An Exotic integral

@Mark Hennings - Nice Solution!!

@Mark Hennings For the case of $n=3$ we split the integral into two integrals each repeated $3$ times thus making a total $3!$ cases which is what we expect when we rearrange all possible variants of the inequality $0<x<y<z<1$ . That is ,

$\displaystyle \int_0^1 \int_0^1 \int_0^1 \left\{\dfrac{x}{y}\right\}\left\{\dfrac{y}{z}\right\}\left\{\dfrac{z}{x}\right\}\; dx\; dy\; dz = 3\underbrace{\int\limits_{0<x<y<z<1}\dfrac{x}{y}\dfrac{y}{z}\left\{\dfrac{z}{x}\right\}\; dx\; dy\; dz}_{\text{1 fractional part}} +3\underbrace{\int_{0<x<z<y<1}\dfrac{x}{y}\left\{\dfrac{y}{z}\right\}\left\{\dfrac{z}{x}\right\}\; dx\; dy\; dz}_{\text{2 fractional parts}}$

Exploiting symmetry for the case $n=4$ in a similar manner we see that there will be three such splits of integrals with different coefficients having $1,2,3$ fractional parts respectively. If we consider $a_n$ to be coefficient of such a split integral having $n$ fractional parts in it's respective region of integration.

For the case $n=3$ it seems that $a_1=a_2=3$ , for the case of 4 integrals $a_1=4,a_2=16,a_3=4$ where ultimately all the $a_i$ 's if added will definitely produce $m!$ where $m$ is the dimension of the integral we are working on since $m$ variables can be rearranged in $m!$ ways $0<x_1<x_2<\cdots<x_m<1$ .

Since it seems quite impossible for me to identify a pattern of those coefficients. For the case of $5$ integrals , the integral will be divided into 4 subintervals where $a_1=5,a_2=55$ and so on. So it comes to me that we need a way to solve for those coefficients. This problem is equivalent to :

How many ways are there to rearrange $0<x_1<_2<\cdots<x_n<1$ so that out of these $n$ fractions $\displaystyle \dfrac{x_1}{x_2},\dfrac{x_2}{x_3},\cdots,\dfrac{x_n}{x_1}$ exactly $k<n$ of them will be $\gt 1$ .

Is there a way to approach this ? I am thinking of this from a pretty long time