how about?

f(1) = 0
f(2) = 1

for n>0,

f(2n+1) = 2^F(n)
f(2n) = 2^F(n-2)

where F(n) is the Fibonacci sequence F(0) = 0, F(1) = 1, F(n) = F(n-1)+F(n-2)

This seems to jive with Lisa's excellent solution