我来个低端做法
大致列一下$a_n:1,2,3,4,5,7,8,9,11,13,15,16,17,19,21,\cdots,$注意到对于有限的前$n$项,多数数字来自集合$A,$不妨当成$a_n=2n-1,$则$S_n>12a_{n+1}$的解是$n\geqslant 25.$首先尝试$n=25,$则$a_{25}=39,a_{26}=41,S_{25}=462<12a_{26}=492,$不满足题意;再尝试$n=26,$则$S_{26}=503<12a_{27}=516,$仍不满足;再尝试$n=27,$则$S_{27}=546>12a_{27}=540,$满足题意.故猜想$n$的最小值为27,只需再证明当$n<27$时,$S_n\leqslant 12a_{n+1}$即可.作为考试,毛想想即可,不需再证明,作为无时间限制的非考试解题,穷举即可 |