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  "body": "\n## Statement\n\nLet $n$ be a natural number, and let $D_m$ be the number of [[derangements|0TH61JR04G]] of an $m$-element set.\nThen\n$$D_{n+2} = (n + 1)\\,(D_{n+1} + D_n).$$\n\n## Remarks\n\nTogether with $D_0 = 1$ and $D_1 = 0$, the recurrence gives all values: $1, 0, 1, 2, 9, 44, 265, \\dots$\nA proof looks at a derangement $\\sigma$ of $\\{1, \\dots, n + 2\\}$ and the value $i = \\sigma(n+2)$. There are $n + 1$ choices of $i$. The case $\\sigma(i) = n + 2$ gives $D_n$ derangements, and the other case gives $D_{n+1}$.\nAnother recurrence is $D_{n+1} = (n + 1) D_n + (-1)^{n+1}$.\n\n## Explanations\n\n### L1 human\n\nLook at where the last element goes, say to position $i$.\nEither $i$ goes back to the last position, and the other $n$ elements form a derangement. Or it does not, and this acts like a derangement of $n + 1$ elements.\n",
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