 # Catalan words avoiding pairs of length three patterns

Catalan words are particular growth-restricted words counted by the eponymous integer sequence. In this article we consider Catalan words avoiding a pair of patterns of length 3, pursuing the recent initiating work of the first and last authors and of S. Kirgizov where (among other things) the enumeration of Catalan words avoiding a patterns of length 3 is completed. More precisely, we explore systematically the structural properties of the sets of words under consideration and give enumerating results by means of recursive decomposition, constructive bijections or bivariate generating functions with respect to the length and descent number. Some of the obtained enumerating sequences are known, and thus the corresponding results establish new combinatorial interpretations for them.

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## 1 Introduction and notation

Catalan words are particular growth-restricted words and they represent still another combinatorial class counted by the Catalan numbers, see for instance [11, exercise 6.19.u, p. 222]. This paper contributes to a recent line of research on classical pattern avoidance on words subject to some growth restrictions (for instance, ascent sequences [2, 5], inversion sequences [4, 9, 13], restricted growth function [3, 8]) by investigating connections between sequences on the On-line Encyclopedia of Integer Sequences  and Catalan words avoiding two patterns of length .

Through this paper we consider words over the set of non-negative integers and we denote such words by sequences (for instance ) or by italicized boldface letter (for instance and ). The word is called Catalan word if

 w1=0 and 0≤wi≤wi−1+1 for i=2,3,…,n.

Catalan words are in bijection with maybe the most celebrated combinatorial class having the same enumerating sequence: Dyck paths111A Dyck path is a path in the first quadrant of the plane which begins at the origin, ends at , and consists of up steps and down steps .. Indeed, in a length Dyck path collecting for the up steps the ordinates of their starting points we obtain a length Catalan word, and this construction is a bijection. See Figure 1 where this bijection is depicted for an example. We denote by the set of length Catalan words and is the th Catalan number

A pattern is a word with the property that if occurs in it, then so does , for any with . A pattern is said to be contained in the word , , if there is a sub-word of , , order-isomorphic with . If does not contain , we say that avoids , see for instance Kitaev’s seminal book  on this topic.
For a pattern , we denote by the set of length Catalan words avoiding , and is the cardinality of and . For example, is the set of Catalan words avoiding , that is, the set of words in such that there are no , and , , with . So, . Likewise, if is the set of patterns , then and denote both the set of length Catalan words avoiding each pattern in ; and and have similar mining as above. A descent in a word is a position , , with . The (ordinary) generating function of a set of pattern avoiding Catalan words is the formal power series

 Cπ(x)=∑n≥0cn(π)xn=∑w∈C(π)x|w|,

where is the length of the word . In our case of generating function approach for counting classes of pattern avoiding Catalan words we consider the descent number as an additional statistic obtaining ‘for free’ the bivariate generating function

 Cπ(x,y)=∑w∈C(π)x|w|ydes(w),

where is the number of descents of . With these notations, the coefficient of in is the number of Catalan words of length avoiding and having descents, and for a set of Catalan words and have similar meaning.

For a word and an integer , we denote by the word obtained from by increasing by each of its entries, that is, the word . In our constructions we will often make use of two particular families of Catalan words: those avoiding (i.e., with no descents) and we call these words weakly increasing (or w.i for short) Catalan words; and those avoiding (and thus necessarily avoiding ) and we call these words strictly increasing (or s.i for short) Catalan words. It is easy to see that for each length there are w.i. Catalan words and one s.i. Catalan word.

The remaining of this paper is structured as follows. In Section 2 we consider classes of Catalan words avoiding both a length two and a length three pattern. In the next sections we discuss Catalan words avoiding two patterns of length three, in increasing order of their complexity: obvious cases (Section 3), cases counted via recurrences (Section 4) and cases counted via generating functions (Section 5); these results are summarized in Table 2. We conclude with some remarks and further research directions.

## 2 Avoiding a length two and a length three pattern

There are three patterns of length two, namely , and , and we have:

###### Proposition 1.

The number of Catalan words avoiding a pattern of length two and a pattern of length three is given by:

 cn(00,π)={0if π=012 and n≥3,1elsewhere; cn(01,π)={0if π=000 and n≥3,1elsewhere; cn(10,π)=⎧⎪⎨⎪⎩Fnif π=000,nif π∈{001,011,012},2n−1elsewhere,

where is the th Fibonacci number defined by and , .

###### Proof.

If , then . Thus, except when and in this case for , and the counting relation for follows.
Similarly, if , then . Thus, except when and in this case for .
Finally, a Catalan word avoids if and only if it avoids . It follows that , witch falls in the case of avoidance of two length 3 patterns and the corresponding proofs are given in the next section, see also Table 2. ∎

## 3 Trivial cases

### Superfluous patterns

If the pattern contains the pattern , then clearly ; but this phenomenon can occur even when and are not related by containment and in this case, following , we say that is a superfluous pattern for . For example, any word in is a binary word, and thus any pattern with at least three different symbols is a superfluous pattern for . In Table 1 are listed all pairs of superfluous patterns of length three. It is worth to mention that superfluousness is a transitive relation, for instance is superfluous for which in turn is superfluous for . So, a pattern can be superfluous for several other ones, for instance is superfluous for each of the patterns . Also it is easy to see that if is superfluous for , then is larger lexicographically than .

### Ultimately constant sequences

It can happen that the number of Catalan words avoiding a pair of length patterns is constant for enough long words. The only two such cases are given below.

and

###### Proof.

If , then where is the word , and the first point follows.
If a Catalan word avoids , then it is a binary word. In addition if its length is larger than it necessarily contains three identical entries, and so for . Considering the initial values of the second point follows. ∎

### Counting sequence n

###### Proposition 3.

If is one of the pairs of patterns , , , , or , then .

###### Proof.

The proof is based on the easy to understand description given below for the corresponding sets of Catalan words:

## 4 Counting via recurrence

### Counting sequence 2(n−1)

###### Proposition 4.

If or , then for (Sequence  in ).

###### Proof.

If and , , then either

1. [parsep=-0.5mm,topsep=0mm]

2. , , with a s.i. Catalan word (of length at least one), or

3. , , with a s.i. Catalan word (of length at least two).

In the first case there are possibilities for and possibilities in the second case, and the result holds. If and , , then either

1. [parsep=-0.5mm,topsep=0mm]

2. , , with a s.i. Catalan word, or

3. , ,

and as previously the result holds. ∎

### Sequences involving 2n

###### Proposition 5.

If , then for ( in ).

###### Proof.

If a Catalan word avoids , then it is unimodal (that is, it can be written not necessarily in a unique way as with a weakly decreasing and weakly decreasing word). In addition, if the word avoids , then its maximal value occurs at most twice, and when it occurs twice this happens in consecutive positions.
We denote by the subset of words in where the maximal entry occurs once and by that where it occurs twice, , , and . Any word with its maximal value occurring twice can be extended into a word in by one of the transformations:

• , and

• ,

and any word with its maximal value occurring once can be extended into a word in by:

• .

Conversely, any word in , , can uniquely be obtained from a word in or in by applying one of the transformations above, so

 dn=2⋅en−1+dn−1,

for .
Reasoning in a similar way we have

 en=2⋅en−2+dn−2,

Thus, for , , and finally

 cn(π)=dn+en=2⋅(dn−1+en−1)=2⋅cn−1(π),

and with the initial conditions and , the result follows. ∎

###### Proposition 6.

If or , then for ( in ).

###### Proof.

If and , , then either

1. [parsep=-0.5mm,topsep=0mm]

2. with , or

3. with , or

4. with a w.i. Catalan word of length , .

The number of words in each of the first two cases is . The number of length w.i. Catalan words is , so the number of words in the last case is

 n−2∑k=12k−1=n−3∑k=02k=2n−2−1.

Thus, , and after calculation the result holds.

If and , , then either

1. [parsep=-0.5mm,topsep=0mm]

2. with , or

3. with , or

4. with a length binary word other than .

The number of words in each of the first two cases is , and the number of words in the last case is . So , and again the result holds. ∎

###### Proposition 7.

If is one of the pairs of patterns , , or , then for ( in ).

###### Proof.

If and , , then either

1. [parsep=-0.5mm,topsep=0mm]

2. is a w.i. Catalan word, or

3. with a w.i. Catalan word of length other than , or

4. where is w.i. binary word ending by and is a w.i. Catalan word of length , .

In the first case the number of words is and in the second case the number of words is . In last case the number of words is

 n−3∑k=1(n−k−2)⋅2k−1=2n−2−(n−1).

Combining these cases and considering the initial values of the result holds.

If and , , then either

1. [parsep=-0.5mm,topsep=0mm]

2. with , or

3. with a w.i. Catalan word of length , .

The number of words in the first case is and the number of words in the second case is . So and after calculation the result follows.

If and , , then either

1. [parsep=-0.5mm,topsep=0mm]

2. with , or

3. with , or

4. with a s.i. Catalan word of length , .

The number of words in each of the first two cases is . For the last case, for each there is exactly one word , so their number is in this case. So and after calculation the result holds.

If and , , then either

1. [parsep=-0.5mm,topsep=0mm]

2. with , or

3. with , or

4. , .

So, again . ∎

###### Proposition 8.

If , then for ( in ).

###### Proof.

If , , then either

1. [parsep=-0.5mm,topsep=0mm]

2. with a length binary word, or

3. with a length w.i. Catalan word, or

4. with a binary word and w.i. Catalan word.

The number of words in each of the first two cases is and the number of words in the last case is

 n−2∑k=12n−k−22k−1=(n−2)2n−3,

and so , which gives the desired result. ∎

If a Catalan word avoids both and , then it has at most one descent. In the second part of the proof of the next proposition we need the following technical lemma which gives the number of Catalan words in with one descent and avoiding the pattern before the descent (note that in this case avoiding is equivalent to avoiding equal consecutive entries). The set of these words is empty for , it is the single word set for , and for .

###### Lemma 1.

Let be the set of words in having one descent and avoiding before the descent. Then .

###### Proof.

A word belongs to , , if and only if it can be written as

 012…(k−1)a(u+a),

with ( is the position of the unique descent in the word), , and is a length w.i. Catalan word over . For each choice of , there are choices for , and for each choice for there are choices for . It follows that

 |Dn|=n−1∑k=2(k−1)⋅(n−k),

and after calculation the statement holds. ∎

###### Proposition 9.

If or , then for ( in ).

###### Proof.

If and , , then either

1. [parsep=-0.5mm,topsep=0mm]

2. with a length binary word, or

3. with a length , , w.i. Catalan word other than and beginning by at least one .

The number of words in the first case is and the number of those in the second case is

 n−1∑k=2(2k−1−1)⋅(n−k)=3⋅2n−1−12(n+1)(n+2)+n,

and combining the two cases the result holds.

If and , then either

1. [parsep=-0.5mm,topsep=0mm]

2. is a w.i. Catalan word, the number of such words is , or

3. , with defined in Lemma 1, or

4. , where is a w.i. Catalan word of length , , is the maximal (rightmost) entry of , and is a word belonging to .

Indeed, the first case corresponds to words with no descents, the second one to those with a descent and no occurrences of before the descent, and the third one to those with both descent and occurrences of before the descent (the rightmost such occurrence is in positions and ). By Lemma 1, the number of words in the third case is

 n−3∑k=12k−1⋅(n−k)⋅(16(n−k)2−12(n−k)+13)=2n−16(n+1)(n2−n+6).

Finally, combining the three previous cases the desired result holds. ∎

### Sequences involving binomial coefficient

In this part we use the notation

 ab⋯cd↑ie⋯f

to denote that the entry is in position in the word .

###### Proposition 10.

If , then for ( in ).

###### Proof.

If , then it has at most one descent.

If has no descents, then it has the form and there are such words.

If has one descent, then it has the form with . It follows that there is a bijection between the family of -element subsets of and the words in with one descent:

 {k,m,ℓ}↦0123⋯(m−1)m⋯↑ℓmk⋯k.

Combining the two cases we have . ∎

###### Proposition 11.

If is one of the pairs of patterns , , , , or , then for ( in ).

###### Proof.

In any of the six cases for , the set is in bijection with the family of subsets of with at most two elements. We give below explicit definitions for such bijections, where the empty set is mapped to by each of them.

If and , then either or for some , with or with , , and the desired bijection is

• [parsep=-2mm,topsep=0.0mm]

• ;

• .

If and , then either or with the maximal entry of , or with , and the desired bijection is

• [parsep=-2mm,topsep=0.0mm]

• ;

• .

If and , then either , or for some , with or with , and the desired bijection is

• [parsep=-2mm,topsep=0.0mm]

• ;

• .

For the next three cases we need the following observation: a Catalan word avoids if and only if it is a binary word (over ) beginning by a .

If and , then either with , or with , , and the desired bijection is

• [parsep=-2mm,topsep=-0.0mm]

• ;

• .

If and , then either with , or with , , and the desired bijection is

• [parsep=-2mm,topsep=0.0mm]

• ;

• .

If and , then either with , or with , , and the desired bijection is

• [parsep=-2mm,topsep=0.0mm]

• ;

• .

### Sequences involving Fibonacci(-like) numbers

As in Proposition 1, we consider the sequence of Fibonacci numbers defined as , and for .

###### Proposition 12.

If or , then for ( in ).

###### Proof.

For the proof is up to a certain point similar to that of Proposition 5. A word belonging to is unimodal and its maximal entry occurs once or twice in consecutive positions. Let denote the subset of words in where the maximal entry occurs once and denote that where it occurs twice. If has its maximal entry , then the insertion of after the rightmost occurrence of in produces a word in , and the insertion of produces a word in . It is easy to see that these transformations induce a bijection between and , and between and , and thus between and . It follows that satisfies a Fibonacci-like recurrence, and by considering the initial values for the result holds.

For , a word is characterized by: is w.i. and does not have three consecutive equal entries. So can be represented by the binary word where iff . And this representation is a bijection between and the set of binary words of length without two consecutive s, which cardinality is the Fibonacci number, see for instance . ∎

If , then for .

###### Proof.

If , , then either

1. [parsep=-0.5mm,topsep=0mm]

2. , or

3. with , or

4. with .

So, satisfies the recurrence for , and solving it we have the desired result. ∎

In the next proposition we will make use of the following relation satisfied by the even index Fibonacci numbers: for .

###### Proposition 14.

If , then for ( in ).

###### Proof.

If , , then either

1. [parsep=-0.5mm,topsep=0mm]

2. with , or

3. with , or

4. with , or

5. with for some , .

In both of the first two cases the numbers of words is and in the third case, this number is . In the last case, the number of words is So, , and with the initial conditions we have . ∎

The sequence of Pell numbers is defined as , and for .

###### Proposition 15.

If , then is the th Pell number for ( in ).

###### Proof.

If , , then either

1. [parsep=-0.5mm,topsep=0mm]

2. with , or

3. with , or

4. with .

In both of the first two cases the numbers of words is and it is in the last case. So, satisfies Pell numbers recurrence and considering its initial values the statement holds. ∎

## 5 Counting via generating function

Here we give bivariate generating functions where the coefficient of is the number of Catalan words of length having descents and avoiding , for each of the remaining pairs of patterns of length . Plugging in we obtain where the coefficient of is the number of Catalan words of length avoiding . All the obtained enumerating sequences are not yet recorded in , except that for: and for (see Corollary 2) and presumably for (see Corollary 14). In almost all the proofs of the next propositions the desired generating function is the solution of a functional equation satisfied by it.

###### Proposition 16.

If , then

 Cπ(x,y)=−x4y+x2y+1x2+x−1.
###### Proof.

Here we need the generating function for the Fibonacci numbers