Consider a $1 \times n$ table with cells labelled by variables $x_1, x_2, \dots, x_n$ from left to right.

| $x_1$ | $x_2$ | ... | $x_n$ |

We assign a polynomial in terms of variables $x_1, x_2, \dots, x_n$ to each tiling of this table by $1 \times 1$ and $1 \times 2$ tiles, as follows. We define the weight of $1 \times 1$ tiles to be the variable of its cell in the table and the weight of a $1 \times 2$ tile to be the sum of squares of variables of its cells. At the end, we associate to each tiling the product of weights of its tiles. For example, for $n = 4$, there are only 5 ways of tiling a $1 \times 4$ table with the mentioned tiles:

Now for each integer $n \ge 0$, we define the polynomial $P_n$ to be the sum of all polynomials corresponding to the tilings of the $1 \times n$ table. For example,
$$P_4(x_1, x_2, x_3, x_4) = (x_1^2+x_2^2)(x_3^2+x_4^2)+(x_1^2+x_2^2)x_3x_4+x_1x_2(x_3^2+x_4^2)+x_1(x_2^2+x_3^2)x_4+x_1x_2x_3x_4$$
Note that there is only the empty tiling in the case $n = 0$ and there is one trivial tiling when $n = 1$. Therefore, $P_0() = 1$, $P_1(x_1) = x_1$.

We claim that for each integer $n \ge 0$, $P_n(x_1, \dots, x_n) = K_n(x_1, \dots, x_n)$. For proving the claim, since $P_0 = K_0$ and $P_1 = K_1$, it suffices to show that $P_n$ satisfies the same recursive relation as $K_n$. This is easy by looking at the last tile of each tiling (the tile covering $x_n$). We have two cases.

1. The last tile of tiling is a $1 \times 2$ tile. The part of $P_n$ corresponding to these tilings is $(x_{n-1}^2 + x_n^2)P_{n-2}(x_1, \dots, x_{n-2})$, because $x_{n-1}^2 + x_n^2$ is the weight of the last tile and the remaining table after removing this tile has $n-2$ cells.

2. The last tile of tiling is a $1 \times 1$ tile. The part of $P_n$ corresponding to these tilings is $x_n P_{n-1}(x_1, \dots, x_{n-1})$ because $x_n$ is the weight of the last tile and the remaining table after removing this tile has $n-1$ cells.

So the proof of the claim is finished. Finally, note that reflection of each tiling with respect to the vertical axis of symmetry of the table is again a tiling. Therefore, if we label the cells by $x_n, x_{n-1}, \dots, x_1$ rather than $x_1, x_2, \dots, x_n$, we have again the polynomial $K_n(x_1, x_2, \dots, x_n)$. So $K_n(x_1, x_2, \dots, x_n) = K_n(x_n, \dots, x_2, x_1)$ as desired.