15. Let $\sigma=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be a permutation of $\{1,2, \cdots, n\}$, and define $F(\sigma)=\sum_{i=1}^{n} a_{i} a_{i+1}, a_{n+1}=a_{1}$, find $\min _{\sigma} F(\sigma)$.
The problem is equivalent to placing $n$ numbers on a circle so that the sum of the products of adjacent numbers is minimized.
Assume $a_{1}=n$, then the number 1 must be adjacent to it. Otherwise, if $a_{j}=1(j \neq 2, n)$, we can change $a_{2}, a_{3}, \cdots, a_{j}$ to $a_{j}, a_{j-1}, \cdots, a_{2}$, then the change in the sum of the products of adjacent numbers is
$$
a_{1} a_{j}+a_{2} a_{j+1}-a_{1} a_{2}-a_{j} a_{j+1}=\left(a_{1}-a_{j+1}\right)\left(a_{j}-a_{2}\right)<0,
$$

Thus, we can determine $a_{2}=1$. Then, we need to show that the number 2 must also be adjacent to the number $n$, i.e., $a_{n}=2$.
In fact, if $a_{k}=2(k \neq n)$, we can change $a_{n}, a_{n-1}, \cdots, a_{k}$ to $a_{k}, a_{k+1}, \cdots, a_{n}$, then the change in the sum of the products of adjacent numbers is
$$
a_{1} a_{k}+a_{n} a_{k-1}-a_{1} a_{n}-a_{k} a_{k-1}=\left(a_{1}-a_{k-1}\right)\left(a_{k}-a_{n}\right)<0 \text {. }
$$

Therefore, when the objective function reaches its minimum value, $a_{1}=n, a_{2}=1, a_{n}=2$. Similarly, we can get $a_{3}=n-1, a_{n-1}=n-2$.
In general, after the two end numbers are arranged, if the remaining numbers are smaller than the two end numbers, we find the two smallest numbers among the remaining numbers and place them in the order of large-small, small-large; if the remaining numbers are larger than the two end numbers, we find the two largest numbers among the remaining numbers and place them in the order of large-small, small-large, as shown in the figure.
By this rule, we get $a_{4}=3, a_{n-2}=4, a_{5}=n-3, a_{n-3}=n-4, \cdots$. Let the minimum value of the objective function be $T_{n}$, and we use the recursive method to calculate $T_{n}$.
Consider $n+2$ numbers, we add 1 to each number in $T_{n}$, and then insert $1, n+2$ between $n+1,2$, i.e., we get $T_{n+2}$.
Thus, $T_{n+2}=T^{\prime}+(n+1)+(n+2)+2(n+2)-2(n+1)=T^{\prime}+2 n+5$, where
$$
\begin{array}{l}
T^{\prime}=\sum_{i=1}^{n}\left(a_{i}+1\right)\left(a_{i+1}+1\right)=\sum_{i=1}^{n}\left(a_{i} a_{i+1}+a_{i}+a_{i+1}+1\right)=T_{n}+2 \sum_{i=1}^{n} a_{i}+n \\
=T_{n}+n(n+1)+n=T_{n}+n^{2}+2 n \Rightarrow T_{n+2}=T_{n}+n^{2}+4 n+5 .
\end{array}
$$

When $n=2 m$,
$$
\begin{array}{l}
T_{2 m}=T_{2}+\left[2^{2}+4^{2}+\cdots+(2 m-2)^{2}\right]+4[2+4+\cdots+(2 m-2)]+5(m-1) \\
=4+4\left[1^{2}+2^{2}+\cdots+(m-1)^{2}\right]+4 m(m-1)+5(m-1) \\
=\frac{2 m(m-1)(2 m-1)}{3}+4 m^{2}+m-1=\frac{n(n-1)(n-2)}{6}+n^{2}+\frac{n}{2}-1 \\
=\frac{n^{3}-3 n^{2}+2 n}{6}+n^{2}+\frac{n}{2}-1=\frac{n^{3}+3 n^{2}+5 n-6}{6}
\end{array}
$$

When $n=2 m-1$,
$$
\begin{array}{l}
T_{2 m-1}=T_{1}+\left[1^{2}+3^{2}+\cdots+(2 m-3)^{2}\right]+4[1+3+\cdots+(2 m-3)]+5(m-1) \\
=1+\left[1^{2}+2^{2}+\cdots+(2 m-2)^{2}\right]-\left[2^{2}+4^{2}+\cdots+(2 m-2)^{2}\right]+4(m-1)^{2}+5(m-1) \\
=\frac{(2 m-2)(2 m-1)(4 m-3)}{6}-\frac{2 m(m-1)(2 m-1)}{3}+(2 m-2)^{2}+5 m-4 \\
=\frac{n(n-1)(2 n-1)}{6}-\frac{n(n-1)(n+1)}{6}+(n-1)^{2}+\frac{5 n-3}{2} \\
=\frac{2 n^{3}-3 n^{2}+n}{6}-\frac{n^{3}-n}{6}+n^{2}+\frac{n-1}{2}=\frac{n^{3}+3 n^{2}+5 n-3}{6} . \\
\text { In summary, } T_{n}=\left\{\begin{array}{l}
\frac{n^{3}+3 n^{2}+5 n-3}{6}, n \text { is odd, } \\
\frac{n^{3}+3 n^{2}+5 n-6}{6}, n \text { is even. }
\end{array}\right.
\end{array}
$$

In summary, $T_{n}=\left\{\begin{array}{l}\frac{n^{3}+3 n^{2}+5 n-3}{6}, n \text { is odd, } \\ \frac{n^{3}+3 n^{2}+5 n-6}{6}, n \text { is even. }\end{array}\right.$