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Given the recurrence relation $a_n=a_{n-1}+(n-1)a_{n-2}$ and the values of $a_0$ and $a_1$. For example, $a_0=a_1=1$.

Can we give an explicit formula of this sequence? If yes, how to achieve that?

(In fact, I find this sequence in Burnside's lemma which describes the number of different positions to put 8 rooks in a chessboard, and each rook should lie in different row and column. )

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  • $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ Commented Feb 4, 2024 at 12:32
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    $\begingroup$ This is sequence A000085 in the OEIS. Looking at the formula section I doubt there is a nice closed form $\endgroup$ Commented Feb 4, 2024 at 12:51
  • $\begingroup$ $$ a_n = n!\sum\limits_{k = 0}^{\left\lfloor {n/2} \right\rfloor } {\frac{{2^{ - k} }}{{k!(n - 2k)!}}} $$ $\endgroup$ Commented Feb 5, 2024 at 12:35

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