Problem 729

Range of periodic sequence

Consider the sequence of real numbers $a_{n}$ defined by the starting value $a_{0}$ and the recurrence $a_{n + 1} = a_{n} - \frac{1}{a_{n}}$ for any $n \geq 0$ .

For some starting values $a_{0}$ the sequence will be periodic. For example, $a_{0} = \sqrt{\frac{1}{2}}$ yields the sequence: $\sqrt{\frac{1}{2}}, - \sqrt{\frac{1}{2}}, \sqrt{\frac{1}{2}}, \dots$

We are interested in the range of such a periodic sequence which is the difference between the maximum and minimum of the sequence. For example, the range of the sequence above would be $\sqrt{\frac{1}{2}} - (- \sqrt{\frac{1}{2}}) = \sqrt{2}$ .

Let $S (P)$ be the sum of the ranges of all such periodic sequences with a period not exceeding $P$ .
For example, $S (2) = 2 \sqrt{2} \approx 2.8284$ , being the sum of the ranges of the two sequences starting with $a_{0} = \sqrt{\frac{1}{2}}$ and $a_{0} = - \sqrt{\frac{1}{2}}$ .
You are given $S (3) \approx 14.6461$ and $S (5) \approx 124.1056$ .

Find $S (25)$ , rounded to $4$ decimal places.

周期数列的极差

考虑实数数列 $a_{n}$ ，初值为 $a_{0}$ ，对于任意 $n \geq 0$ 的递推式为 $a_{n + 1} = a_{n} - \frac{1}{a_{n}}$ 。

对于某些初值 $a_{0}$ ，该数列为周期数列。例如，若 $a_{0} = \sqrt{\frac{1}{2}}$ ，则数列为： $\sqrt{\frac{1}{2}}, - \sqrt{\frac{1}{2}}, \sqrt{\frac{1}{2}}, \dots$

我们希望研究这类周期数列的极差，也即数列中的最大值和最小值之差。例如，在上述数列中，极差是 $\sqrt{\frac{1}{2}} - (- \sqrt{\frac{1}{2}}) = \sqrt{2}$ 。

记 $S (P)$ 为所有周期不超过 $P$ 的周期数列的极差之和。
例如， $S (2) = 2 \sqrt{2} \approx 2.8284$ ，对应的周期数列初值分别是 $a_{0} = \sqrt{\frac{1}{2}}$ 和 $a_{0} = - \sqrt{\frac{1}{2}}$ 。
已知 $S (3) \approx 14.6461$ ， $S (5) \approx 124.1056$ 。

求 $S (25)$ ，并保留 $4$ 位小数。