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I use notation $D_n$ for the number of derangements of an $n$ element set. As in title I seek a proof of $$nD_{n-1}+(-1)^n=D_n. \tag{1}$$ If this is added to its version with $n$ replaced by $n+1$ then the powers of $-1$ cancel and after subtracting $D_n$ from each side one gets another identity $$n(D_{n-1}+D_n)=D_{n+1}. \tag{2}$$ This consequence of $(1)$ can be shown by considering how the number $(n+1)$ can be placed somehow with a derangement of $\{1,2,\cdots,n\}$ to produce a derangement of $\{1,2,\cdots,n+1\}.$ [I can give details but I think this consequence of $(1)$ is known.]

So I seek preferably some proof of $(1)$ which concretely enumerates the two sides of it. In particular I'm curious why one adds $1$ or $-1$ (according to the parity of $n$) to $nD_{n-1}$ to get $D_n.$

Added after answer below: The answer of @David shows how to get from $(2)$ to $(1)$ algebraically. Note I still am curious whether there is a direct enumeration-type proof for $(1)$ if someone could do that.

coffeemath
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  • "This consequence of (1) can be shown by considering how the number (n+1) can be placed somehow with a derangement of {1,2,⋯,n} to produce a derangement of {1,2,⋯,n+1}" I would imagine that this would give us a proof by "enumeration" of the recursion by using suitable wording. – Asinomás Apr 08 '21 at 15:18
  • @HereToRelax : What you quote in your comment is just about the consequence of the relation I'm asking about, which is the second displayed statement in the question. I already have a proof by enumeration of that consequence. What I'm looking for is an enumeration proof for (1). – coffeemath Apr 08 '21 at 17:22
  • Oooh, I'm sorry you're completely right. My bad. I think this is a good question by the way. – Asinomás Apr 08 '21 at 17:33
  • There are at least five purely combinatorial proofs of $(1)$, one going back to $1983$, but the simplest that I’ve found is Sergi Elizalde, A simple bijective proof of a familiar derangement recurrence, arXiv:2005.11312, which has references to the others. – Brian M. Scott May 22 '21 at 23:24
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    A combinatorial proof of $(2)$ is given in $(1)$ from this answer. Then it is shown that $(1)$ is a simple consequence of $(2)$ in $(7)$ and $(8)$ of the aforementioned answer. – robjohn May 23 '21 at 03:48
  • Thanks for that link, @robjohn .. I'll enjoy going through it. – coffeemath May 23 '21 at 10:18

1 Answers1

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As you say, your consequence $n(D_{n-1}+D_n)=D_{n+1}\tag 2$ has a natural enumerative proof. Using (2), $$D_{n+1}-(n+1)D_n=nD_n+nD_{n-1}-(n+1)D_n= -(D_n-nD_{n-1}),$$ which equals $(-1)^{n+1}$ by induction because $D_2=1.$ This proves (1) with a proof whose essential part is enumeration. I think that induction explained the $\pm 1.$

David
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  • Thanks for this answer... I will look more closely soon. [was trying to get from (2) to my (1) somehow but no luck] – coffeemath Apr 08 '21 at 20:14
  • +1 for this answer, which shows indeed how to arrive at what I called (1) in the question, from the consequence (2) of your answer. I'm still looking for a direct enumeration type proof of eqn (1) of my question. – coffeemath Apr 08 '21 at 22:28
  • You're welcome. I called your enumeration proof natural because I thought it was something like this: If $n+1$ is in a 2-cycle, then delete its two edges to give a point plus an $(n-1)$-derangement. For each $n$-derangement, break up ${1,\dots,n}$ into cycles and replace each edge $a\to b$ by $a\to n+1 \to b.$ (Edges and cycles are directed.) I think that with (2) you found the natural setting for an enumerative proof. But that's just my opinion! – David Apr 08 '21 at 23:00
  • David: Yes it was indeed what you (and now I in my slightly edited question) call (2) which has an enumerative proof. But I had no luck going for a directly enumerative proof for (1) of my question. From one point of view my outlined algebraic proof of (2) using (1) together with your algebraic derivation of (1) from (2) shows these two relations are equivalent, so no real need for a direct enumerative proof of (1). Still i am curious if such could be done. – coffeemath Apr 09 '21 at 00:10