A new record for 1324-avoiding permutations

Abstract

We prove that the class of $1324$-avoiding permutations has exponential growth rate at most $13.74$.

1 Introduction

1.1 Definitions and open questions

The theory of pattern avoiding permutations has seen tremendous progress during the last two decades. The key definition is the following. Let $k\le n$, let $p=p_1p_2\ldots p_n$ be a permutation of length $n$, and let $q=q_1q_2\ldots q_k$ be a permutation of length $k$. We say that $p$ avoids $q$ if there are no $k$ indices $i_1<i_2<\cdots <i_k$ such that for all $a$ and $b$, the inequality $p_{i_a}<p_{i_b}$ holds if and only if the inequality $q_a<q_b$ holds. For instance, $p=2537164$ avoids $q=1234$ because $p$ does not contain an increasing subsequence of length four. See [3] for an overview of the main results on pattern avoiding permutations.

A spectacular result of Adam Marcus and Gábor Tardos [6] shows that for all patterns $q$, there exists a constant $c_q$ such that $S_n(q)\le c_q^n$ for all $n$, and a short argument [1] then shows that this implies the existence of $L(q)=\lim _{n\rightarrow \infty } \root n \of {S_n(q)}$.

The shortest pattern for which even some of the most basic questions are open is $q=1324$, a pattern that has been studied for at least $2$0 years. For instance, there is no known exact formula for the number $S_n(1324)$ of permutations of length $n$ (or, in what follows, $n$-permutations) avoiding $1324$. Even the value of $L(1324)=\lim _{n\rightarrow \infty } \root n \of {S_n(1324)}$ is unknown. It is also known that no pattern of length four is easier to avoid than the pattern $1324$, that is, for any pattern $q$ of length four, the inequality $S_n(q)\le S_n(1324)$ holds. The inequality is sharp unless $q=4231$. See [3, Chapter 4] for a treatment of the series of results leading to these inequalities.

Recently, there has been some progress in the very challenging problem of determining $L(1324)$. First, in 2011, Anders Claesson, Vit Jelínek and Einar Steingrímsson have proved that $L(1324)\le 16$ [5]. A year later, in July of 2012, the present author improved that upper bound [4] showing that $L(1324)\le 7+4\sqrt{3}<13.93$. As far as lower bounds go, in 2005 Albert et al. proved that $L(1324)\ge 9.42$, which has recently been improved to $L(1324)\ge 9.81$ by David Bevan [2].

In the present paper, we further refine the counting argument of [4], leading to the improved upper bound $L(1324)\le 13.73718$. While this improvement is numerically not huge, the way in which it is obtained is interesting as we are able to exploit some dependencies between two words $w(p)$ and $z(p)$ related to the $1324$-avoiding permutation $p$ that have previously been counted as though they were independent. A key element of our improved enumeration method is that we will consider pairs of words $(w,z)$ in which $w$ and $z$ do not have to be of the same length.

2 Words over a finite alphabet

2.1 Coloring

The starting point of our proof is the following decomposition of $1324$-avoiding permutations, given in [5] in a slightly different form, and then given in the present form in [4].

Let $p=p_1p_2\ldots p_n$ be a $1324$-avoiding permutation, and let us color each entry of $p$ red or blue as we move from left to right, according the following rules.

• (I) If coloring $p_i$ red would create a $132$-pattern with all red entries, then color $p_i$ blue, and

• (II) otherwise color $p_i$ red.

It is then proved in [5] that the red entries form a $132$-avoiding permutation and the blue entries form a $213$-avoiding permutation. The following fact will be useful in the next section.

Proposition 2.1

[4] If an entry $p_i$ is larger than a blue entry on its left, then $p_i$ itself must be blue.

If $p=p_1p_2\ldots p_n$ is a permutation, then we say that $p_i$ is a left-to-right minimum if $p_j>p_i$ for all $j<i$. That is, $p_i$ is a left-to-right minimum if it is smaller than all entries on its left. Similarly, we say that $p_k$ is a right-to-left maximum if $p_k>p_\ell$ for all $\ell >k$. In other words, $p_k$ is a right-to-left maximum if it is larger than all entries on its right. It is easy to see that $p_1$ and $1$ are always left-to-right minima, $p_n$ and $n$ are always right-to-left maxima, and both the sequence of left-to-right minima and the sequence of right-to-left maxima are decreasing.

In [4], the decomposition of $1324$-avoiding permutations given by rules (I) and (II) above was refined as follows. Let us color each entry of the $1324$-avoiding permutation $p=p_1p_2\ldots p_n$ red or blue as above. Furthermore, let us mark each entry of $p$ with one of the letters $A, B, C$, or $D$ as follows.

• (a) Mark each red entry that is a left-to-right minimum in the partial permutation of red entries by $A$,

• (b) mark each red entry that is not a left-to-right minimum in the partial permutation of red entries by $B$,

• (c) mark each blue entry that is not a right-to-left maximum in the partial permutation of blue entries by $C$, and

• (d) mark each blue entry that is a right-to-left maximum in the partial permutation of blue entries by $D$.

Call entries marked by the letter $X$ entries of type $X$. Let $w(p)$ be the $n$-letter word over the alphabet $\{A,B,C,D\}$ defined above. In other words, the $i$th letter of $w(p)$ is the type of $p_i$ in $p$. Let $z(p)$ be the $n$-letter word over the alphabet $\{A,B,C,D\}$ whose $i$th letter is the type of the entry of value $i$ in $p$. In other words, $w(p)$ is obtained by reading the letter marking the entries of $p$ from left to right, while $z(p)$ is obtained by reading these letters from the bottom up.

Example 2.2

Let $p=3612745$. Then the subsequence of red entries of $p$ is $36127$, the subsequence of blue entries of $p$ is $45$, so $w(p)=ABABBCD$, while $z(p)=ABACDBB$. See Fig. 1 for an illustration.

Fig. 1

Colors and labels for $p=3612745$ (Color figure online)

The following important fact is proved in [4]. Let $\mathrm{NO}_n(CB)$ be the set of all words of length $n$ over the alphabet $\{A,B,C,D\}$ that contain no $CB$-factors, that is, that do not contain a letter $C$ immediately followed by a letter $B$. Let $\mathrm{Av}_n(1324)$ denote the set of all $1324$-avoiding permutations of length $n$.

Theorem 2.3

The map $f:\mathrm{Av}_n(1324)\rightarrow \mathrm{NO}_n(CB) {\times } \mathrm{NO}_n(CB)$ given by $f(p)=(w(p),z(p))$ is an injection.

Consequently, $S_n(1324)=|\mathrm{Av}_n(1324)| \le | \mathrm{NO}_n(CB)|^2$, leading to the upper bound

$$\begin{aligned} S_n(1324)\le (7+4\sqrt{3})^n . \end{aligned}$$

(1)

3 A connection between $w(p)$ and $z(p)$

3.1 A small modification of the encoding

One wasteful element of the method used to prove the upper bound (1) that we sketched in the last section is that it treats $w(p)$ and $z(p)$ as two independent words for enumeration purposes, which leads to the bound $|\mathrm{Av}_n(1324)|\le |\mathrm{NO}_n(CB)|^2$. The full extent of the dependencies between $w(p)$ and $z(p)$ is difficult to describe, but in this paper, we describe it well enough to improve the upper bound for the exponential growth rate of $S_n(1324)$.

First, we modify the letter encoding of $1324$-avoiding permutations a little bit by adding the following last rule to the existing rules (a)–(d).

(d$'$):

If an entry is a right-to-left maximum in all of $p$, but not a left-to-right minimum in $p$, then color it blue and mark it $D$, regardless what earlier rules said.

In the rest of this paper, we will use rules (a)–(d) together with (d$'$). With a slight abuse of notation, for the sake of brevity, we denote the pair of words encoding $p$ by these rules by $w(p)$ and $z(p)$, just as we did before rule (d$'$) was added. This will not create confusion, since the older, narrower set of rules will not be used. The advantage of adding rule (d$'$) is that it ensures that every letter $C$ is eventually followed by a letter $D$ so that the entry corresponding to that $D$ is larger than the entry corresponding to that given $C$.

Note that if an entry changes its letter code because of this last rule, then that entry was previously a letter $B$, and now it is a $D$. It is straightforward to check that it is still true that the string of red entries forms a $132$-avoiding permutation (since their string simply lost a few entries) and the string of blue entries forms a $213$-avoiding permutation. Indeed, if $u$ is a right-to-left minimum in $p$ that has just become blue because of rule (d$'$), then $u$ could only play the role of $3$ in a blue $213$-pattern $yxu$. However, that would mean that before rule (d$'$) was applied, the red entry $u$ was on the right of the smaller blue entries $x$ and $y$, contradicting Proposition 2.1. It is also clear that if $p\in \mathrm{Av}_n(1324)$, then neither $w(p)$ nor $z(p)$ contains a $CB$-factor (since the set of letters $B$ shrank), and it is straightforward to show that the new map $p\mapsto (w(p),z(p))$ is still injective. Indeed, given $w(p)$ and $z(p)$, we can first use the letters $A$ and $B$ in these words to recover the “red part” of $p$, then we can use the letters $C$ and $D$ in these words to recover the ”blue part” of $p$. See [4] for details.

3.2 A connection between $w(p)$ and $z(p)$

We are now ready to describe some connections between $w(p)$ and $z(p)$. We define a segment in a finite word $v$ over the alphabet $\{A,B,C,D\}$ is a subword of consecutive letters that starts with an $A$ and ends immediately before the next letter $A$, or at the end of $v$. For instance, the word $v=ABBDCACDB$ has two segments. The key observation is the following.

Lemma 3.1

Let $p$ be a $1324$-avoiding permutation, and let $(w(p),z(p))$ be its image under the coloring defined by rules (a)–(d) and $(\mathrm{d}')$. Then for all $i$, the following holds. If the $i$th letter $A$ from the right in $w(p)$ is in the middle of a $CAB$-factor, then the $i$th segment of $z(p)$ from the left must contain a letter $B$.

Proof

Let us assume that the statement does not hold, that is, there is $i$ that provides a counterexample. Let $a$ be the $i$th smallest left-to-right minimum in $p$. That means that $a$ corresponds to the $i$th letter $A$ from the left in $z(p)$, and to the $i$th letter $A$ from the right in $w(p)$. Let us say that the letters corresponding to the entries $cab$ of $p$ form a $CAB$-factor in $w(p)$. Find the closest entry $d$ on the right of $b$ that corresponds to a letter $D$, and the closest entry $a'$ on the left of $c$ that corresponds to a letter $A$. Then the entries $a'cbd$ will form a $1324$-pattern in $p$ unless $b<a'$. However, that would mean that $a<b<a'$, that is, there is a letter $B$ in the $i$th segment of $z(p)$ after all, contradicting the assumption that $i$ is a counterexample. $\square$

The following simple proposition will be useful for us.

Proposition 3.2

Let $s_n$ be the number of segments of length $n$ that do not contain a $CB$-factor. Let $S(x)=\sum _{n\ge 1} s_nx^n$. Then

$$\begin{aligned} S(x)=\frac{x}{x^2-3x+1}. \end{aligned}$$

Proof

By definition, $s_0=0$ (since any segment must contain a letter $A$ at its front), and $s_1=1$. If $n\ge 2$, then $s_n=3s_{n-1}-s_{n-2}$, since we get a segment of length $n$ if we affix a letter $B, C$, or $D$ at the end of any segment of length $n-1$, except in the $s_{n-2}$ cases when this results in a $CB$-factor at the end. $\square$

We are now ready to announce our main tool of this section. Let $h_n$ be the number of the pairs of words $(w,z)$ over the alphabet $\{A,B,C,D\}$ that satisfy the following requirements.

• (i) Both $w$ and $z$ start with the letter $A$, and $|w|+|z|=n$,

• (ii) the words $w$ and $z$ contain the same number of letters $A$,

• (iii) neither $w$ nor $z$ contains a $CB$-factor, and

• (iv) for all $i$, if the $i$th letter $A$ from the right in $w$ is in the middle of a $CAB$-factor, then the $i$th segment of $z$ (from the left) contains a letter $B$.

Note that the words $w$ and $z$ do not have to have the same length.

Theorem 3.3

Let $h_n$ be as above and $H(x)=\sum \nolimits _{n\ge 2} h_n x^n$. Then

$$\begin{aligned} H(x)=\sum _{n\ge 2}h_nx^{n}= \frac{x^2(1-2x)}{x^6-5x^5+14x^4-26x^3+22x^2-8x+1}. \end{aligned}$$

Proof

If a pair of words $(w,z)$ enumerated by $H(x)$ contains a total of two letters $A$, then both $w$ and $z$ must consist of a single segment, since each segment contains a letter $A$. Let us now assume that $(w,z)$ contains more than two letters $A$. Removing the last segment of $w$ and the first segment of $z$, we get another, shorter pair $(w',z')$ of words enumerated by $H(x)$. On the other hand, inserting a new last segment $S_1$ at the end of $w'$ and a new first segment $S_2$ at the front of $z'$, we get a new pair of words enumerated by $H(x)$ except when the newly inserted last letter $A$ in $w$ is in the middle of a $CAB$-factor, and the new first segment of $z$ does not contain any letters $B$.

This leads to the generating function identity

$$\begin{aligned} H(x)=S^2(x)+S^2(x)H(x) - \frac{x}{1-2x} \cdot \frac{x^2}{1-3x+x^2} \cdot xH(x). \end{aligned}$$

(2)

The last summand of the right-hand side is justified as follows. The generating function for $S_2$ is $x^2/(1-2x)$ since $S_2$ starts with a letter $A$, and then consists of letters $C$ and $D$ with no restrictions, while the generating function for $S_1$ is $x^2/(1-3x+x^2)$, since this segment starts with $AB$, and then consists of a string over the alphabet $\{B,C,D\}$ with no $CB$-factors, so it is just a segment with a letter $B$ inserted into its second position. Finally, the rest of the pair $(w,z)$ is just a word counted by $H(x)$ where a letter $C$ is inserted at the end of $w$.

Solving (2) for $H(x)$, we get the statement of theorem. $\square$

Corollary 3.4

There exists a positive constant $c$ such that $h_n\le c{ \cdot } 3.709381^n$.

Proof

This is routine after noticing that the denominator of $H(x)$ as given in Theorem 3.3 has a unique root of smallest modulus, namely $\alpha \ge 0.2695867676$, and computing $\beta =1/\alpha \le 3.709381$. $\square$

Now we can prove the main result of this section.

Theorem 3.5

$L(1324)\le 13.76$.

Proof

This is immediate from the fact that if $p\in \mathrm{Av}_n(1324)$, then the pair $w(p),z(p)$ is in the set that is counted by $h_{2n}$ and is defined by rules (i)–(iv). Indeed, $w(p)$ and $z(p)$ are both words of length $n$ over the alphabet $\{A,B,C,D\}$, they contain the same number of letters of each type, do not contain any $CB$-factors as shown in Theorem 2.3, and the pair $(w(p),z(p))$ satisfies condition (iv) by Lemma 3.1. Therefore, $S_n(1324)\le h_{2n}$, and our result follows from Corollary 3.4 by computing $\beta ^2$. $\square$

4 Extending the reach of our method

There are several other constraints that the pair $(w(p),z(p))$ must satisfy if $p$ is a $1324$-avoiding permutation. The problem is that it is difficult to count pairs that satisfy all these constraints at once. On the other hand, taking only a few constraints into account results only in a small improvement of the result of Theorem 3.5.

For instance, Lemma 3.1 can be generalized in the following way. A $CAB^k$-factor in a word is a subword of $k+2$ letters in consecutive positions, the first of which is a letter $C$, the second of which is an $A$, and all the others are $B$.

Lemma 4.1

Let $p$ be a $1324$-avoiding permutation, and let $(w(p),z(p))$ be its image under the coloring defined by rules (a)–(d) and $(\mathrm{d}')$. Then for all $i$, and all $k$, the following holds. If the $i$th letter $A$ from the right in $w(p)$ is in the middle of a $CAB^k$-factor, then the $i$th segment of $z(p)$ from the left must contain at least $k$ letters $B$.

Proof

The proof is analogous to that of Lemma 3.1. The only change is that the role of $b$ in that proof must be played by the $k$th letter $B$ in the $CAB^k$-factor of the lemma. In order for $p$ to be $1324$-avoiding, $b<a'$ must hold (keeping the notation of the proof of Lemma 3.1). As $b$ is clearly the largest of the $k$ letters $B$ immediately following $a$, this implies that there are at least $k$ letters $B$ between $a$ and $a'$ in $z(p)$. $\square$

Let us now check what numerical improvement we obtain if we enforce the constraint of Lemma 4.1 for $k=2$.

Theorem 4.2

Let $k_n$ be the number of pairs of words $(w,z)$ over the alphabet $\{A,B,C,D\}$ that satisfy all of the rules (i)–(iv), and also the following rule.

• (v) For all $i$, if the $i$th letter $A$ from the right in $w$ is in the middle of a $CABB$-factor, then the $i$th segment of $z$ $($from the left$)$ contains at least two letters $B$.

Let $K(x)=\sum \limits _{n\ge 2}k_nx^n$. Then

$$\begin{aligned} K(x)=\frac{x^2(1-2x)^2}{1-10x+38x^2-70x^3+66x^4-33x^5+12x^6-6x^7+4x^8-x^9}. \end{aligned}$$

Proof

If a pair of words $(w,z)$ enumerated by $K(x)$ contains a total of two letters $A$, then both $w$ and $z$ must consist of a single segment. Otherwise, removing the last segment of $w$ and the first segment of $z$, we get another, shorter pair $(w',z')$ of words enumerated by $K(x)$. On the other hand, inserting a new last segment $S_1$ at the end of $w'$ and a new first segment $S_2$ at the front of $z'$, we get a new pair of words enumerated by $H(x)$ except in two disjoint cases. The first such case is when the newly inserted last letter $A$ in $w$ is in the middle of a $CAB$-factor, and the new first segment of $z$ does not contain any letters $B$. The second such case is when the newly inserted last letter $A$ in $w$ is in the middle of a $CABB$-factor, and the new first segment of $z$ contains exactly one letter $B$. This leads to the functional equation

$$\begin{aligned} K(x)&=\frac{x^2}{(1-3x+x^2)^2} \cdot (1+K(x)) - \frac{x}{1-2x} \cdot \frac{x^2}{1-3x+x^2} \cdot xK(x)\\&\qquad \qquad -\frac{x^3}{1-3x+x^2}\cdot x\cdot \biggl (\frac{x}{1-2x}+1\biggr )\cdot x\cdot \frac{1}{1-2x} \cdot xK(x). \end{aligned}$$

The first and second summand of the right-hand side can be explained just as in the proof of Theorem 3.3. The third summand counts the pairs $(w,z)$ described as the second case in the previous paragraph. In such pairs, $S_2$ is a segment containing exactly one letter $B$. Such segments are counted by the generating function

$$\begin{aligned} x\cdot \biggl (\frac{x}{1-2x}+1\biggr )\cdot x\cdot \frac{1}{1-2x}, \end{aligned}$$

since in such segments, a letter $A$ is followed by a (possibly empty) sequence of $C$s and $D$s that ends in a letter $D$ if it is not empty, then comes a letter $B$, and then comes a sequence of $C$s and $D$s again. On the other hand, in such pairs, $S_1$ is just a segment into which two letters $B$ are inserted right after the first letter, leading to the generating function $x^3/(1-3x+x^2)$. Finally, the rest of the pair $(w,z)$ is just a pair counted by $K(x)$, with a letter $C$ inserted at the end of the first word of the pair. Solving the last displayed equation for $K(x)$, we get the statement of the theorem. $\square$

Corollary 4.3

There exists a constant $c$ such that $k_n\le c{\cdot } 3.70672^n$ holds for all $n$.

Proof

This follows from the fact that $K(x)$ is a rational function whose denominator has a unique root of smallest modulus, namely $0.26978$. Taking its reciprocal, our claim is proved. $\square$

As $S_n(1324)\le k_{2n}$ we have

Theorem 4.4

$L(1324)\le 13.73977$.

It goes without saying that further improvements are possible if we enforce the restriction of Lemma 4.1 not only for $k=2$, but for larger values of $k$ as well. However, these improvements will be minuscule. Indeed, let us replace restrictions (iv) and (v) by the following general restriction.

• (vi) For all $i\ge 1$ and all $k\ge 1$, if the $i$th letter $A$ from the right in $w$ is in the middle of a $CAB^k$-factor, then the $i$th segment of $z$ (from the left) contains at least $k$ letters $B$.

Theorem 4.5

Let $t_n$ be the number of pairs $(w,z)$ that satisfy restrictions (i)–(iii) and (vi). Let $T(x)=\sum _{n\ge 2}t_nx^n$. Then

$$\begin{aligned} T(x)=\frac{x^2(1-2x-x^2+x^3)}{1-8x+21x^2-19x^3-2x^4+11x^5-6x^6+x^7}. \end{aligned}$$

Proof

This follows from the functional equation

$$\begin{aligned} T(x)=\frac{x^2}{(1-3x+x^2)^2} \cdot (1+T(x)) - x\cdot T(x) \cdot \frac{x^3}{(1-3x+x^2)(1-2x-x^2+x^3)}, \end{aligned}$$

which can be proved in a way that is analogous to the proofs of Theorems 3.3 and 4.2. Note that the term $1-2x-x^2+x^3$ in the denominator is obtained from the infinite sum

$$\begin{aligned} \sum _{k\ge 0} \biggl (x^2 \cdot \biggl (1+\frac{x}{1-2x}\biggr ) \biggr )^{\!k} = \sum _{k\ge 0} \biggl ( \frac{x^2-x^3}{1-2x} \biggr )^{\!k} =\frac{1-2x}{1-2x-x^2+x^3}, \end{aligned}$$

which is explained by the enforcement of restriction (vi) for all $k$. $\square$

Corollary 4.6

$L(1324) \le 13.73719$.

Proof

This is immediate if we notice that $S_n(1324)\le t_{2n}$, and find the root of smallest modulus of the denominator of $T(x)$. $\square$

Other minor improvements can be obtained if one considers $CAAB$-factors in $w(p)$.

A more substantial improvement could possibly be obtained in the following way. The restrictions involving $CAB$-factors that we described have dual versions that involve $CDB$-factors. If one could find a way to count pairs of words that satisfy both kinds of restrictions at once, (that is, restrictions involving $CDB$-factors and restrictions involving $CAB$-factors), then a somewhat better upper bound could be obtained.