China Southeastern Mathematical Olympiad

By $a_1=2$, $a_2=\min \{\lambda \mid \frac{1}{a_1}+\frac{1}{\lambda }<1,\lambda \in N^{\ast }\}$, consider $\frac{1}{a_1}+\frac{1}{\lambda }<1$, then $\frac{1}{\lambda }<1-\frac{1}{2}=\frac{1}{2}$, $\lambda >2$, hence $a_2=3$. So the conclusion is true for $n=1$.

Suppose that the conclusion is true for all integer $n\le k-1$ ($k\ge 2$). If $n=k$, then $a_{k+1}=\min \{\lambda \mid \frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_k}+\frac{1}{\lambda }<1,\lambda \in N^{\ast }\}$.

Considering $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_k}+\frac{1}{\lambda }<1,$$ that is, $0<\frac{1}{\lambda }<1-\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_k}\right)$, we have $$\lambda >\frac{1}{1-\frac{1}{a_1}-\frac{1}{a_2}-\cdots -\frac{1}{a_k}}.$$ In the following, we show that $$\frac{1}{1-\frac{1}{a_1}-\frac{1}{a_2}-\cdots -\frac{1}{a_k}}=a_k(a_k-1).$$ By the induction hypotheses, for $2\le n\le k$, $a_n=a_{n-1}(a_{n-1}-1)+1$, we have $$\frac{1}{a_n-1}=\frac{1}{a_{n-1}(a_{n-1}-1)}=\frac{1}{a_{n-1}-1}-\frac{1}{a_{n-1}},$$ therefore $\frac{1}{a_{n-1}}=\frac{1}{a_{n-1}-1}-\frac{1}{a_n-1}$. By taking the sum, we have $$\sum _{i=2}^k\frac{1}{a_{i-1}}=1-\frac{1}{a_k-1},\text{ that is}$$ $$\sum _{i=1}^k\frac{1}{a_i}=1-\frac{1}{a_k-1}+\frac{1}{a_k}=1-\frac{1}{a_k(a_k-1)}.$$ Consequently, $\frac{1}{1-\frac{1}{a_1}-\frac{1}{a_2}-\cdots -\frac{1}{a_k}}=a_k(a_k-1)$. Therefore, $$\begin{array}{rl}{a}_{k+1}& =\min \left\{\lambda \mid \frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_k}+\frac{1}{\lambda }<1,\lambda \in {\mathbb{N}}^{\ast }\right\}\\ & =a_k(a_k-1)+1.\end{array}$$

By induction on $n$, for all positive integer $n$, we have $a_{n+1}=a_n^2-a_n+1$. $\square$