$\; \; \; \; \; $There are $2n$ balls with $n$ different colors, two for each color. (Balls of the same color are the same), and there are $n$ identical boxes. Now, we put all the balls into all the boxes, and each box happens to have two balls. Regardless of the order of the boxes and the order of the balls inside the boxes, how many different placement methods are there?
When $n = 2$, there are $2$ methods: $1.AA\; BB$$2.AB\; AB$.
When $n = 3$, there are $5$ methods: $1.AA\; BB\; CC$$2.AB\; AB\; CC$$3.AC\; BB\; AC$$4.AA\; BC\; BC$$5.AB\; BC\; AC$.
$\; \; \; \; \; $At first, I thought it was simple. Isn't this the order of a permutation group? Later, I discovered that $\begin{pmatrix} 1 & 2 &3 \\ 2 & 3 &1 \end{pmatrix}$ and $\begin{pmatrix} 1 & 2 &3 \\ 3 & 1 &2 \end{pmatrix}$ are the same,all equal to $\begin{pmatrix} 1 & 1 &2 \\ 2 & 3 &3 \end{pmatrix}$,so there are only $5$ situations when $n=3$.$\begin{pmatrix} 1 & 2 &3 \\ 1 & 2 &3 \end{pmatrix}$$\begin{pmatrix} 1 & 1 &3 \\ 2 & 2 &3 \end{pmatrix}$$\begin{pmatrix} 1 & 1 &2 \\ 3 & 3 &2 \end{pmatrix}$$\begin{pmatrix} 1 & 2 &2 \\ 1 & 3 &3 \end{pmatrix}$$\begin{pmatrix} 1 & 1 &2 \\ 2 & 3 &3 \end{pmatrix}$.
I calculated the number of $n \le 5$, which is $1, 2, 5, 17, 63$. Then I typed it into OEIS, but couldn't find the answer I wanted. Did I make a mistake or did I discover a new sequence?