Restricted partitions and q-Pell numbers Toufik Mansour Department of Mathematics, University of Haifa, 31905 Haifa, Israel
[email protected]
Mark Shattuck Department of Mathematics, University of Tennessee, Knoxville, TN 37996
[email protected]
Abstract In this paper, we provide new combinatorial interpretations for the Pell numbers pn in terms of finite set partitions. In particular, we identify six classes of partitions of size n, each avoiding a set of three classical patterns of length four, all of which have cardinality given by pn . By restricting the statistic recording the number of inversions to one of these classes, and taking it jointly with the statistic recording the number of blocks, we obtain a new polynomial generalization of pn . Similar considerations using the comajor index statistic yields a further generalization of the q-Pell number studied by Santos and Sills.
Keywords: pattern avoidance, inversion, major index, Pell number, q-generalization. 2010 Mathematics Subject Classification: 05A18, 05A15.
1
Introduction
We’ll use the following notational conventions: N := {0, 1, 2, . . . }, P := {1, 2, . . . }, [0] := ∅, and [n] := {1, . . . , n} for n ∈ P. Empty sums take the value 0 and empty products the value 1, with 00 := 1. The letter q denotes an indeterminate, with 0q := 0, nq := 1 + q + · · · + q n−1 for n ∈ P, nq ! 0q !:=1, nq ! := 1q 2q · · · nq for n ∈ P, and nk q := kq !(n−k) for n ∈ N and 0 6 k 6 n. The q! n binomial coefficient k is equal to zero if k is a negative integer or if 0 6 n < k.
Let pn denote the Pell number defined by the recurrence pn = 2pn−1 + pn−2 if n ≥ 2, with the initial conditions p0 = 0 and p1 = 1. By convention, we will take p−1 = 1. The pn count, most notably, the square and domino tilings of length n − 1 in which the squares come in two varieties or, equivalently, the lattice paths from (0, 0) to the line x = n − 1 consisting of U = (1, 1), D = (1, −1) and H = (2, 0) steps. They also enumerate, among other items, the 132-avoiding two-stack sortable permutations of size n as well as the walks of length n from one vertex of a triangle to another to which a loop has been added. See, e.g., A000129 of [12] for further information on these numbers as well as the papers of [4] and [2] for various identities. Here, we provide new interpretations of pn in terms of finite set partitions, the first such examples that we have been able to find. In particular, they arise as enumerators of certain pattern-avoidance classes. A partition of [n] is any collection of nonempty, disjoint subsets, called blocks, whose union is [n]. A partition with k blocks is also called a k-partition and is denoted by B1 /B2 / . . . /Bk , where the blocks are arranged in the standard order: min(B1 ) < · · · < min(Bk ). The set of k-partitions of [n] will be denoted by Pn,k , and the set of all partitions of [n] by Pn . In what follows, we will represent Π = B1 /B2 / . . . /Bk ∈ Pn,k , equivalently, by the canonical sequential form π = π1 π2 · · · πn wherein j ∈ Bπj , 1 ≤ j ≤ n, and in such case we will write Π = π. Note 1
that the word π = π1 π2 · · · πn is a restricted growth function from [n] onto [k] (see e.g., [14] for details). For example, the partition Π = 1, 7/2, 4, 5/3/6 ∈ P7,4 has the canonical sequential form π = 1232241. A generalized pattern τ is a member of [ℓ]m which contains all of the letters in [ℓ]. We say that a word σ ∈ [k]n contains a generalized pattern τ if σ contains a subsequence isomorphic to τ (i.e., one that contains all of the same pairwise comparisons as τ ). Otherwise, we say that σ avoids τ . For example, a word σ = a1 a2 · · · an avoids the pattern 132 if it has no subsequence ai aj ak with i < j < k and ai < ak < aj and avoids 1212 if it has no subsequence ai aj ak aℓ with i < j < k < ℓ and ai = ak < aj = aℓ . The first consideration of pattern avoidance began with that of permutations avoiding a pattern τ of length 3. Knuth [8] found that for any τ ∈ S3 , there are Cn members of Sn which avoid τ , where Cn denotes the nth Catalan number. This result was later generalized by Simion and Schmidt [11] who determined the number of elements of Sn avoiding the patterns in any subset of S3 . More recently, there has been comparable work done on pattern avoidance in set partitions; see, e.g., the papers by Klazar [7], Sagan [9], and Jel´ınek and Mansour [6] and the references therein. In what follows, we will consider the problem of avoidance of certain generalized patterns in set partitions, represented as words in their canonical sequential forms. If {w1 , w2 , . . .} is a set of generalized patterns, then let Pn (w1 , w2 , . . .) denote the subset of Pn which avoid all of the patterns. In this paper, we identify six classes of partitions each avoiding a set of three patterns of length four and each enumerated by the Pell number pn . This extends recent work concerning the avoidance of three-letter patterns (see [5] and [9]), and partially addresses the question of avoiding more than one pattern of length four mentioned in Goyt (see first paragraph of final section of [5]). For one of the aforementioned avoidance classes, we enumerate its members jointly according to the statistics which record the number of blocks and the number of inversions, where partitions are represented as words in their canonical sequential forms. For another class, similar considerations are made using instead the comajor index statistic. In the former case, this leads to a new two-variable x, q-generalization of pn which reduces to pn when x = q = 1. In the latter, we obtain an x, q-generalization, which when x = q reduces to a variant of the q-Pell numbers introduced recently by Santos and Sills [10]. In both cases, we derive explicit formulas for the resulting polynomials which reduce to the familiar Pell number formula pn =
⌊ n−1 2 ⌋
X j=0
n − 1 − j n−1−2j 2 j
when x = q = 1. Using these formulas, we are able to derive simple explicit expressions for the total number of inversions and the total comajor index value of all of the members of the respective avoidance classes. Recall that the inversion (denoted by inv ) and comajor index (comaj ) statistics for the word w = w1 w2 · · · wm in some alphabet consisting of positive integers are given, respectively, by inv(w) := |{(i, j) : 1 ≤ i < j ≤ m and wi > wj }| and comaj(w) :=
X
i∈A(w)
i,
where A(w) := {1 ≤ i ≤ m − 1 : wi < wi+1 }.
For example, if m = 8 and w = 12322132, then inv (w) = 1 + 1 + 4 + 2 = 8, A(w) = {1, 2, 6}, and comaj (w) = 1 + 2 + 6 = 9.
2
Pattern avoidance and Pell numbers
The following result identifies six classes of partitions of [n] each avoiding three patterns of length four and each having cardinality pn . 2
Theorem 2.1. If n ≥ 1, then |Pn (w, w′ , w′′ )| = pn for the following sets (w, w′ , w′′ ): (1) (1112, 1212, 1213) (4) (1211, 1212, 1213)
(2) (1121, 1212, 1213) (5) (1211, 1221, 1231)
(3) (1121, 1221, 1231) (6) (1212, 1222, 1232).
Proof. We prove all but the fourth and sixth cases, which we consider later. Throughout, we denote Pn (w, w′ , w′′ ) by Rn and let rn = |Rn | for the particular set of patterns under consideration. Also, given π ∈ Pn , we denote by inc(π) the member of Pn+1 gotten by adding 1 to each letter of π and then writing a 1 in front; for example, if π = 12123 ∈ P5 , then inc(π) = 123234 ∈ P6 .
Proof of (1): For the first case, note that any non-empty member λ ∈ Pn (1112, 1212, 1213) where n ≥ 3 must be of one of the following three forms: (i) |11 {z · · · 1}, (ii) 1π ′ 11 · · · 1}, or (iii) | {z n
a
11π ′ 11 · · · 1}, where 0 ≤ a ≤ n − 2, 0 ≤ b ≤ n − 3, and π ′ is any non-empty partition on the | {z b
letters {2, 3, . . .} avoiding the three patterns.
Thus, all members of Rn may be obtained by one of the following: (a) performing inc on σ ′ ∈ Rn−1 , (b) writing a 1 just after a member of Rn−1 , or (c) writing two 1’s just before σ ′′ ∈ Rn−2 and then adding 1 to each letter of σ ′′ . Note that steps (a) and (c) yield all the members of Rn not ending in a 1, while step (b) yields all those that do. This implies that rn satisfies rn = 2rn−1 + rn−2 , n ≥ 3, (1) with r1 = 1 and r2 = 2, whence rn = pn for all n ≥ 1.
Proof of (2): For this case, we first make some observations concerning the members λ of Rn = Pn (1121, 1212, 1213). If λ starts with two or more 1’s, then λ can have no other occurrences of 1 since it avoids 1121. If λ starts with a single 1, then any additional 1’s can occur only as a run at the end of λ since it avoids both 1212 and 1213. So if n ≥ 3, the members of Rn · · · 1}, where 2 ≤ a ≤ n, must be of one of the following two forms: (i) |11 {z · · · 1} π ′ , or (ii) 1π ′′ 11 | {z a
b
0 ≤ b ≤ n − 2, and π ′ , π ′′ denote, respectively, possibly empty and non-empty partitions on the letters {2, 3, . . .} which avoid all three patterns.
Thus members of Rn , n ≥ 3, may be formed, recursively, by (a) performing inc on σ ′ ∈ Rn−1 , (b) writing a 1 at the beginning and at the end of σ ′′ ∈ Rn−2 and then adding 1 to each letter of σ ′′ , or (c) either writing a 1 after a member of Rn−1 if it ends in a 1 or writing a 1 just before a member of Rn−1 if it does not. Note that step (a) and the second part of step (c) yields all members of Rn having form (i) above, while step (b) and the rest of step (c) yields those having form (ii). Thus, recurrence (1), along with its initial values, is satisfied by |Rn |, which implies |Rn | = pn in this case as well.
Proof of (3): Note first that members of Rn , n ≥ 3, must be of one of the following two forms: (i) |11 {z · · · 1} π ′ , or (ii) 12 |11 {z · · · 1} π ′′ , where 2 ≤ a ≤ n, 0 ≤ b ≤ n − 2, π ′ is a possibly a
b
empty partition on the letters {2, 3, . . .} which avoid all three patterns, and π ′′ is such that 2π ′′ is a non-empty partition on {2, 3, . . .} avoiding the patterns. The members of Rn , n ≥ 3, may then be obtained by (a) performing inc on σ ′ ∈ Rn−1 , (b) writing a 1 just after the right-most 1 within a member of Rn−1 , or (c) writing a 1 just before and just after the first letter of σ ′′ ∈ Rn−2 and then adding 1 to each letter of σ ′′ . This implies |Rn | satisfies (1) and completes the third case. Proof of (5): Note that members of Rn = Pn (1211, 1221, 1231), n ≥ 3, must be of one of the following two forms: (i) |11 {z · · · 1} π ′ , or (ii) |11 {z · · · 1} 21π ′′ , where 1 ≤ a ≤ n, 1 ≤ b ≤ n − 2, and a
b
π ′ , π ′′ are as in the prior case. The proof is completed as in the prior case, except for step (b), where we would instead write a 1 just before the first letter of a member of Rn−1 . 3
3
Counting by inv
In this section, we count the members of An := Pn (1211, 1212, 1213) according to the number of inversions, where all partitions are represented by their canonical sequential forms. First note that each member π of An can be decomposed as either 11 · · · 1} π ′ or |11 {z · · · 1} π ′′ 1, where | {z a
b
1 ≤ a ≤ n, 1 ≤ b ≤ n − 2, and π ′ , π ′′ denote, respectively, possibly empty and non-empty partitions on {2, 3, . . .} which avoid the three patterns. Therefore, members of An , n ≥ 3, may be formed in one of the following three ways: (a) performing inc on π ′ ∈ Rn−1 , (b) writing a 1 just before a member of An−1 , or (c) writing a 1 just before and just after π ′′ ∈ Rn−2 and adding 1 to each letter of π ′′ . Let an (x, q) :=
X
xν(π) q inv(π) ,
π∈An
n ≥ 1,
with a0 (x, q) := 1, where ν(π) denotes the number of blocks of π and inv(π) denotes the number of inversions of π, when represented in its canonical sequential form. For example, if n = 3, then An = {111, 112, 122, 121, 123} and a3 (x, q) = x + 2x2 + x2 q + x3 . Then the total weights of the members of An formed in steps (a), (b), and (c) above are, respectively, xan−1 (x, q), an−1 (x, q), and xq n−2 an−2 (x, q); note that in step (c), the 1 written just after π ′′ contributes n − 2 additional inversions since we are to add 1 to each letter of π ′′ . This implies the following recurrence relation: an (x, q) = (x + 1)an−1 (x, q) + xq n−2 an−2 (x, q), with a1 (x, q) = x and a2 (x, q) = x + x2 .
n ≥ 3,
(2)
Remark 1 : Taking x = q = 1 in (2) gives |An | = an (1, 1) = pn , n ≥ 1. Taking x = 1 and q = −1 in (2) gives an (1, −1) = 2an−1 (1, −1) + (−1)n an−2 (1, −1), n ≥ 3,
with a1 (1, −1) = 1 and a2 (1, −1) = 2. Considering even and odd cases for n, we obtain from this recurrence by induction, an (1, −1) = Fr , n ≥ 1, where r = ⌊ 3n 2 ⌋ and Fn denotes the Fibonacci sequence defined by Fn = Fn−1 + Fn−2 if n ≥ 3, with F1 = F2 = 1. Therefore, for all n ≥ 1, the members of An having an even number of inversions outnumber those members having an odd number by Fr . Remark 2 : By (2) and induction, one has for all n ∈ N, a2n+2 (x, q) = (1 + x)
n X
xn−i q n(n+1)−i(i+1) a2i+1 (x, q),
i=0
and 2
a2n+1 (x, q) = xn+1 q n + (1 + x)
n X
2
xn−i q n
−i2
a2i (x, q).
i=1
Define the ordinary generating function a(y) = a(y; x, q) by X a(y) := an (x, q)y n . n≥0
n
Multiplying both sides of (2) above by y and summing over n ≥ 3 implies
a(y) − 1 − xy − (x + x2 )y 2 = (x + 1)y(a(y) − 1 − xy) + xy 2 (a(qy) − 1),
or
1 − y − xy 2 xy 2 + a(qy). 1 − (x + 1)y 1 − (x + 1)y Iterating this recurrence an infinite number of times yields the following result. a(y) =
4
Proposition 3.1. We have a(y) =
X xj (1 − q j y − xq 2j y 2 )q j(j−1) y 2j . (1 − (1 + x)y) · · · (1 − (1 + x)q j y)
(3)
j≥0
2
1−y−y Note that a(y) = 1−2y−y 2 when x = q = 1 in (3), which is the generating function for the m P m tn sequence (δn,0 + pn )n≥0 . Using the fact (1−t)(1−tq)···(1−tq (see, e.g., [1]), one n) = m≥0 n q t may expand the denominators in the expression above for a(y) to obtain X xj (1 − q j y − xq 2j y 2 )q j(j−1) y j+i i . a(y) = (1 + x)j−i j q i,j≥0
This implies that the coefficient of y n in a(y) is given by X X 2 n−j n−1−j xj q j(j−1) (1 + x)n−2j − xj q j (1 + x)n−1−2j j j q q j≥0 j≥0 X n−2−j − xj+1 q j(j+1) (1 + x)n−2−2j j q j≥0 X X j j(j−1) n − j n−2j j j2 n − 1 − j = x q (1 + x) − x q (1 + x)n−1−2j j j q q j≥0 j≥0 X n − 1 − j − xj q j(j−1) (1 + x)n−2j j−1 q j≥1 " # X n−1−j n−j n−1−j qj j j(j−1) n−2j = − − x q (1 + x) 1+x j j j−1 q q q j≥0 " # X n−1−j qj n−1−j = xj q j(j−1) (1 + x)n−2j q j − j 1+x j q q j≥0 X 2 n−1−j = xj+1 q j (1 + x)n−1−2j , j q j≥0
using the fact
n−j j q
−
n−1−j j−1 q
Theorem 3.2. If n ≥ 1, then
an (x, q) =
n−1−j . j q
= qj
⌊ n−1 2 ⌋
X
This yields the following explicit formula.
n−1−j q (1 + x)n−1−2j , j q
j+1 j 2
x
j=0
(4)
with a0 (x, q) = 1. P⌊ n−1 ⌋ n−1−j n−1−2j Note that (4) reduces to the well-known formula pn = j=02 2 when x = q = 1. j Consider the companion sequence qn defined by qn = pn+1 + pn−1 if n ≥ 1. They count the number of circular tilings of length n consisting of squares and dominos in which the squares P⌊ n2 ⌋ n n−j n−2j come in two varieties, and are given, equivalently, by qn = j=0 2 . See A002203 n−j j of [12] for further information on these numbers. The following explicit formula reduces to qn when x = q = 1. Corollary 3.3. If n ≥ 2, then n
an+1 (x, q) + xan−1 (x, q) =
⌊2⌋ X j=0
(n − j)q + q 1−2j jq n − j q (1 + x)n−2j . (n − j)q j q
j+1 j 2
x
5
Proof. By (4), we have an+1 (x, q) + xan−1 (x, q) n
⌊2⌋ X
=
n−j q (1 + x)n−2j j q
j+1 j 2
x
j=0
⌊ n−2 2 ⌋
n−2−j x q (1 + x)n−2−2j +x j j=0 " # ⌊n 2⌋ X n−j n j+1 j 2 1−2j n − 1 − j = x(1 + x) + x q +q (1 + x)n−2j j j − 1 q q j=1 X
n
=
⌊2⌋ X
(n − j)q + q 1−2j jq n − j q (1 + x)n−2j , (n − j)q j q
j+1 j 2
x
j=0
by virtue of the fact
n−j j q
j+1 j 2
=
(n−j)q n−1−j jq j−1 q .
Recall the familiar result that the total number of inversions occurring within all of the members of the symmetric group Sn is given by the formula 12 n2 n! if n ≥ 1 (see, e.g., [13]). We conclude this section by deriving a comparable result for the total number of inversions taken over all of the members of An , using (4) above. Theorem 3.4. If n ≥ 1, then the total number of inversions occurring within all the members of An is given by n−1 (n(pn−1 + pn−2 ) − pn ) . (5) 8 In particular, the average number of inversions is approximately large.
√ √ 2− 2 2 3− 2 1 8 n − 8 n+ 8
as n grows
n d n j n−1 Proof. Let fn,j = dq + n−1 , we have fn,j = fn−1,j + j q |q=1 . From the fact j q = q j j−1 q q Pn n−1 fn−1,j−1 + j j . Define Fn (t) = j=0 fn,j tj . Then Fn (t) = (1 + t)Fn−1 (t)+ (n− 1)t(1 + t)n−2 , with F0 (t) = F1 (t) = 0. Therefore, by induction on n, we obtain Fn (t) = n2 t(1 + t)n−2 , which implies fn,j = n2 n−2 j−1 for all j = 0, 1, . . . , n. Now we can use Theorem 3.2 to find the total number of inversions occurring within all of the members of An : n−1
⌊ 2 ⌋ X d n−1−j n−1−j n−3−j an (x, q) |x=q=1 = j2 + 2n−1−2j , dq j 2 j − 1 j=0
which is equivalent to n−1
⌊ 2 ⌋ X d j(n − 1 − 2j) n − 1 − j 2 n−1−j an (x, q) |x=q=1 = j + 2n−1−2j dq j 2 j j=0 n−1
⌊ 2 ⌋ n−1 X n − 1 − j n−1−2j = j 2 . 2 j j=0 To complete the proof of the first statement, we need to show bn :=
⌊ n−1 2 ⌋
X j=0
j
n − 1 − j n−1−2j 1 2 = (n(pn−1 + pn−2 ) − pn ), j 4 6
n ≥ 1.
(6)
To do so, observe that for n ≥ 3, bn =
⌊ n−1 2 ⌋
X j=1
=
⌊ n−3 2 ⌋
X j=0
n − 2 − j n−1−2j (n − 1 − j) 2 j−1 ⌊ n−3 2 ⌋ X n − 3 − j n−3−2j n − 3 − j n−3−2j (n − 2 − j) 2 = (n − 2) 2 − bn−2 , j j j=0
so that bn = (n − 2)pn−2 − bn−2 , from which (6) follows by induction (note that both sides of (6) are zero when n = 1 and n = 2). The second statement is an√immediate consequence of the first upon dividing through by pn and pn−1 noting limn→∞ pn = 2 − 1, which follows from the Binet formula √ √ √ 2 (1 + 2)n − (1 − 2)n , pn = 4
4
n ≥ 0.
Counting by comaj
In this section, we count the members of Bn := Pn (1212, 1222, 1232) according to the comajor index. We first make some preliminary observations concerning the members of Bn . Suppose π = 1w1 2w2 · · · kwk ∈ Bn for some k ≥ 2, where wi denotes a word in [i]. Since π avoids 1212, the wi cannot increase. Since π avoids 1232, no wi can contain a letter j with 1 < j < i, which implies wi must be of the form |ii {z · · · }i 11 · · · 1} if i ≥ 2 for some a, b ≥ 0. Since π avoids 1222, we | {z a
b
must have a = 0 or a = 1, i.e., each wi must have the form i 11 · · · 1} or |11 {z · · · 1}. | {z b
b
Suppose π = 1w1 2w2 . . . kwk for some k ≥ 1, where each wi is a word in [i]. If π ∈ Bn−1 , then either write k + 1 just after π or write 1 just after π to obtain two members of Bn and if π ∈ Bn−2 , then write the letter k + 1 twice just after π to obtain a member of Bn . Note that all members of Bn , n ≥ 3, arise exactly once in this way, upon allowing k to vary. Let
bn (x, q) :=
X
xν(π) q maj(π) ,
π∈Bn
n ≥ 1,
with b0 (x, q) := 1, where ν(π) denotes the number of blocks of π and comaj(π) denotes the comajor index of π, represented in its canonical sequential form. Observe that the total weights of the members of Bn formed by the above three steps are, respectively, xq n−1 bn−1 (x, q), bn−1 (x, q), and xq n−2 bn−2 (x, q), which implies the following recurrence relation: bn (x, q) = (xq n−1 + 1)bn−1 (x, q) + xq n−2 bn−2 (x, q),
n ≥ 3,
(7)
with b1 (x, q) = x and b2 (x, q) = x + x2 q. Remark: Taking x = q in (7) yields the recurrence satisfied by the Pell polynomials Pn (q) introduced in [10], but with the initial conditions of q and q + q 3 replacing those of Pn (q) of 1 and 1 + q. See [3] for a combinatorial approach to Pn (q) based on square and domino tilings. P Let b(y) := n≥0 bn (x, q)y n and bo (y) := b(y) − 1. Multiplying both sides of (7) by y n and summing over n ≥ 3 implies bo (y) =
xy xy(1 + y) o + b (qy). 1−y 1−y
Iterating this relation an infinite number of times yields the following result. 7
Proposition 4.1. We have b(y) = 1 +
X
xn q ( 2 ) y n n
n≥1
(1 + y) · · · (1 + q n−2 y) . (1 − y) · · · (1 − q n−1 y)
(8)
Using the facts (see [1]) X m tn = tm (1 − t)(1 − tq) · · · (1 − tq n ) n q m≥0
and (1 + t)(1 + qt) · · · (1 + q
n−1
n X
j n ( ) 2 t) = q tj , j q j=0
one may rewrite bo (y) as X X n n k m m n−1 xn q ( 2 ) y m+1 (1 + y) · · · (1 + q n−2 y) = xn q ( 2 )+( 2) y m+k+1 . n−1 q n−1 q k q
m≥0, n≥1
k,m≥0, n≥1
The coefficient of y r , r ≥ 1, is then given by r−1 X
m+1 X
xn q ( 2 )+( n
r−1−m 2
)
m=0 n=r−m
n−1 q r−1−m q m+1 r−1 X X r−1−m n 2m + 1 − r m , xn q ( 2 ) = q( 2 ) r − 1 − m q n=r−m n+m−r q m=0 m n−1
by virtue of the identity ab q bc q = ac q n by n + m + 1 implies the coefficient is r−1 X
m=0
q( 2 ) m
a−c b−c q .
Replacing m by r − 1 − m and then replacing
r−m X n r−1−m r − 2m − 1 xn q ( 2 ) m n−m−1 q q n=m+1 r−2m−1 r−1 X X m n+m+1 r−1−m r − 2m − 1 = q( 2 ) xn+m+1 q ( 2 ) m n q q m=0 n=0 r−1 r−2m−1 X X n 2 r−1−m r − 2m − 1 = xm+1 q m (xq m+1 )n q(2) m n q q m=0 n=0 r−1 r−2m−2 X Y 2 r−1−m = xm+1 q m (1 + xq i+m+1 ), m q m=0 i=0
by the q-binomial theorem [1] and the fact following result.
n+m+1 2
=
n 2
+
m+1 2
+ n(m + 1). This yields the
Theorem 4.2. If n ≥ 1, then bn (x, q) =
⌊ n−1 2 ⌋
X j=0
xj+1 q j
2
n−j−1 Y n−1−j (1 + xq i ), j q i=j+1
(9)
with b0 (x, q) = 1. Using this result, one can obtain a simple closed form expression for the total comajor index of all the members of Bn . 8
Theorem 4.3. If n ≥ 1, then the total comajor index of all the members of Bn is given by 1 2n2 pn − n(3pn − pn−1 ) + pn . 8
In particular, the average comajor index value is approximately large. Proof. By Theorem 4.2 and the proof of Theorem 3.4, we have n−j−1 ⌊ n−1 2 ⌋ X Y 2 d j n−1−j d bn (x, q) |x=q=1 = q (1 + q i ) dq dq j q i=j+1 j=0
(10) 1 2 4n
−
√ 4− 2 8 n
+
1 8
as n grows
q=1
⌊ n−1 2 ⌋
⌊ n−1 2 ⌋ X X n−1−j n − 3 − j n−1−2j n−1−2j 2 n−1−j = j 2 + 2 j 2 j−1 j=0 j=0 +
⌊ n−1 2 ⌋
X
n − 1 − j n−2−2j n−j j+1 2 − j 2 2
j=0
=
⌊ n−1 2 ⌋
=
⌊ n−1 2 ⌋
X j=0
X j=0
n−j j+1 n − 1 − j n−2−2j 2j + j(n − 1 − 2j) + − 2 2 2 j 2
n n − 1 − j n−2−2j −j 2 2 j
⌊ n−1 ⌊ n−1 2 ⌋ 2 ⌋ n − 1 − j n−1−2j 1 X n − 1 − j n−1−2j 1 n X 2 − j 2 = 2 2 j=0 j 2 j=0 j 1 n 1 = pn − (n(pn−1 + pn−2 ) − pn ) 2 2 8 1 = 2n2 pn − n(3pn − pn−1 ) + pn . 8
References [1] G. E. Andrews, The Theory of Partitions, Encyclopedia of Mathematics and its Applications, Vol. 2, Addison-Wesley, Reading, Mass. 1976. [2] A. T. Benjamin, S. P. Plott, and J. A. Sellers, Tiling proofs of recent sum identities involving Pell numbers, Ann. Comb. 12 (2008), 271–278. [3] K. S. Briggs, D. P. Little, and J. A. Sellers, Tiling proofs of various q-Pell identities via tilings, to appear in Ann. Comb. [4] J. L. Diaz–Barrero and S. F. Santana, Some properties of sums involving Pell numbers, Missouri Journal of Mathematical Sciences 18, No. 1 (2006), 33–40. [5] A. Goyt, Avoidance of partitions of a three element set, Adv. Appl. Math. 41 (2008), 95–114. [6] V. Jel´ınek and T. Mansour, On pattern-avoiding partitions, Electron. J. Combin. 15:1 (2008), #R39.
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[7] M. Klazar, On abab-free and abba-free set partitions, Europ. J. Combin. 17 (1996), 53–68. [8] D. E. Knuth, The Art of Computer Programming, Vol’s. 1 and 3, Addison-Wesley, Reading, Mass. 1968, 1973. [9] B. E. Sagan, Pattern avoidance in set partitions, Ars Combin. 94 (2010), 79–96. [10] J. O. Santos and A. V. Sills, q-Pell sequences and two identities of V. A. Lebesgue, Discrete Math. 257 (2002), 125–142. [11] R. Simion and F. W. Schmidt, Restricted permutations, Europ. J. Combin. 6 (1985), 383– 406. [12] N. J. Sloane, The On-Line Encyclopedia of Integer Sequences, (2006), published electronically at http://oeis.org/. [13] R. P. Stanley, Enumerative Combinatorics, Vol. I, Wadsworth and Brooks/Cole, Monterrey 1986. [14] D. Stanton and D. White, Constructive Combinatorics, Springer, New York 1986.
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