Question

数列 $(a_i)$ 由递推式 $a_{n+2} = \frac{a_n + 2009}{1 + a_{n+1}}$ 定义，其中 $n \ge 1$ ，且各项都是正整数。求 $a_1 + a_2$ 的最小可能值。

The terms of the sequence $(a_i)$ defined by $a_{n+2} = \frac{a_n + 2009}{1 + a_{n+1}}$ for $n \ge 1$ are positive integers. Find the minimum possible value of $a_1 + a_2.$

答案

答案：90

知识点：递推极端原理质因数分解

难度评级：3060

解答：

清除分母得，对所有 $n \ge 1$ 有 $a_{n+2}(1 + a_{n+1}) = a_n + 2009$ 。将每个式子与下一个式子相减，得到 $\begin{aligned} &a_{n+2} - a_n \\ &= (a_{n+2} + 1)(a_{n+3} - a_{n+1}). \end{aligned}$

如果某个差 $a_{n+2} - a_n$ 非零，那么之后每一个这样的差也都非零；又因为每个 $a_{n+2} + 1 \ge 2$ ，上式会迫使 $|a_3 - a_1|$ $\gt |a_4 - a_2|$ $\gt |a_5 - a_3|$ $\gt \cdots$ ，形成一个无限严格递减的正整数序列，矛盾。因此对所有 $n$ 都有 $a_{n+2} = a_n$ ：奇数项全相等，偶数项全相等，而且任意这样的正整数选择都可行。

此时递推式变为 $a_1(1 + a_2) = a_1 + 2009$ ，所以 $a_1 a_2 = 2009 = 7^2 \cdot 41$ 。在 $2009$ 的因数对中，和最小的是 $41 \cdot 49$ ，得 $41 + 49 = 90$ 。

Clearing denominators, $a_{n+2}(1 + a_{n+1}) = a_n + 2009$ for all $n \ge 1.$ Subtracting each instance from the next gives $\begin{aligned} &a_{n+2} - a_n \\ &= (a_{n+2} + 1)(a_{n+3} - a_{n+1}). \end{aligned}$

If some difference $a_{n+2} - a_n$ were nonzero, then every later difference would be nonzero as well, and since each $a_{n+2} + 1 \ge 2,$ the identity would force $|a_3 - a_1|$ $\gt |a_4 - a_2|$ $\gt |a_5 - a_3|$ $\gt \cdots,$ an infinite strictly decreasing sequence of positive integers — impossible. Hence $a_{n+2} = a_n$ for all $n:$ the odd-indexed terms are all equal and the even-indexed terms are all equal, and any such choice of positive integers works.

The recursion then reads $a_1(1 + a_2) = a_1 + 2009,$ so $a_1 a_2 = 2009 = 7^2 \cdot 41.$ Among the factor pairs of $2009,$ the sum is smallest for $41 \cdot 49,$ giving $41 + 49 = 90.$

2009 AIME I 第 13 题

答案：90

难度评级：3060

解答：

其他年份的第 13 题