证明 lower bound
Iterating recurrence 对于 f(2n+1) f(2^{n+1})f(2n+1) 且applying pairing lemma 到resulting block gives f(2n+1)>2nf(2n−1) f(2^{n+1})>2^n f(2^{n-1})f(2n+1)>2nf(2n−1). Starting 从 f(2)=2 f(2)=2f(2)=2 且 f(4)=4 f(4)=4f(4)=4, inducti上yields f(2n)>2n2/4 f(2^n)>2^{n^2/4}f(2n)>2n2/4 对每个 n≥3 n\ge3n≥3.