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Denote $S_n$ by the symmetric group of order $n$. For a permutation $\sigma\in S_n$ we define the sign of $\sigma$ as $\def\sgn{\operatorname{sgn}} \sgn(\sigma)=1$ if it is an even permutation and $\sgn(\sigma)=-1$ if it is an odd permutation. Now, we would like to find this sum $$\sum_{\sigma\in\mathsf{AV}_n(123)}\sgn(\sigma),$$ where $\mathsf{AV}_n(123)$ is the set of all $123$-avoiding permutations of length $n$. A permutation $\sigma$ of length $n$ is called $123$-avoiding as long as there do not exist three indices $1\le i<j<k\le n$ such that $\sigma(i)<\sigma(j)<\sigma(k)$.

By Sage, I get the sequence from $n\geqslant 3$,

$$-1,0,2,0,-5,0,14,0,-42,\ldots$$ So, I guess this answer from the above pattern is $$(-1)^{\frac{n-1}{2}}C_{\frac{n-1}{2}},$$ where $C_n$ is so-called the Catalan number. But I have no ideal for this combinatorial proof. If there is any reference I would appreciate.

RobPratt
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user1992
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    What is the total cardinality of 123-avoiding permutations of length $n$? – Akiva Weinberger Mar 27 '25 at 21:20
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    @AkivaWeinberger: It is Catalan number. Please see https://en.wikipedia.org/wiki/Catalan_number. – user1992 Mar 28 '25 at 01:13
  • I'm not sure why my comment was deleted (?) since it pointed to the corresponding wikipage in which the Simion and Schmidt paper is cited... – Benjamin Dickman Mar 28 '25 at 20:00
  • It was flagged as "no longer needed" by a trusted user, @BenjaminDickman. – Shaun Mar 29 '25 at 00:34
  • @Shaun Interesting; my own feeling is that a comment should remain if it contains the paper/relevant info described in an answer. Like a sort of citation integrity. (I know that comments aren't supposed to be answers: in this case, it was truly a comment tho – I didn't write up the content!) But, I suspect mods are better versed in the actual policies for the site... Anyway, my 2¢. – Benjamin Dickman Mar 29 '25 at 00:45
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    @BenjaminDickman OK, I was the trusted user who flagged you. You were linking to the Wikipedia page for permutation patterns, which cited 38 references. Your comment was analogous to saying "Here's this haystack containing the needle you are looking for!", so it seemed like an unhelpful comment that was more clutter than useful. Certainly, at the time you wrote the comment, you did not KNOW that the 38 references contained a helpful paper, right? Otherwise, you would have just cited the correct paper, instead of the Wikipedia article about the general subject. – Mike Earnest Mar 29 '25 at 05:27
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    @MikeEarnest Googling the title of this post yields (as the first non-MSE answer, now) the wikipage you mention, but see the subsection on "Enumerative Origins" in which the first two paragraphs suggest they contain the answer or at least relevant background missing from the question. (Before googling, I didn't know that the "AV" notation wasn't original to the poster.) Moreover, the "no longer needed" designation arriving after the reference to Simion and Schmidt strikes me [as I said above] as a "lack of citation integrity." I don't see the benefit of comment removal but it's all good! – Benjamin Dickman Mar 29 '25 at 05:45
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    @BenjaminDickman You were right, and I was wrong. Your comment was a good one. I wouldn't have flagged your comment if your comment was more specific, pointing to the Simion and Schmidt paper specifically instead of just pointing to the Wikipedia subsection. – Mike Earnest Mar 29 '25 at 06:08

1 Answers1

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This result is established in the paper "Restricted Permutations" by Simion and Schmidt, available for free online. Specifically, Proposition 2 on page 386 of their paper proves that, when $n$ is odd, $$ E_n(123) - O_n(123) = (-1)^{\binom{n}{2}} C_{(n-1)/2} =(-1)^{(n-1)/2}C_{(n-1)/2}.$$ where $E_n(123)$ and $O_n(123)$ are the counts of even and odd permutations in $\mathsf{AV}_n(123)$, respectively. Their proof is entirely combinatorial.

Mike Earnest
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