证明： f(n)=⌊(5+12)n⌋ f(n) = \Bigl\lfloor\left(\frac{\sqrt{5}+1}{2}\right)n\Bigr\rfloor f(n)=⌊(2√5+1)n⌋ 为初等函数。
f(n)=maxx≤2n.x≤5+12nf(n) = \max x \leq 2n. x \leq \frac{\sqrt{5}+1}{2}nf(n)=maxx≤2n.x≤2√5+1n
f(n)=maxx≤2n.2x≤5n+nf(n) = \max x \leq 2n. 2x \leq \sqrt{5} n+nf(n)=maxx≤2n.2x≤√5n+n
f(n)=maxx≤2n.2x−¨n≤5nf(n) = \max x \leq 2n. 2x \ddot{-}n \leq \sqrt{5}nf(n)=maxx≤2n.2x−¨n≤√5n
f(n)=maxx≤2n.(2x−¨n)2≤5n2f(n) = \max x \leq 2n. (2x \ddot{-}n)^2 \leq 5n^2f(n)=maxx≤2n.(2x−¨n)2≤5n2
f(n)=maxx≤2n.4x2≤4n2+4xnf(n) = \max x \leq 2n. 4x^2 \leq 4n^2 + 4xnf(n)=maxx≤2n.4x2≤4n2+4xn
f(n)=maxx≤2n.x2−˙(n2+xn)∈EFf(n) = \max x \leq 2n. x^2 \dot{-} (n^2 + xn) \in \mathcal{EF}f(n)=maxx≤2n.x2−˙(n2+xn)∈EF