译文：Record monotonicity
The recurrence shows th在 f f f 是nondecreasing. Iterating f(2r)=f(2r−2)+f(r) f(2r)=f(2r-2)+f(r)f(2r)=f(2r−2)+f(r) 译文： gives f(2N)=∑j=0Nf(j) f(2N)=\sum_{j=0}^{N}f(j)f(2N)=∑j=0Nf(j). 因此 f(2n+1)<(2n+1)f(2n) f(2^{n+1})<(2^n+1)f(2^n)f(2n+1)<(2n+1)f(2n) 对于 n≥2 n\ge2n≥2, 因为和 has 2n+12^n+12n+1 terms 且所有terms except harmless j=0 j=0j=0 是在most f(2n) f(2^n)f(2n).