f(n1+n2)=f(n1)f(n2),又f(2)=4
f(2)=f(1+1)=[f(1)]^2
f(n)>0
f(1)=2
f(2)=4
f(3)=f(1+2)=f(1)f(2)=8
f(4)=f(1+3)=f(1)f(3)=16
……
f(n)=2^n