942

已知数列 ${\displaystyle \{a_n\}}$ 满足: ${\displaystyle a_1=0,a_{n+1}=a_n^2+c(n\in {\mathbb{N}}^{\ast },c\in \mathbb{R})}$.
当 ${\displaystyle n\ge 2}$ 时，记 ${\displaystyle M_n=\{c|\mathrm{\forall }i\in \{1,2,\cdots ,n\},|a_i|\le 2\},M=\{c|\mathrm{\forall }i\in {\mathbb{N}}^{\ast },|a_i|\le 2\}}$. 给出如下4个结论:

1. ${\displaystyle M_3=[-2,1]}$;
2. 当 ${\displaystyle {\displaystyle c>\frac{1}{4}}}$, 数列 ${\displaystyle \{a_n\}}$ 是递增数列;
3. 当 ${\displaystyle 0<c\le {\displaystyle \frac{1}{4}}}$ 时，存在正数 ${\displaystyle t}$ 使得 ${\displaystyle \mathrm{\exists }m\in {\mathbb{N}}^{\ast },\mathrm{\forall }n>m,|a_n-t|<2^{-100}}$;
4. 集合${\displaystyle M={\displaystyle [0,\frac{1}{4}]}}$.

其中正确命题的序号是_____.

1. 令 ${\displaystyle \{\begin{array}{rl}& |a_2|=|c|\le 2,\\ & |a_3|=|c^2+c|\le 2\end{array}\Rightarrow c\in [-2,1],}$ 故正确.
2. 作差得$$\begin{array}{r}{A}_{n+1}-a_n={(a_n-\frac{1}{2})}^2+c-\frac{1}{4}>0,\end{array}$$
   故正确.
3. ${\displaystyle n=2}$ 时 ${\displaystyle a_n=c<{\displaystyle \frac{1}{2}}}$, 假设 ${\displaystyle n=k}$ 时 ${\displaystyle a_k<{\displaystyle \frac{1}{2}}}$, 则$$\begin{array}{r}{A}_{k+1}=a_k^2+c<\frac{1}{4}+\frac{1}{4}<\frac{1}{2},\end{array}$$
   这样根据归纳原理恒有 ${\displaystyle {\displaystyle a_n<\frac{1}{2}}}$. 当 ${\displaystyle c={\displaystyle \frac{1}{4}}}$, 类似 2，有 ${\displaystyle \{a_n\}}$ 单增，且有界，故由单调有界收敛定理知 ${\displaystyle \{a_n\}}$ 收敛 (设 ${\displaystyle t={\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n}}$, 则由递推式 ${\displaystyle t=t^2+{\displaystyle \frac{1}{4}\Rightarrow t=\frac{1}{2}}}$.) 当 ${\displaystyle 0<c<{\displaystyle \frac{1}{4}}}$, 记 ${\displaystyle t={\displaystyle \frac{1-\sqrt{1-4c}}{2}\in (0,\frac{1}{2})\Rightarrow c=t-t^2}}$, 则 $$\begin{array}{r}|a_{n+1}-t|=|a_n^2+c-t|=|a_n^2-t^2|=|a_n+t||a_n-t|<(\frac{1}{2}+t)\cdot |a_n-t|,\end{array}$$
   则$$\begin{array}{r}|a_n-t|<{(\frac{1}{2}+t)}^{n-1}|a_1-t|=t{(\frac{1}{2}+t)}^{n-1}\to 0,n\to \mathrm{\infty },\end{array}$$
   这样我们证明了无论 ${\displaystyle c}$ 取何值，${\displaystyle \{a_n\}}$ 收敛，故正确.
4. 注意到 ${\displaystyle -2\in M}$, 因为 ${\displaystyle c=-2}$ 时，${\displaystyle a_n=\{\begin{array}{rl}& -2,n=2,\\ & 2,n\ge 3,\end{array}}$. 故 ${\displaystyle |a_n|\le 2,\mathrm{\forall }n\in {\mathbb{N}}^{\ast }}$. 这样明显错误.

综上，答案是 123.