=2
f(2)=f(f(1))=3
f(3)=f(f(2))=6
f(6)=f(f(3))=9
又f(x)在(0,正无穷)单调递增,f(n)属于N+
所以f(4)=7 (f(3)=6,f(4)=7,f(5)=8,f(6)=9)