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

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 $k\in \{1,\dots ,n\}$ contains exactly one of $(x_1,x_2,\dots ,x_n)$.

Define $F(n)$ to be the minimal value of the sum $x_1+x_2+\cdots +x_n$ of a tuple $(x_1,x_2,\dots ,x_n)$ 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.