Log-Convexity and Fixed Points of Permutations

Abstract

We provide a combinatorial proof for the fact that the numbers of derangements of length n form a log-convex sequence for $n\ge 3$. A minor modification of our proof yields the same result for permutations of length $n\ge 4$ that have exactly one fixed point. We explain why these statements do not generalize for permutations with exactly k fixed points where $k\ge 3$.

1. Introduction

A derangement is a permutation with no 1-cycles. Derangments are well-studied objects; see Chapter 6 of [1] or Chapter 3 of [2] for basic facts about derangements. Let $D\left(n\right)$ be the number of derangements of length n. Our first goal in this paper is to provide a combinatorial proof of the fact that the infinite sequence $D\left(2\right), D\left(3\right), D\left(4\right),\cdots$ is log-convex, that is, $D{\left(n\right)}^2\le D(n-1)D(n+1)$ for all $n\ge 3$. (The sequence $\left\{a_n\right\}$ is called log-convex if $a_n^2\le a_{n-1}{a}_{n+1}$ for all n where those numbers are defined.) A computational proof was given by Lily L. Liu and Yi Wang in [3].

As derangements are defined using the fact that permutations are products of cycles, it is perhaps surprising that they are equinumerous to another class of permutations that are defined using the fact that permutations are also linear orders. This is surprising because the two definitions look unrelated at first sight, yet the classes of the two permutations they define prove to be closely related. This surprising fact turns these permutations called desarrangements into our main tool in this paper. If $p=p_1{p}_2\cdots p_n$ is a permutation written in the one-line notation, then i is called an ascent of p if $p_i<p_{i+1}$. Let $aa\left(p\right)$ be the smallest of all the ascents of p, if p has any ascents, and let $aa\left(p\right)=n$ otherwise, that is, when p is the decreasing permutation. A desarrangement is a permutation for which $aa\left(n\right)$ is even. In other words, the decreasing permutation is a desarrangment if and only if n is even. For example, 213, 54, 312, and 21 are desarrangements. Jacques Désarménien introduced desarrangements in [4]. (See [5] for a source in English.) Let $J\left(n\right)$ be the number of desarrangements of length n. It is straightforward to prove computationally that

$$J\left(n\right)=D\left(n\right)$$

(1)

for all positive integers n. See [4] or [5] for a bijective proof.

Because of (1), proving that sequence of the numbers $D\left(n\right)$ is log-convex is equivalent to proving that the sequence of the numbers $J\left(n\right)$ is log-convex. This is our goal in Section 2.

Let $F\left(n\right)$ be the number of permutations of length n that have exactly one fixed point. It is well-known that for all positive integers n, the equality

$$F\left(n\right)=D\left(n\right)+{(-1)}^{n-1}$$

(2)

holds. This fact is straightforward to prove computationally. A combinatorial proof is not obvious, but Herb Wilf gave one in [6]. In Section 3, we will use (2) and a minor modification of the notion of desarrangements to obtain a combinatorial proof of the log-convexity of the sequence ${\left\{F\left(n\right)\right\}}_{n\ge 3}$.

We then study permutations in which there are exactly k fixed points for some $k>1$, from the perspective of log-convexity. This shows that the results that we prove for $k=0$ and $k=1$ do not generalize for $k>2$ for every n, but they do hold for n if n is large enough.

2. Desarrangements

Let $n\ge 3$, and let $\mathcal{J}\left(n\right)$ be the set of all desarrangements of length n. We are going to define a map

$$f_n:\mathcal{J}\left(n\right)\times \mathcal{J}\left(n\right)\to \mathcal{J}(n-1)\times \mathcal{J}(n+1).$$

We will then prove that this map is injective, which will imply that the sequence $J\left(2\right), J\left(3\right),\cdots$ is log-convex.

Here is how we define $f_n(p,q)$, where $(p,q)\in \mathcal{J}\left(n\right)\times \mathcal{J}\left(n\right)$.

• If n is odd, simply move the last entry $p_n$ of p to the end of q, increasing each entry of q that is at least as large as $p_n$ by 1. Decrease all entries of p larger than $p_n$ by 1. An example is shown below.

  $$f_7(2135764,6321475)⟶(213465,73215864).$$

  Note that if $f_n(p,q)=(p^{\prime },q^{\prime })$, then $q^{\prime }$ never ends in $n+1$.
• If n is even, and p is not the decreasing permutation, proceed in the same way as in the preceding paragraph. Again, $q^{\prime }$ will not end in $n+1$.
• If n is even, and p is the decreasing permutation, then we have to proceed slightly differently because we cannot remove the last entry of p without destroying the desarrangement property. In this case, we set $p^{\prime }=(n-1)(n-3)(n-2)\cdots 321$, and we set $q^{\prime }$ to be q with $n+1$ attached to the end. Therefore, in this case, $q^{\prime }$ always ends with $n+1$.

Also note that in Rules 1 and 2, as shown above, removing the last entry of p never destroys the desarrangement property of p. Indeed, the only case in which that removal influences the leftmost ascent of p is when n is odd and the leftmost descent is in position $n-1$, but then the removal of $p_n$ results in a decreasing permutation of even length, and that is a desarrangement. The addition of a new entry at the end of q also does not destroy the desarrangement property.

Theorem 1.

The map $f_n$ defined above is an injection. Therefore, sequence ${\left\{J\left(n\right)\right\}}_{n\ge 3}$ is log-convex.

Proof. 

If n is odd, then we can recover the unique preimage $(p^{\prime },q^{\prime })\in \mathcal{J}(n-1)\times \mathcal{J}(n+1)$ (if it exists) as follows. If the last entry of $q^{\prime }$ is $n+1$, the preimage does not exist. Otherwise, remove that last entry $q_{n+1}^{\prime }$, and decrease all entries of $q^{\prime }$ less than that entry by 1 to obtain q. In order to obtain p, insert $q_{n+1}^{\prime }$ to the end of p, increasing all entries of p that are larger than or equal to $q_{n+1}^{\prime }$ by 1.

If n is even, then the procedure is the same if $q_{n+1}^{\prime }\ne n+1$. If $q_{n+1}^{\prime }=n+1$, and we also have $p^{\prime }=(n-1)(n-3)(n-2)\cdots 321$, then p is the decreasing permutation of length n, and $q_n$ is recovered by removing the last entry of $q_{n+1}^{\prime }$. Finally, if $q_{n+1}^{\prime }=n+1$, and $p^{\prime }\ne (n-1)(n-3)(n-2)\cdots 321$, then the preimage of $(p^{\prime },q^{\prime })$ does not exist.

As $f_n\mathcal{J}\left(n\right)\times \mathcal{J}\left(n\right)\to \mathcal{J}(n-1)\times \mathcal{J}(n+1)$ is an injection, it follows that the size of its domain is at most as large as the size of the set of its images, that is, $J{\left(n\right)}^2\le J(n-1)J(n+1)$, which shows that the sequence $J\left(2\right), J\left(3\right),\cdots$ is log-convex. □

3. Permutations with One Fixed Point

Recall from the introduction that $F\left(n\right)$ denotes the number of permutations of length n that have exactly one fixed point, and that for all positive integers n, the equality $F\left(n\right)=D\left(n\right)+{(-1)}^{n-1}$ holds. It is therefore easy to modify the definition of desarrangements to suit our purposes. Let $\mathcal{K}\left(n\right)$ be the set of permutations of length n whose first ascent is in an even position. If p is the decreasing permutation, then let $p\in \mathcal{K}\left(n\right)$ if and only if p is odd. It follows from this definition, (1) and (2) that

$$K\left(n\right):=\left|\mathcal{K}\right(n\left)\right|=F\left(n\right).$$

Therefore, proving that sequence of the numbers $F\left(n\right)$ is log-convex (for $n\ge 3$) is equivalent to proving that the sequence of the numbers $K\left(n\right)$ is log-convex.

Let $n\ge 4$. We define a map

$$g_n:\mathcal{K}\left(n\right)\times \mathcal{K}\left(n\right)\to \mathcal{K}(n-1)\times \mathcal{K}(n+1).$$

We will now prove that this map is injective, and therefore, the sequence ${\left\{K\left(n\right)\right\}}_{n\ge 4}$ is log-convex.

The definition of $g_n$ is similar to that of $f_n$ in the last section, but there are some differences. Let $(p,q)\in \mathcal{J}\left(n\right)\times \mathcal{J}\left(n\right)$.

1.

If n is even, simply move the last entry $p_n$ of p to the end of q, increasing each entry of q that is at least as large as $p_n$ by 1. Decrease all entries of p larger than $p_n$ by 1. The obtained pair is $(p^{\prime },q^{\prime })=g_n(p,q)$. Note that $q^{\prime }$ does not end in $n+1$.

2.

If n is odd, and p is not the decreasing permutation, and the leftmost ascent of p is not in position $n-1$, proceed as in the preceding paragraph. Again, the last entry of $q^{\prime }$ is not $n+1$.

3.

If n is odd, and p is the decreasing permutation, then we set $p^{\prime }=(n-1)(n-3)(n-2)\cdots 321$, and we set $q^{\prime }$ to be q, with $n+1$ attached to the end. So in this case, $q^{\prime }$ always ends with $n+1$.

4.

If n is odd, and the leftmost descent of $p=p_1{p}_2\cdots p_n$ is in position $n-1$,

4.a

and the rightmost entry of p is not n, then remove $p_1$ (note that, necessarily, $p_1=n$), and then simply reverse $p_2\cdots p_n$ to obtain $p^{\prime }$. Simply add $n+1$ to the end of q to obtain $q^{\prime }$. For example,

$$g_5(54213,32451)=(3124,324516).$$

4.b

If the rightmost entry of p is n, that is, if $p=(n-1)(n-2)\cdots 1n$, then let $p^{\prime }$ be the decreasing permutation of length $n-1$. Simply add $n+1$ to the end of q to obtain $q^{\prime }$. So $q^{\prime }$ again ends in $n+1$. Note that this is the only case in which $q^{\prime }$ ends in $n+1$ and $p^{\prime }$ is the decreasing permutation.

Theorem 2.

The map $g_n$ defined above is an injection. Therefore, the sequence ${\left\{K\left(n\right)\right\}}_{n\ge 4}$ is log-convex.

Proof. 

Let $(p^{\prime },q^{\prime })\in \mathcal{K}(n-1)\times \mathcal{K}(n+1)$. If $q^{\prime }$ does not end in $n+1$, then we can recover the unique preimage $g_n^{-1}(p^{\prime },q^{\prime })$ in the proof of Theorem 1.

If $q^{\prime }$ ends in $n+1$, and n is even, then $g_n^{-1}(p^{\prime },q^{\prime })$ does not exist. If n is odd, then we look at $p^{\prime }$.

(3)

If $p^{\prime }=(n-1)(n-3)(n-2)\cdots 321$, then Rule 3 was used, so $p=n(n-1)\cdots 21$, and q is obtained by removing the last entry $(n+1)$ of $q^{\prime }$.

(4.a)

If $p^{\prime }\ne (n-1)1324\cdots (n-2)$, then Rule 4.a was used, so p is obtained by reversing $p^{\prime }$ and prepending it with n. As in the previous case, q is by removing the last entry $(n+1)$ of $q^{\prime }$.

(4.b)

If $p^{\prime }$ is the decreasing permutation, then Rule 4.b was used, so $p=(n-1)(n-2)\cdots 1n$, and q is obtained by removing the last entry $(n+1)$ of $q^{\prime }$.

□

4. Permutations with $\mathit{k}$ Fixed Points

The following theorem of Désarménien and Wachs [5] generalizes the connection between derangements and desarrangements.

Theorem 3.

Let $F_{n,k}$ be the number of n-permutations with k fixed points. Let $E_{n,k}$ be the number of n-permutations $p=p_1{p}_2\cdots p_n$ with the following properties:

• The chain of inequalities $p_1<p_2<\cdots <p_k$ holds;
• The string $p_{k+1}{p}_{k+2}\cdots p_n$ is a desarrangement.

Then, the equality $F_{n,k}=E_{n,k}$ holds.

Remark 1.

It would be tempting to think that Theorem 3 can be used to prove that for any fixed k, the sequence ${\left\{E_{n,k}\right\}}_{n\ge k+2}$ is log-convex, but that log-convexity does not hold in general. Indeed, for $k=3$, we obtain a counterexample at $n=7$, since

$$E_{6,3}·E_{8,3}=40·2464=98560<99225=E_{7,3}^2={315}^2.$$

Similarly, we obtain a counterexample for $k=4$ at $n=8$.

The injective maps of the preceding two sections do not generalize, because in the case when $n-k$ is even, the string $p_{k+1}<p_{k+2}<\cdots <p_n$ may be decreasing, and it can have many (exactly $({\displaystyle \genfrac{}{}{0pt}{}{n}{k}})$ to be precise) different underlying sets. In the case of $n-k$ being odd, the inequalities defining log-convexity still hold, as is proved by the first step of the relevant maps (just move the last entry of p to the end of q).

This leaves the question of what happens for $k=2$. The answer is that the sequence ${\left\{F_{n,2}\right\}}_{n\ge 4}$ is log-convex, as can be proved computationally using the formula $E_{n,2}=\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2}}\right)D(n-2)$, and the approximation that $D\left(n\right)$ is the integer closest to $n!/e$.

The phenomenon that we have seen—that for a fixed k, the sequence ${\left\{F_{n,k}\right\}}_{n\ge k+2}$ is log-convex if $k=0$, $k=1$, or $k=2$, but not for larger k—can be explained as follows. It is well-known that $D\left(n\right)$ is the integer closest to $n!/e$, and that the sequence $n!$ is log-convex. It is routine to prove that for a large n, the number $F_{n,k}$ is approximately $n!/(k!e)$, so, for a large n, the log-convex condition will hold. However, for large values of k, the error terms can be significant for small values of n, and that leads to our counterexamples. Also note that while (for fixed k), the sequence $D(n-k)$ is log-convex, the sequence $({\displaystyle \genfrac{}{}{0pt}{}{n}{k}})$ is log-concave, and it takes time for log-convexity to prevail in the product of the two sequences.