一道关于函数的问题。。请教高手。。高分

f(9) = fff(3) = 3f(3) = 3fff(1) =9f(1)
f(96) = fff(32)=3f(32)

f(1)属于N*
若f(1)=1
则ff(1) = f(1) = 3矛盾
若f(1)>=3
f(f(1)) = 3 <= f(1)
由单调性知f(1) <= 1 矛盾
故只有f(1)=2
f(2)=ff(1)=3
f(3)=ff(2)=6
f(6)=ff(3)=9
由严格单调性6=f(3)<f(4)<f(5)<f(6)=9
=>
f(4)=7,f(5)=8
=>

f(9)=ff(6)=18
f(12)=ff(7)=21
同理根据18=f(9)<f(10)<f(11)<f(12)=21
知：f(10)=19,f(11)=20
=>
f(19)=ff(10)=30
f(20)=ff(11)=33
=>
f(30)=ff(19)=57
f(33)=ff(20)=60
知：f(31)=58,f(32)=59

所以原式 = 10f(1) + 3f(32) = 10*2 + 3*59 = 197