Width-$k$ Generalizations of Classical Permutation Statistics

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by  Robert Davis

WIDTH-k GENERALIZATIONS OF CLASSICAL PERMUTATION
STATISTICS
ROBERT DAVIS
7 Abstract. Weintroducenewnaturalgeneralizationsofthe classicaldescentandinversion
1 statisticsforpermutations,calledwidth-k descents andwidth-k inversions. Thesevariations
0 inducegeneralizationsoftheexcedanceandmajorstatistics,providingaframeworkinwhich
2
the most well-known equidistributivity results for classical statistics are paralleled. We

n explore additional relationships among the statistics providing specific formulas in certain
a special cases. Moreover, we explore the behavior of these width-k statistics in the context
J
of pattern avoidance.
7
1

]
O
1. Introduction
C
h. Let S denote the set of permutations σ = a ···a of [n], and let S = S ∪S ∪.... A
n 1 n 1 2
at function st : Sn → N is called a statistic, and the systematic study of permutation statistics
m is generally accepted to have begun with MacMahon [6]. In particular, four of the most
[ well-known statistics are the descent, inversion, major, and excedance statistics, defined

1 respectively by
v
8 desσ = |{i ∈ [n−1] | a > a }|
i i+1
8
invσ = |{(i,j) ∈ [n]2 | i < j and a > a }|
7 i j
4
majσ = i
0
.
1 i∈Desσ
X
0 excσ = |{i ∈ [n] | a > i}|
i
7
1 where Desσ = {i ∈ [n−1] | a > a }.
: i i+1
v Given any statistic st, one may form the generating function
i
X
Fst(q) = qstσ.
r n
a σX∈Sn
A famous result due to MacMahon [6] states that Fdes(q) = Fexc(q), and that both equal
n n
are to the Eulerian polynomial A (q). The Eulerian polynomials themselves may be defined
n
via the identity
A (q)
(1+j)nqj = n .
(1−q)n+1
j≥0
X
Anotherwell-knownresult, duetoMacMahon[5]andRodrigues[7],statesFinv(q) = Fmaj(q) =
n n
[n] ! Foata famously provided combinatorial proofs of the equidistributivity of each; see [4]
q
for a treatment of these. Thus any statistic st for which Fst(q) = A (q) is called Eulerian,
n n
Department of Mathematics, Michigan State University, East Lansing, MI 48824-1027, USA. E-mail:
[e-mail].
1
2 ROBERTDAVIS
and if Fst(q) = [n] ! then st is called Mahonian. These four statistics have many general-
n q
izations; in this article, we discuss new variations, induced from a simple generalization of
des.
For each of the following definitions, we assume n ∈ Z , k ∈ [n−1], ∅ 6= K ⊆ [n−1],
>0
and σ = a a ...a ∈ S . We define a width-k descent of σ to be an index i ⊆ [n−k] for
1 2 n n
which a > a . Thus the width-1 descents are the usual descents of a permutation. Denote
i i+k
the set of all width-k descents of σ by
Des (σ) = {i ∈ [n−k] | a > a },
k i i+k
and set
des (σ) = |Des (σ)|.
k k
If one is interested in descents of σ of various widths, first let K ⊆ [n−1] denote the set
of widths under consideration. Now, we define Des (σ) to be the multiset Des (σ),
K k∈K k
and des (σ) = |Des (σ)|.
K K
S
Now, define a width-k inversion of σ to be a pair (i,j) for which a > a and j −i = mk
i j
for some positive integer m. Denote the set of width-k inversions by
Inv (σ) = {(i,j) ∈ [n]2 | a > a and j −k = mk,k ∈ Z}
k i j
and set
inv (σ) = |Inv (σ)|.
k k
Again, one may beinterested in width-k inversions for multiple values of k, so for K ⊆ [n−1]
define Inv (σ) = Inv (σ). Additionally, let inv (σ) = |Inv (σ)|.
K k∈K k K K
Example 1.1. If σ = 4136572, then
S
Des (σ) = {1,4,5} and Inv (σ) = {(1,3),(1,7),(3,7),(4,7),(5,7)}.
{2,3} {2,3}
Thus, des (σ) = 3, inv (σ) = 5.
{2,3} {2,3}
As we will see in the next section, the above definitions motivate us to generalize exc
and maj in such a way that well-known known relationships among des,inv,maj,exc are
paralleled. However, it is not very convenient to work with exc and maj directly. So, this
K K
paper will focus much more on des and inv .
K K
The main focus of this paper is to explore these new statistics and their relationships
among each other. While they are well-behaved for special cases of K ⊆ [n− 1], formulas
for more general cases have been more elusive. Section 3 continues this exploration by
considering the same statistics for classes of permutations avoiding a variety of patterns.
2. Main Results
We begin with simple expressions for des (σ) and inv (σ) for a fixed σ ∈ S . First, we
K K n
point out that if K ⊆ [n−1], j ∈ K and ij ∈ K for some positive integer i, then for any
σ ∈ S , inv (σ) = inv (σ). This occurs because for any k ∈ [n−1], inv (σ) counts all
n K K\{ij} k
descents whose widths are multiples of k. Thus ij is already accounted for in inv (σ).
K\{j}
This follows quickly from the definitions of inv and des .
k k
Proposition 2.1. For any nonempty K ⊆ [n−1],
inv (σ) = (−1)|K′|+1inv (σ),
K lcm(K′)
∅(K′⊆K
X
where we set inv (σ) = 0 if lcm(K′) ≥ n.
lcm(K′)
WIDTH-k GENERALIZATIONS OF CLASSICAL PERMUTATION STATISTICS 3
Proof. We first consider the case where K = {k}. The elements of Inv (σ) are pairs of the
k
form (i,i+jk) for some positive integer j. Such an element exists if and only if there is a
width-(jk) descent of σ at i. Thus, inv (σ) simply counts the number of width-jk descents
k
of σ for all possible j. This leads to the equality
inv (σ) = des (σ).
k jk
j≥1
X
The formula for general K then follows from inclusion-exclusion: by adding the number of all
width-k inversions for all k ∈ K, we are twice counting any instances of a width-lcm(k ,k )
1 2
inversion since such an inversion is also of widths k and k . If K contains three distinct
1 2
elements k ,k ,k , then lcm(k ,k ,k ) would have been added three times (once for each
1 2 3 1 2 3
inv (σ)) and subtracted three times (once for each inv (σ), i < j), so it must be added
ki {ki,kj}
again for the sum. Extending this argument to range over larger subsets of K results in the
(cid:3)
claimed formula.
Example 2.2. Let us returnto σ = 4136572. Wesaw fromExample 1.1that inv (σ) = 5,
{2,3}
where four inversions have width 2 and two have width 3. But since the inversion (1,7) has
width both 2 and 3, it must also have width lcm(2,3) = 6. So, it contributes two summands
of 1 and one summand of −1.
We come now to a function which helps demonstrate the interactions among the width-k
statistics. Let n and k be positive integers for which n = dk + r for some d,r ∈ Z with
0 ≤ r < k. To each σ = a ...a ∈ S we may then associate the set of disjoint substrings
1 n n
β (σ) = {β1 (σ),...,βk (σ)} where
n,k n,k n,k
a a a ...a if i ≤ r
βi (σ) = i i+k i+2k i+dk
n,k
a a a ...a if r < i < k
( i i+k i+2k i+(d−1)k
Now, define
φ : S → Sr ×Sk−r
n d+1 d
by setting φ(σ) = (stdβ1 (σ),...,stdβk (σ)), where std is the standardization map, that is,
n,k n,k
the permutation obtained by replacing the smallest element of σ with 1, the second-smallest
element with 2, etc. Note in particular that each stdβi (σ) is a permutation of [d +1] or
n,k
[d].
The first of the identities in the following proposition was originally established in [8],
though with slightly different notation. We provide an alternate proof and extend their
result to width-k inversions.
Theorem 2.3. Let n and k be positive integers such that n = dk+r, where 0 ≤ r < k, and
let A (q) denotethe ith Eulerian polynomial. Also let M denote themultinomial coefficient
i n,k
n
M =
n,k (d+1)r,dk−r
(cid:18) (cid:19)
where ij indicates i repeated j times. We then have the identities
Fdesk(q) = M Ar (q)Ak−r(q)
n n,k d+1 d
Finvk(q) = M [d+1]r![d]k−r!
n n,k q q
4 ROBERTDAVIS
Proof. Let k ∈ [n − 1] and consider φ defined above. Note that φ is an M -to-one func-
n,k
tion since, given (σ ,...,σ ) ∈ Sr × Sk−r, there are M ways to partition [n] into the
1 k d+1 d n,k
subsequences βi (σ) such that stdβi (σ) = σ for all i. Also note that
n,k n,k i
k
des (σ) = des(stdβi (σ)).
k n,k
i=1
X
Thus,
Finvk(q) = qdeskσ
n
σX∈Sn
= M qdesσ1...qdesσk
n,k 
(σ1,...,σk)X∈Srd+1×Skd−r 
= M Ar (q)Ak−r(q). 
n,k d+1 d
This proves the first identity.
The second identity follows completely analogously, with the main modification being that
an element of Inv (σ) corresponds to a usual inversion in some unique stdβj (σ). (cid:3)
k n,k
Given des and inv , one must wonder what the corresponding generalizations of exc and
K K
maj are whose relationships with des and inv parallel that of the classical statistics. To
K K
do this, we define the multiset
k
Exc (σ) = {{j ∈ [n−1] | ⌈j/k⌉ ∈ Exc(stdβi (σ))}}
K n,k
k∈Ki=1
[ ]
and set exc (σ) = |Exc (σ)|, and also set
K K
i
maj (σ) = .
K k
kX∈Ki∈DXesk(σ)(cid:24) (cid:25)
We could equivalently state that
k k
maj(stdβi (σ)) and exc (σ) = exc(stdβi (σ)).
n,k K n,k
i=1 i=1
X X
These are exactly the definitions needed in order to obtain identities that parallel Fdes(q) =
n
Fexc(q) and Finv(q) = Fmaj(q), as we will soon see.
n n n
One important distinct to make between exc and exc is the following. If σ = a a ...a
K 1 2 n
and τ = b b ...b , then even if a = b for some i, one cannot say i ∈ Exc (σ) if and only
1 2 n i i K
if i ∈ Exc (τ). For example, if σ = 4136572, then 1 ∈ Exc (σ), but if τ = 4153627 then
K 2
1 ∈/ Exc (τ).
2
Example 2.4. Again let σ = 4136572. We then have
exc (σ) = |{1,3,1,2}| = 4
{2,3}
and
1 5 4
maj (σ) = + + = 6.
{2,3} 2 2 3
(cid:24) (cid:25) (cid:24) (cid:25) (cid:24) (cid:25)
WIDTH-k GENERALIZATIONS OF CLASSICAL PERMUTATION STATISTICS 5
k G G (q) G (q)
6,k 8,k 9,k
1 6A (q) 8A (q) 9A (q)
5 7 9
2 180A (q)2 1120A (q)2 9q−1A (q)
2 3 9
3 6! 8q−2A (q) 45360A (q)3
7 2
4 180q−2A (q)2 8! 9q−3A (q)
2 9
5 6q−4A (q) 8q−4A (q) 9q−4A (q)
5 7 9
6 1120q−4A (q)2 45360q−3A (q)3
3 2
7 8q−6A (q) 9q−6A (q)
7 9
8 9q−7A (q)
9
Table 1. The polynomials G (q) for n = 6,8,9 and 1 ≤ k ≤ n.
n,k
By constructing a nearly identical argument as in the proof Theorem 2.3, and using the
facts that Fdes(q) = Fexc(q) and Finv(q) = Fmaj(q), we have the following corollary.
n n n n
Corollary 2.5. The identities Fndesk(q) = Fnexck(q) and Fninvk(q) = Fnmajk(q) hold.
Now that we have established the analogous parallels between the four classical statistics
and their width-k counterparts, we wish to explore what other structure is present. A simple
proposition relates des and inv when k is large.
k k
Corollary 2.6. For all k ≥ n/2, Fdesk(q) = Finvk(q).
n n
Proof. Since k ≥ n/2, the sets βi (σ) contain at most two elements. So, width-k descents
n,k
and width-k inversions of σ are identical. (cid:3)
We now show that interesting behavior occurs when considering the function
G (q) = qdesk(σ)−desn−k(σ).
n,k
σX∈Sn
According to computational data, the following conjecture holds for all n ≤ 9 and 1 ≤ k < n
for which gcd(k,n) = 1.
Conjecture 2.7. If gcd(k,n) = 1, then G (q) = nq1−kA (q).
n,k n−1
Several illustrative polynomials are given in Table 2. This does not hold more generally,
and it would be interesting to determine if a general formula exists.
Question 2.8. For which values of n,k does there exist a closed formula for G (q), and
n,k
what is the formula?
3. Pattern Avoidance
We say that a permutation σ ∈ S contains the pattern π ∈ S if there exists a sub-
n m
sequence σ′ of σ such that std(σ′) = π. If no such subsequence exists, then we say that σ
avoids the pattern π. If Π ⊆ S, then we say σ avoids Π if σ avoids every element of Π. The
set of all permutations of S avoiding Π is denoted Av (Π). In a mild abuse of notation, if
n n
Π = {π}, we will write Av (π).
n
6 ROBERTDAVIS
In this section, we consider the functions
Fst(Π;q) = qstσ,
n
σ∈AXvn(Π)
which specializes to Fst(q) if Π = ∅. In most instances, Fdesk will be the main focus, but
n n
FdesK and Finvk will also make appearances.
n n
An important concept within pattern avoidance is that of Wilf equivalence. Two sets
Π,Π′ ⊂ S are said to be Wilf equivalent if |Av (Π)| = |Av (Π′)| for all n. In this case,
n n
we write Π ≡ Π′ to denote this Wilf equivalence, which is indeed an equivalence relation.
For example, it is known [3, 6] that whenever π,π′ ∈ S , then |Av (π)| = |Av (π′)| = C ,
3 n n n
where
1 2n
C =
n
n+1 n
(cid:18) (cid:19)
is the nth Catalan number.
Proving whether Π ≡ Π′ is often quite difficult, and their Wilf equivalence does not imply
that Fst(Π;q) = Fst(Π′;q). However, in some instances, the problems of establishing these
n n
identities have straightforward solutions by applying basic transformations on the elements
of the avoidance classes. Given π = a ...a ∈ S , let πr denote its reversal and πc denote
1 n m
its complement, respectively defined by
πr = a a ...a and πc = (m+1−a )(m+1−a )...(m+1−a ).
m m−1 1 1 2 m
Similarly, given Π ⊆ S, we let
Πr = {πc | π ∈ Π} and Πc = {πr | π ∈ Π}
be the reversal and complement of Π, respectively.
Our results of this section begin with a multivariate generalization of Fst(Π;q), and with
n
showing how its specializations describe relationships among Π,Πr, and Πc.
Definition 3.1. Fix Π ⊆ S. Define
n−1
T (Π;t ,...,t ) = tdesk(σ).
n 1 n−1 k
σ∈AXvn(Π)Yk=1
Several conclusions quickly follow. This function specializes to FdesK(Π;q) by setting
n
t = q for i ∈ K and t = 1 for all i ∈/ K. We can also recover FinvK(Π;q) from T by setting
i i n n
t = q whenever i ∈ [n−1]∩kZ for some k ∈ K, and setting t = 1 otherwise. Additionally,
i i
when Π = {π}, a nice duality appears, providing a mild generalization of Lemma 2.1 from
[1].
Proposition 3.2. For any Π ⊆ S, let Π′ denote either Πr or Πc. We then have
T (Π′;t ,...,t ) = tn−1tn−2···t T (Π;t−1,...,t−1 ).
n 1 n−1 1 2 n−1 n 1 n−1
Consequently,
T ((Πr)c;t ,...,t ) = T (Π;t ,...,t ).
n 1 n−1 n 1 n−1
Proof. It is enough to prove the claim when the set of patterns being avoided is {π} for
some π ∈ S , since the full result follows by applying the argument to all elements of Π
m
simultaneously.
WIDTH-k GENERALIZATIONS OF CLASSICAL PERMUTATION STATISTICS 7
First consider when π′ = πc and σ ∈ Av (π). Because σ ∈ Av (π) if and only if σc ∈
n n
Av (πc), we have that for each k, i ∈ Des (σ) if and only if i ∈/ Des (σc). This implies
n k k
Des (σc) = [n−k]\Des (σ), hence des (σc) = n−k −des (σ). So,
k k k k
n−1
T (πc;t ,...,t ) = tdesk(σ)
n 1 n−1 k
σ∈AXvn(πc)kY=1
n−1
= tdesk(σc)
k
σ∈AXvn(π)kY=1
n−1
= tn−k−desk(σ)
k
σ∈AXvn(π)kY=1
= tn−1tn−2···t T (π;t−1,...,t−1 ).
1 2 n−1 n 1 n−1
Proving that the result holds for π′ = πr follows similarly.
The second identity in the proposition statement holds by applying the first identity twice:
first for Πr and then for Πc. (cid:3)
The above identities significantly reduce the amount of work needed to study FdesK(Π;q)
n
for all Π ⊆ S . For the remainder of this paper, we systematically approach Π for |Π| ≤ 2.
n
3.1. Avoiding singletons. By Proposition 3.2, we immediately get
Fdesk(123;q) = qn−kFdesk(321;q−1)
n n
and
Fdesk(132;q) = Fdesk(213;q) = qn−kFdesk(231;q−1) = qn−kFdesk(312;q−1).
n n n n
So, studying Fdesk(π;q) for π ∈ S reduces to studying the function for a choice of one
n 3
pattern from {123,321} and one from the remaining patterns. For some choices of Π, the
permutations in Av (Π) are especially highly structured, which leads to similar arguments
n
throughout the rest of this paper.
Webeginwithπ = 312. Noticethatifa ...a ∈ Av (312)anda = 1,thenstd(a ...a ) ∈
1 n n i 1 i−1
Av (312), std(a ...a ) ∈ Av (312), and
i−1 i+1 n n−i
max{a ,...,a } < min{a ,...,a }.
1 i−1 i+1 n
Proposition 3.3. For all n,
k n
Fdesk(312;q) = C Fdesk(312;q)+ qFdesk(312;q)Fdesk(312;q)
n i−1 n−i i−1 n−i
i=1 i=k+1
X X
where C is the ith Catalan number.
i
Proof. First consider when σ = a ...a ∈ Av (312) and a = 1 for some i ≤ k. By the
1 n n i
discussion preceding thisproposition, j ∈/ Des (σ)foranyj ≤ i. So, noneoftheC possible
k i−1
permutations that make up std(a ...a ) contribute to des (σ). The only contributions to
1 i−1 k
des (σ) come from std(a ...a ) ∈ Av (312;q). The overall contribution to Fdesk(312;q)
k i+1 n n−i n
is the first summand of the identity.
Now suppose a = 1 for some i > k. Each choice of a ...a , contributes to des (σ) as
i 1 i k
usual, but there will be an additional width-k descent produced at i − k. The elements
a ...a contribute to des (σ) as before. The overall contribution to Fdesk(312;q) is the
i+1 n k n
8 ROBERTDAVIS
first summand of the identity. Since we have considered all possible indices i for which we
could have a = 1, we add the two cases together and are done. (cid:3)
i
Note that when we set q = 1, the above recursion specializes to the well-known recursion
n
for Catalan numbers C = C C .
n+1 i=0 i n−i
Wecanusetheprevious propositioninconjunction withProposition3.2to produceformu-
laeforFdesk(Π;q)andFinvk(ΠP;q) whenever Πisa single element ofS other than123or 321.
n n 3
The only nontrivial work required, then, is to compute the degrees of the two polynomials.
Corollary 3.4. For all n,
n−k
n−i
degFdesk(312;q) = n−k and degFinvk(312;q) = .
n n k
i=1 (cid:22) (cid:23)
X
Proof. The degrees of the above polynomials are given by identifying a permutation in
Av (312) with the most possible descents. This is satisfied by n(n − 1)...21 ∈ Av (312)
n n
which has n−k descents of width k. Determining the number of width-k inversions in this
(cid:3)
permutation is done similarly.
Although 321 is Wilf equivalent to 312, it is not so obvious how to construct a recurrence
relation for Fdesk(321;q). To help describe the elements of Av (321), first let σ = a ...a ∈
n n 1 n
S and call a a left-right maximum if a > a for all j < i. Thus σ ∈ Av (321) if and
i i j n
only if its set of non-left-right maxima form an increasing subsequence of σ. Indeed, if the
non-left-right maxima did contain a descent, then there would be some left-right maximum
preceding both elements, which violates the condition that σ avoid 321. Despite such a
description, using it to reveal Fdesk(321;q) has thus far been unsuccessful. So, we leave the
n
following as an open question.
Question 3.5. Is there a closed formula or simple recursion for Fdesk(321;q) (equivalently,
n
for Fdesk(123;q))?
n
3.2. Avoiding doubletons. At this point, we begin studying the functions Fdesk(Π;q)
n
when avoiding doubletons from S . Recall that, by the Erd˝os-Szekeres theorem [2], there
3
are no permutations in S for n ≥ 5 that avoid both 123 and 321. Thus, we will not consider
n
this pair. Additionally, the functions Finvk(Π;q) are quite unwieldy, so we will not consider
n
these either.
For our first nontrivial example, we begin with {123,132}. Permutations a ...a ∈
1 n
Av (123,132) have the following structure: for any i = 1,...,n, if a = n, then the substring
n i
a a ...a is decreasing and consists of the elements from the interval [n − i + 1,n − 1].
1 2 i−1
Additionally, the substring a ...a is an element of Av (123,132). This structure makes
i+1 n n−i
it easy to show that there are 2n−1 elements of Av (123,132) [9].
n
Proposition 3.6. For all n and 1 ≤ k ≤ n−1,
k
Fdesk(123,132;q) = qmin(i,n−k)Fdesk(123,132;q)
n n−i
i=1
X
n−k
+ qmin(i−1,n−k−1)Fdesk(123,132;q)
n−i
i=k+1
X
+2n−max(k+1,n−k+1)qn−k−1.
WIDTH-k GENERALIZATIONS OF CLASSICAL PERMUTATION STATISTICS 9
Proof. Let σ = a ...a ∈ Av (123,132;q). If a = n for i ≤ k, then it is clear from the
1 n n i
preceding description of the elements in Av (123,132) that all of 1,...,min(i,n − k) are
n
elements of Des (σ). If j > n − k for some j, then j ∈/ Des (σ) since a does not exist.
k k j+k
The remaining elements of Des (σ) arise as a width-k descent of a ...a , which accounts
k i+1 n
for the factor of Fdesk(123,132;q) in the first summand.
n−i
Now, if k +1 ≤ n−k and a = n for i = k +1,...,n−k, then every element of 1,...,i
i
except i−k is anelement ofDes (σ). If k+1 > n−k, thenthesecond summand is empty and
k
nothing is lost by continuing to the case of a = n for i ≥ max(k+1,n−k+1). Again, the
i
remaining elements of Des (σ) are the width-k descents of a ...a , hence the additional
k i+1 n
factor of Fdesk(123,132;q). This accounts for the second summand.
n−i
Finally, let m = max(k +1,n−k + 1). If a = n for i ≥ m, then all of 1,...,n−k are
i
descents except i−k, and these are the only possible width-k descents. In particular, none
of i+1,i+2,...,n can be the index for a width-k descent. This leads to the sum
n n−1
qn−k−1|Av (123,132)| = 1+ 2n−i−1 qn−k−1 = 2n−mqn−k−1,
n−i
!
i=m i=m
X X
(cid:3)
which accounts for the third summand.
Next, we consider {123,312}. We proceed similarly as before but with some minor differ-
ences, reflecting the new structure we encounter. If σ = a ...a ∈ Av (123,312) and a = 1
1 n n i
for some i < n, then σ is of the form
σ = i(i−1)...21n(n−1)...(i+2)(i+1),
since neither subsequence a ...a anda ...a may contain anascent. If a = 1, though,
1 i−1 i+1 n n
then std(a ...a ) ∈ Av (123,312).
1 n−1 n−1
Proposition 3.7. For all n and 1 ≤ k < n,
k n−1
Fdesk(123,312;q) = qmax(0,n−k−i) + qmax(n−2k,i−k)
n
i=1 i=k+1
X X
+qFdesk(123,312;q)
n−1
Proof. Suppose a = 1 for some i ≤ k. Using the description of elements in Av (123,312),
i n
we know there are n−k −i width-k descents. Since there is only one such permutation for
each i, we simply add all of the qn−k−i, so long as n−k −i ≥ 0. If this inequality does not
hold, then there are no width-k descents in this range. This accounts for the first summand.
The second summand is computed very similarly to that of the second summand in
Proposition 3.6. The final summand is a direct result of noting that when a = 1, then
n
(a −1)...(a −1) may be any element of Av (123,312;q), which accounts for the fac-
1 n−1 n−1
tor Fdesk(123,312;q). For each of these choices, we know n−k ∈ Des (σ), hence the factor
n−1 k
of q. Adding the sums results in the identity claimed. (cid:3)
Now we consider {132,231}. Note that if a ...a ∈ Av (132,231), then a 6= n for any
1 n n i
1 < i < n. If a = n, then std(a ...a ) ∈ Av (132,231), and similarly if a = n. Once
1 2 n n−1 n
again, this allows us to quickly compute that |Av (132,231)| = 2n−1.
n
10 ROBERTDAVIS
Proposition 3.8. Let K = {k ,...,k } be a nonempty subset of [n−1] whose elements are
1 l
listed in increasing order. We then have
l+1
FdesK(132,231;q) = (1+qi−1)ki−ki−1
n
i=1
Y
where k = 1 and k = n.
0 l+1
Proof. It follows from the description of elements in Av (132,231) that there are two sum-
n
mands in a recurrence for FdesK(132,231;q): one corresponding to a = n and one corre-
n 1
sponding to a = n. When a = n, then there are |K| = l copies of 1 ∈ Des (σ); if a = n,
n 1 K n
thena makes no contribution toDes (σ). Thus, by deleting nfromσ, we get therecurrence
n K
FdesK(132,231;q) = (1+ql)FdesK(132,231;q).
n n−1
Repeating this, a factor of 1+ql appears until we get to
FdesK(132,231;q) = (1+ql)n−klFdesK(132,231;q).
n k
l
At this point, note that
FdesK(132,231;q) = FdesK\{kl}(132,231;q).
k k
l l
(cid:3)
Repeating the previous argument ends up with the identity claimed.
Next, consider {132,213}. If σ = a ...a ∈ Av (132,213), then if a = n for any i, then
1 n n i
a ...a must be increasing in order to avoid 213. Moreover,
1 i−1
max(a ,...,a ) < a
i+1 n 1
in order to avoid 132, and std(a ...a ) ∈ Av (132,213).
i+1 n n−i
Proposition 3.9. For all n and 1 ≤ k < n,
k
Fdesk(132,213;q) = qmin(i,n−k)Fdesk(132,213;q)
n n−i
i=1
X
n−k
+ qmin(k,n−i)Fdesk(132,213;q)
n−i
i=k+1
X
n
+ qn−iFdesk(132,213;q).
n−i
i=max(k+1,n−k+1)
X
Proof. The proof of this is entirely analogous to the proof of Proposition 3.6. (cid:3)
Next, we consider {132,312}. For each i = 1,...,n−1, either a = max{a ,...,a }+1
i+1 1 i
or a = min{a ,...,a }−1. This doubleton often results in especially pleasant formulas,
i+1 1 i
and our results are no exception.
Proposition 3.10. Let K = {k ,...,k } be a nonempty subset of [n− 1] whose elements
1 l
are listed in increasing order. We then have
l+1
FdesK(132,312;q) = (1+qi−1)ki−ki−1
n
i=1
Y
where k = 1 and k = n.
0 l+1