Problem 794

Seventeen Points

This problem uses half open interval notation where $[a,b)$ represents $a \le x \lt b$.

A real number, $x_1$, is chosen in the interval $[0,1)$.

A second real number, $x_2$, is chosen such that each of $[0,\frac{1}{2})$ and $[\frac{1}{2},1)$ contains exactly one of $(x_1, x_2)$.

Continue such that on the $n$-th step a real number, $x_n$, is chosen so that each of the intervals $[\frac{k-1}{n}, \frac{k}{n})$ for $k \in {1, …, n}$ contains exactly one of $(x_1, x_2, …, x_n)$.

Define $F(n)$ to be the minimal value of the sum $x_1 + x_2 + … + x_n$ of a tuple $(x_1, x_2, …, x_n)$ chosen by such a procedure. For example, $F(4) = 1.5$ obtained with $(x_1, x_2, x_3, x_4) = (0, 0.75, 0.5, 0.25)$.

Surprisingly, no more than $17$ points can be chosen by this procedure.

Find $F(17)$ and give your answer rounded to $12$ decimal places.

十七个点

本题中使用的左闭右开区间记号$[a,b)$指的是满足$a \le x \lt b$的区间。

在区间$[0,1)$中选择第一个实数$x_1$。

再选择第二个实数$x_2$，满足在区间$[0,\frac{1}{2})$和$[\frac{1}{2},1)$上各包含$(x_1, x_2)$中的恰好一个实数。

继续这一选择过程，并始终满足在第$n$步时选择实数$x_n$，使得对于$k \in {1, …, n}$，每个区间$[\frac{k-1}{n}, \frac{k}{n})$都各包含$(x_1, x_2, …, x_n)$中的恰好一个实数。

记$F(n)$为根据上述流程选择的元组$(x_1, x_2, …, x_n)$之和$x_1 + x_2 + … + x_n$的最小值。例如，$F(4) = 1.5$，其对应的元组为$(x_1, x_2, x_3, x_4) = (0, 0.75, 0.5, 0.25)$。

令人惊奇的是，上述流程最多只能选择$17$个实数。

求$F(17)$，并将你的答案保留$12$位小数。