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Problem 729

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Range of periodic sequence

Consider the sequence of real numbers $a_n$ defined by the starting value $a_0$ and the recurrence $\displaystyle a_{n+1}=a_n-\frac 1 {a_n}$ for any $n \ge 0$.

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.

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周期数列的极差

考虑实数数列$a_n$，初值为$a_0$，对于任意$n \ge 0$的递推式为$\displaystyle a_{n+1}=a_n-\frac 1 {a_n}$。

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。

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