The k-Extensions of some new Mahonian statistics

Robert J. Clarke^{∗} Pure Mathematics Department University of Adelaide Adelaide, South Australia 5005

Einar Steingrímsson^{†} Matematiska institutionen CTH & GU 412 96 Göteborg, Sweden

Jiang Zeng^{‡} Département de mathématique Université Louis-Pasteur 67084 Strasbourg Cedex, France

October 19, 1996

∗This work was carried out while the author was Visiting Professor at the Université Louis-Pasteur in Strasbourg during summer and autumn of 1995.

†Partially supported by a grant from The Icelandic Council of Science and partially supported by the EU Network in Algebraic Combinatorics while visiting Université Louis-Pasteur in Strasbourg.

‡Partially supported by the EU Network in Algebraic Combinatorics.

Abstract

In previous work by the authors, new Mahonian statistics env, mad and mak were defined on words and it was shown that env is equal to the classical statistic inv and that the triple statistics (des, mak, mad) and (exc, den, env) are equidistributed over any rearrangement class of words. Here, exc and des are the classical Eulerian statistics, while den is Denert’s statistic. In addition, a bijection between the symmetric group and sets of weighted Motzkin paths was used to give a continued fraction expression for the generating function of (exc, inv) or (des, mad) on the symmetric group. These results are extended to the case in which the letters of the alphabet used are divided into two classes, large and small, with corresponding changes to the definitions of the above statistics.

1 Introduction

Let c = (c_{1}, c_{2}, …, c_{m}) be a sequence of non-negative integers with n = c_{1} + c_{2} + ⋯ + c_{m}. Denote by R(c) the set of all rearrangements of the (nondecreasing) word 1^{c_{1}}2^{c_{2}} … m^{c_{m}}. In [3], some new Mahonian statistics mad, mak and env on R(c) were introduced and the following results were proved.

Theorem 1 For all w ∈ R(c), we have env w = inv w.

Theorem 2 The triples (des, mad, mak) and (exc, inv, den) are equidistributed on R(c).

In fact, a bijection Φ_{W} was constructed on R(c) such that for all w ∈ R(c),

(des, mad, mak) w = (exc, inv, den) Φ_{W}(w).

In [1, 2, 4, 5], the statistics des, exc, maj, den and inv were extended to the case in which the letters in [m] := {1, 2, …, m} are divided into two classes. Namely, let k be a non-negative integer such that k ≤ m and put ℓ = m − k. The letters 1, 2, …, ℓ are called small and the letters ℓ + 1, …, m are called large. In counting descents and excedances, one counts strict inequalities between small letters, but equalities and inequalities between large letters—see below for the details. Then k-extended statistics des_{k}, exc_{k}, maj_{k}, den_{k} and inv_{k} are defined, which reduce to their more familiar Eulerian and Mahonian counterparts in the case k = 0. Further, the pairs (des_{k}, maj_{k}) and (exc_{k}, den_{k}) are equidistributed on R(c). It is the purpose of this paper to k-extend the results of [3]. We will define statistics mad_{k}, mak_{k} and env_{k} and prove the following results.

Theorem 3 For all w ∈ R(c), we have env_{k} w = inv_{k} w.

Theorem 4 The triples (des_{k}, mad_{k}, mak_{k}) and (exc_{k}, inv_{k}, den_{k}) are equidistributed on R(c).

We will begin by recalling the definitions of k-descent and k-excedance from [1]. Let 1 ≤ i ≤ m and w = a_{1}a_{2} … a_{n} ∈ R(c). Let \overline{w} = b_{1}b_{2} … b_{n} be the non-decreasing rearrangement of w. It is convenient to introduce a new small letter ∗ such that ℓ < ∗ < ℓ + 1, and to put w∗ = a_{1}a_{2} … a_{n}∗. Thus we write a_{n+1} = ∗. (We use this convention throughout the paper.) We also introduce the following partial ordering on [m]∗ = [m] ∪ {∗}.

Definition 1 If a, b ∈ [m] then a ≺ b means either a < b or a = b with b large. Thus a ⊀ b means either b < a or a = b with b small.

Definition 2 Let w = a_{1}a_{2} … a_{n} be a word with \overline{w} = b_{1}b_{2} … b_{n}. A k-descent in the word w is a triple (i, a_{i}, a_{i+1}), where 1 ≤ i ≤ n, such that a_{i+1} ≺ a_{i}. Here i is called the k-descent place, a_{i} is called the k-descent top and a_{i+1} is called the k-descent bottom. A k-excedance in w is a triple (i, a_{i}, b_{i}), where 1 ≤ i ≤ n, such that b_{i} ≺ a_{i}. Here i is called the k-excedance place, a_{i} is called the k-excedance top and b_{i} is called the k-excedance bottom. The numbers of k-descents and k-excedances in w are denoted by des_{k} w and exc_{k} w respectively.

The k-descent set of w, D_{k}(w), is the set of k-descent places. The kexcedance set of w, E_{k}(w), is the set of k-excedance places.

This is a slight change of wording from [1].

Definition 3 The k-major index of w is

maj_{k} w = ∑_{i∈D_{k}(w)} i.

Recall the k-extensions of Imv and inv from [4] and [5]:

Definition 4

inv_{k} w = #{ (i, j) | i < j, a_{j} ≺ a_{i} } + #{ i | a_{i} > ℓ};
Imv_{k} w = #{ (i, j) | i < j, a_{i} ⊀ a_{j} }.

Here, #S denotes the number of elements in the set S.

Let i_{1} < i_{2} < ⋯ < i_{e} be the increasing sequence of elements of E_{k}(w) and let j_{1} < j_{2} < ⋯ < j_{n−e} be the increasing complementary sequence in [n]. We form the excedance subword w_{E} = a_{i_{1}}a_{i_{2}} … a_{i_{e}} and the non-excedance subword w_{N} = a_{j_{1}}a_{j_{2}} … a_{j_{m−e}} of w. Put

Ebot_{k} w = ∑_{i∈E_{k}(w)} i,

that is, Ebot_{k} w is the sum of the k-excedance places in w.

Remark 1 Note that Ebot_{k} w is equal to the sum of the values of the kexcedance bottoms in w̄ if there are no large letters in w as in [3], but this is no longer true if there are large letters in w, because of the ∗ and because we code equal large letters from right to left (See Section 2.1 below). However, we use the notation Ebot_{k} to be compatible with [3].

The following definition of den_{k} w, from [5], is equivalent to the definition of [2].

Definition 5

den_{k} w = Ebot_{k} w + Imv_{k} w_{E} + inv_{k} w_{N}.

We can write the definition of den_{k} w in a slightly different way, using the following definition.

Definition 6 Let w = a_{1}a_{2} … a_{n} be a word. If i is a k-excedance place in w then the k-imversion bottom number of i is the number of places j such that 1 ≤ j < i and a_{j} ⊀ a_{i}. If i is not a k-excedance place in w then the k-inversion top number of i is the number of places j such that i < j ≤ n + 1 and a_{j} ≺ a_{i}. The k-side number s_{i} of i (in w) is the k-imversion bottom number or k-inversion top number of i as appropriate.

Clearly, the sum of the k-imversion bottom numbers for the letters in w_{E} equals Imv_{k} w_{E}, while the sum of the k-inversion top numbers for the letters in w_{N} equals inv_{k} w_{N}. We write

Ine_{k} w = Imv_{k} w_{E} + inv_{k} w_{N}
= s_{1} + ⋯ + s_{n}.

Thus den_{k} w = Ebot_{k} w + Ine_{k} w.

Example 1 Let

w = 2 4 4 1 3 1 1 3, so that \overline{w} = 1 1 1 2 3 3 4 4.

Take k = 2. Thus 1 and 2 are small, while 3 and 4 are large. There are 2-descent places at i = 2, 3, 5 and 8. Thus des_{2} w = 4 and maj_{2} w = 2 + 3 + 5 + 8 = 18.

There are 2-exedance places at i = 1, 2, 3 and 5. Thus exc_{2} w = 4 and Ebot_{2} w = 11. Now, one readily checks that the sequence of side numbers of w is (0, 0, 0, 0, 2, 0, 0, 1). Hence Ine_{2} w = 3 and den_{2} w = 14.

2 Further Mahonian statistics

2.1 Height and value

For each letter a, we define the height of a in w as

h_{k}(a) = h_{k,w}(a) = 1 + #{ j | 1 ≤ j ≤ n, a_{j} ≺ a }.

It is easy to see that

i ∈ E_{k}(w) if and only if i < h_{k,w}(a_{i}).

If 1 ≤ i ≤ n + 1, we define the value of the i-th letter in w by

v_{k,i} = v_{k,i}(w) = #{ j | j ≤ n + 1, a_{j} < a_{i} }

+ \{ \begin{matrix}#\{ j | 1 ≤ j ≤ i, a_{j} = a_{i} \}, & if a_{i} small; \\ #\{ j | i ≤ j ≤ n, a_{j} = a_{i} \}, & if a_{i} large. \end{matrix}

Thus the height of a letter in a word is a function of the letter, not of its position in the word, whereas the value of a letter does depend on its position in the word. Hence we can strictly only speak of the value of a position in a word.

We will normally supress the argument w to v_{k} and h_{k}.

Using the relation ≺ we can write

v_{k,i} = 1 + #{ j | i < j ≤ n + 1, a_{j} ≺ a_{i} } + #{ j | 1 ≤ j < i, a_{i} ⊀ a_{j} }. (1)

Notice that the word

v_{k}(w) = v_{k,1}v_{k,2} … v_{k,n}v_{k,n+1}

is a permutation. In fact, our definition of v_{k}(w) amounts to “coding” w∗ to a permutation by coding equal small letters from left to right and coding equal large letters from right to left. The equation

D_{k}(w) = D(v_{k}(w)),  (2)

which is easily verified, is the motivation for this coding. Note that, if w ∈ S_{n} is a permutation and 1 ≤ i ≤ n, then

v_{k,i} = h_{k}(a_{i}) = \{ \begin{matrix}a_{i}, & if 1 ≤ a_{i} ≤ ℓ, \\ a_{i} + 1, & if ℓ < a_{i} ≤ n. \end{matrix}

Further, v_{k,n+1} = ℓ + 1.

Example 2 For the word w = 2 4 4 1 3 1 1 3 considered in example 1 with k = 2, we have v_{2}(w) = 4 9 8 1 7 2 3 6 5. Also h_{2}(1) = 1, h_{2}(2) = 4, h_{2}(3) = 7 and h_{2}(4) = 9.

2.2 The statistic env_{k}

Now we put

Edif_{k} w = ∑_{i∈E_{k}(w)}(h_{k}(a_{i}) − i)

and we define a new statistic env_{k} w by

Definition 7

env_{k} w = Edif_{k} w + Imv_{k} w_{E} + inv_{k} w_{N}.

We can also write env_{k} w = Edif_{k} w + Ine_{k} w. For k = 0, when there are no large letters, we have env_{k} = env, the statistic defined in [3].

Example 3 For the word w = 2 4 4 1 3 1 1 3 considered above, we have Edif_{2} w = (4−1)+(9−2)+(9−3)+(7−5) = 18 and env_{2} w = 18+3 = 21.

One can easily check that inv_{2} w = 21.

Lemma 5 For any word w we have

env_{k} w = ∑_{i∈E_{k}(w)}(v_{k,i} − i) + inv_{k} w_{E} + inv_{k} w_{N} − δ,

where δ is the number of large letters in w_{E}.

Proof: Suppose that w_{E} contains d_{a} letters equal to a, for 1 ≤ a ≤ m. Then one can easily verify that

Imv_{k} w_{E} − inv_{k} w_{E} = ∑_{a=1}^{l} \binom{d_{a}}{2} − ∑_{a=l+1}^{m} \binom{d_{a} + 1}{2}

and

∑_{i∈E_{k}(w)}(v_{k,i} − h_{k}(a_{i})) = ∑_{a=1}^{l} \binom{d_{a}}{2} − ∑_{a=l+1}^{m} \binom{d_{a}}{2},

bearing in mind that the d_{a} letters ‘a’ occurring in w_{E} will be the left-most d_{a} letters ‘a’ that occur in w. Hence the result follows.

We can now prove that env_{k} and inv_{k} are identical. Proof of Theorem 3: From Lemma 5 we must show that

∑_{i∈E_{k}(w)}(v_{k,i} − i) = δ + #{ (i, j) | i < j ≤ n, a_{j} ≺ a_{i}, i ∈ E_{k}(w), j ∉ E_{k}(w) }
= δ + ∑_{i∈E_{k}(w)} #{ j | i < j ≤ n, a_{j} ≺ a_{i}, j ∉ E_{k}(w) }
= δ + ∑_{i∈E_{k}(w)} (#{ j | i < j ≤ n, a_{j} ≺ a_{i} }
−#{ j | i < j ≤ n, a_{j} ≺ a_{i}, j ∈ E_{k}(w) }).  (3)

Now, since i = #{ j | j ≤ i }, it follows from equation (1) that

∑_{i∈E_{k}(w)}(v_{k,i} − i) = ∑_{i∈E_{k}(w)}(#{ j | n + 1 ≥ j > i, a_{j} ≺ a_{i} }
−#{ j | j < i, a_{i} ≺ a_{j} })
= δ + ∑_{i∈E_{k}(w)}(#{ j | n ≥ j > i, a_{j} ≺ a_{i} }
−#{ j | j < i, a_{i} ≺ a_{j} }).  (4)

Hence, comparing equations (3) and (4), we need only to show that

∑_{i∈E_{k}(w)} #{ j | i < j ≤ n, a_{j} ≺ a_{i}, j ∈ E_{k}(w) } = ∑_{i∈E_{k}(w)} #{ j | j < i, a_{i} ≺ a_{j} }.

But each of the sums in the above equation is inv_{k} w_{E} − δ.

2.3 The statistics mad_{k} and mak_{k}

Let w = a_{1}a_{2} … a_{n} be a word, with a_{n+1} as usual. By analogy with Ebot_{k} w and Edif_{k} w we put

Dbot_{k} w = ∑_{i∈D_{k}(w)} v_{k,i+1}

and

Ddif_{k} w = ∑_{i∈D_{k}(w)}(h_{k}(a_{i}) − v_{k,i+1}).

Definition 8 Let w = a_{1}a_{2} … a_{n} be a word. For 1 ≤ i ≤ n, the (right) k-embracing number e_{i} in w is

e_{i} = \{ \begin{matrix}#\{ j | i < j ≤ n, a_{j+1} < a_{i} ≤ a_{j} \}, & if a_{i} is small; \\ #\{ j | i < j ≤ n, a_{j+1} ≤ a_{i} < a_{j} \}, & if a_{i} is large. \end{matrix}

The (right) k-embracing sum of w is

Res_{k} w = e_{1} + ⋯ + e_{n}.

Alternatively, we may if we wish define e_{i} for both small and large letters a_{i} by

e_{i} = #{ j | i < j ≤ n, a_{j+1} ≺ a_{i}, a_{j} ⊀ a_{i} }.

Definition 9 Let w = a_{1}a_{2} … a_{n} be a word. The k-descent (j, a_{j}, a_{j+1}) embraces a_{i} if a_{j+1} ≺ a_{i} and a_{j} ⊀ a_{i}.

So e_{i} is the number of k-descents in w to the right of a_{i} that embrace a_{i}. The statistics mad_{k} and mak_{k} can now be defined as follows:

Definition 10

mad_{k} w = Ddif_{k} w + Res_{k} w;
mak_{k} w = Dbot_{k} w + Res_{k} w.

For k = 0 we have mad_{k} = mad and mak_{k} = mak as defined in [3]. Note that, for any word w, it follows from Equation (2) and from

Res_{k} w = Res_{k} v_{k}(w)  (5)

that

mak_{k} w = mak_{k} v_{k}(w).

However, it is not in general true that mad_{k} w = mad_{k} v_{k}(w).

In Section 3, we will produce a bijection Φ_{k} on S_{n} such that, for all π ∈ S_{n}, we have

(des_{k}, mad_{k}, mak_{k}) π = (exc_{k}, inv_{k}, den_{k})Φ_{k}(π).

In Section 4, we extend the definition of Φ_{k} to words, thereby proving Theorem 4.

Example 4 Consider the word

w = 3 1 4 1 2 1 3 4,

with k = 2. Then

v_{2}(w) = 7 1 9 2 4 3 6 8 5

and h_{2}(1) = 1, h_{2}(2) = 4, h_{2}(3) = 7, h_{2}(4) = 9. Thus Dbot_{2} w = 1 + 2 + 3 + 5 = 11 and Ddif_{2} w = (7−1)+(9−2)+(4−3)+(9−5) = 18. The sequence of 2-embracing numbers of w is (2, 0, 0, 0, 0, 0, 1, 0, 0), so Res_{2} w = 3. Therefore mad_{2} w = 18 + 3 = 21 and mak_{2} w = 11 + 3 = 14.

3 The bijection for permutations

Let π = a_{1}a_{2} … a_{n} ∈ S_{n}. Then the set of k-descents in π equals the set of descents in π∗. Hence Dbot_{k} π = Dbot π∗ and Ddif_{k} π = Ddif π∗. Moreover, if (e_{1}, e_{2}, …, e_{n}) is the sequence of right k-embracing numbers of π, the sequence of right embracing numbers of π∗ is (e_{1}, e_{2}, …, e_{n}, 0). Thus Res_{k} π = Res π∗. Therefore

(des_{k}, mad_{k}, mak_{k}) π = (des, mad, mak) π ∗ .  (6)

Further, since exc_{k} π = exc_{k} π̃, where

π̃ = ( \begin{matrix}1 & 2 & … & ℓ & ℓ + 1 & … & n \\ a_{1} & a_{2} & … & a_{ℓ} & a_{ℓ+1} & … & a_{n} \end{matrix} )

and exc π∗ = exc π̃∗, where

π̃∗ = ( \begin{matrix}1 & 2 & … & ℓ & ∗ & ℓ + 1 & … & n − 1 & n \\ a_{1} & a_{2} & … & a_{ℓ} & a_{ℓ+1} & a_{ℓ+2} & … & a_{n} & ∗ \end{matrix} ) ,

it follows that there is a one-to-one correspondence between the k-excedances of π and the excedances of π∗. For, if a_{i} = i > ℓ then a_{i} > i − 1. Thus the set of k-excedance places of π equals the set of excedance places of π∗ and the set of k-excedance tops of π equals the set of excedance tops of π∗. (The corresponding equality does not hold for the excedance bottoms, but they play no explicit part in our statistics.) So Ebot_{k} π = Ebot π∗ and Edif_{k} π = Edif π∗. Similarly, we have that Ine_{k} π = Ine π∗, so that

(exc_{k}, env_{k}, den_{k}) π = (exc, env, den) π ∗ .  (7)

In [3], a bijection Φ was constructed on S_{n} such that, for all π ∈ S_{n},

(des, mad, mak) π = (exc, env, den) Φ(π).  (8)

In order to prove Lemma 6 below, we briefly describe the bijection Φ. Let F, F′, G and G′ denote respectively the sets of descent bottoms, descent tops, non descent bottoms and non descent tops for π. Let f and g be the non-decreasing words formed from the letters of F and G respectively. Let f′ and g′ be the words formed from the letters of F′ and G′ respectively, whose imversion bottom and inversion top numbers respectively are the embracing numbers of those letters in π. Form the biword

α = ( \begin{matrix}f & | & g \\ f′ & | & g′ \end{matrix} ) .

Then the columns to the left of the bar are excedances, while those to the right are non-excedances. Finally, rearrange the columns of α so that its first row is increasing and read off the permutation Φ(π) from the second row.

The bijection Φ has the following property.

Lemma 6 Let π = a_{1}a_{2} … a_{n} ∈ S_{n} and let Φ(π) = b_{1}b_{2} … b_{n}. Then b_{n} = a_{n}. (That is, the bijection Φ fixes the last letter of a permutation.)

Proof: Let π = a_{1}a_{2} … a_{n} ∈ S_{n}. Clearly, a_{n} is a non-descent top and its (right) embracing number is zero. Hence a_{n} will be a letter in g′ and all the non-descent top letters smaller than a_{n} will be to the left of a_{n} in g′. It remains to show that any non-descent top letter larger than a_{n} will not be the rightmost letter in g′. If a_{i} (i < n) is a non-descent top letter larger than a_{n}, then there is a letter a_{j} (i < j < n) such that a_{i} < a_{j} (hence a_{j} > a_{n}), for, otherwise, a_{i} would be a descent top. Let a_{j} be the smallest such letter. Thus a_{i} > a_{j+1} and the embracing number of a_{i} must be larger than or equal to 1. This proves that a_{i} cannot be the rightmost letter in g′.

Hence we define a bijection Φ_{k} on S_{n} by

π ↦ π∗ ↦ Φ(π∗) = π′∗ ↦ π′,

where π′ is a permutation. It follows from Equations (6), (7) and (8) and Theorem 3 that

(des_{k}, mad_{k}, mak_{k}) π = (exc_{k}, inv_{k}, den_{k}) Φ_{k}(π).  (9)

This proves Theorem 4 in the case of permutations, that is, if all c_{i} ≤ 1.

We note for future reference that Φ_{k} satisfies a stronger property than that specified in equation (9), namely that the set of k-excedance places of Φ_{k}(π) equals the set of k-descent bottoms of π, the set of k-excedance tops of Φ_{k}(π) equals the set of k-descent tops of π and the sequence of k-side numbers s_{1}, s_{2}, …, s_{n} of Φ_{k}(π) is a permutation of the sequence of k-embracing numbers e_{1}, e_{2}, …, e_{n} of π.

Example 5 Consider the permutation π = 6 1 8 2 4 3 5 7 with k = 4. Thus π∗ = 6 1 8 2 4 3 5 7 ∗, where 4 < ∗ < 5. We may more conveniently write π∗ as τ = 7 1 9 2 4 3 6 8 5, a permutation on {1, …, 9}. Now, applying the bijection of [3] gives Φ(τ) = τ′ = 4 8 9 1 7 2 3 6 5. Thus π′∗ = 4 7 8 1 6 2 3 5 ∗ and Φ_{2}(π) = π′ = 4 7 8 1 6 2 3 5.

Although we can decompose Φ_{k} as the composition of two bijections between the symmetric group S_{n} and the weighted Motzkin paths as in [3], this decomposition does not seem to yield an interesting continued fraction expansion. Here we just record some partial results in this direction.

Let

A_{m}^{k}(t, q) = ∑_{π∈S_{n}} t^{exc_{k} π}q^{INV_{k} π}

and

A_{m}(x, y, t, q) = ∑_{k=0}^{m} \binom{m}{k}x^{k}y^{m−k}A_{m}^{k}(t, q).

The first values of the polynomials A_{m}(x, y, t, q) are as follows:

A_{1}(x, y, t, q) = y + xtq;
A_{2}(x, y, t, q) = y^{2}(1 + tq) + 2xyt(q + q^{2}) + x^{2}(tq^{3} + t^{2}q^{2});
A_{3}(x, y, t, q) = y^{3}(1 + (2q + q^{2} + q^{3})t + t^{2}q^{2})
+3xy^{2}((q + q^{2} + q^{3} + q^{4})t + (q^{2} + q^{3})t^{2})
+3x^{2}y((q^{3} + q^{4})t + (q^{2} + q^{3} + q^{4} + q^{5})t^{2})
+x^{3}(q^{5}t + (2q^{4} + q^{5} + q^{6})t^{2} + q^{3}t^{3}).

It has been noted (see [1]) that

∑_{m≥0} \frac{u^{m}}{m!} A_{m}(x, y, t, 1) = \frac{(1 − t) exp(uy(1 − t))}{1 − t exp(u(x + y)(1 − t))}.

Therefore, by applying the addition formula of Rogers – Stieltjes (see [6]) we can derive the following result.

Proposition 7 The ordinary generating function of A_{m}(x, y, t, 1) has the following Jacobi continued fraction expansion:

∑_{m≥0} A_{m}(x, y, t, 1)u^{m} = \frac{1}{1 − b_{0}u − \frac{λ_{1}u^{2}}{\frac{...}{1 − b_{n}u − \frac{λ_{n+1}u^{2}}{...}}}} ,

where for n ≥ 0,

b_{n} = (tx + y)(n + 1) + (x + ty)n
λ_{n+1} = (n + 1)^{2}(x + y)^{2}t.

A combinatorial proof of the above result will eventually produce a qanalogue of Proposition 7. More precisely, set [n]_{q} = 1 + q + ⋯ + q^{n−1} for n ≥ 1 and [0]_{q} = 0. Since

A_{m}^{k}(1, q) = ∑_{π∈S_{n}} q^{INV_{k} π} = q^{k}A_{m}^{0}(1, q),

hence

A_{m}(x, y, 1, q) = ∑_{k=0}^{m} \binom{m}{k}(qx)^{k}y^{m−k}A_{m}^{0}(1, q) = (qx + y)^{m}A_{m}^{0}(1, q).

We derive then from [3, Theorem 10] the following result.

Proposition 8 The ordinary generating function of A_{m}(x, y, 1, q) has the following Jacobi continued fraction expansion:

∑_{m≥0} A_{m}(x, y, 1, q)u^{m} = \frac{1}{1 − b_{0}u − \frac{λ_{1}u^{2}}{\frac{...}{1 − b_{n}u − \frac{λ_{n+1}u^{2}}{...}}}} ,

where for n ≥ 0,

b_{n} = q^{n}([n + 1]_{q} + [n]_{q})(qx + y),
λ_{n+1} = q^{2n+1}[n + 1]_{q}^{2}(qx + y)^{2}.

However, the series ∑_{m≥0} A_{m}(x, y, t, q)u^{m} does not seem to have a nice Jacobi continued fraction expansion. In fact, suppose

∑_{m≥0} A_{m}(x, y, t, q)u^{m} = \frac{1}{1 − b_{0}u − \frac{λ_{1}u^{2}}{\frac{...}{1 − b_{n}u − \frac{λ_{n+1}u^{2}}{...}}}} ,

then we find b_{0} = tqx + y, λ_{1} = q (qx + y)^{2} t and

b_{1} = q(y^{3}q + y^{3}t + y^{3} + x^{3}q^{3} + tq^{3}x^{3}
+x^{3}tq^{4} − xy^{2} + 3 qxy^{2} + 3 xy^{2}q^{2}
+xy^{2}t + 3 xy^{2}tq + x^{2}yq + 3 yq^{2}x^{2}
+3 tq^{2}x^{2}y + 3 x^{2}ytq^{3} − x^{2}ytq)/(qx + y)^{2}.

4 Working with words

Let w = a_{1}a_{2} … a_{n} ∈ R(c) and let π∗ = v_{k}(w) = v_{1}v_{2} … v_{n}v_{n+1} as defined in Section 2.1. (Here, we are identifying ∗ with v_{n+1}.) Apply the bijection Φ_{k} of Section 3 to π to obtain a permutation π′ = v′_{1}v′_{2} … v′_{n}. Finally, decode π′ by replacing each letter x of π′ by a letter a_{i} of w such that v_{k,i}(w) = x, to obtain a word w′ = a′_{1}a′_{2} … a′_{n}. The mapping Φ_{k,W} is then defined on R(c) by Φ_{k,W}(w) = w′.

We introduce two pieces of notation to aid in the proof of Theorem 4. Firstly, let the rightmost occurence of the largest small letter ℓ in w be coded to the letter L. Then for any letter a_{i} in w, a_{i} is small if and only if v_{i} ≤ L. Secondly, write θ as the decoding map. Thus a_{i} = θ(x) if v_{k,i}(w) = x.

Proof of Theorem 4: We shall show that Φ_{k,W} is a bijection on R(c) such that for all w ∈ R(c), we have

(des_{k}, mad_{k}, mak_{k}) w = (exc_{k}, inv_{k}, den_{k}) Φ_{k,W}(w).  (10)

By equation (2), we have D_{k}(π) = D_{k}(w), and moreover the sequence of k-embracing numbers of π equals the sequence of k-embracing numbers of w. By the remark following the proof of equation (9), it follows that the set of k-excedance places of π′ equals the set of k-descent bottoms of w, the set of k-excedance tops of π′ equals the set of k-descent tops of w and that the sequence of k-side numbers of π′ is a permutation of the sequence of k-embracing numbers of w.

To complete the proof of equation (10), we must look in more detail at the construction of Φ_{k,W}, derived from the construction of Φ given above. Let F, F′, G and G′ denote respectively the sets of descent bottoms, descent tops, non descent bottoms and non descent tops for π∗. We note that each of the above sets is the image under v_{k} of the corresponding k-set for w∗. Let f and g be the non-decreasing words formed from the letters of F and G respectively. Let f′ and g′ be the words formed from the letters of F′ and G′ respectively, whose imversion bottom and inversion top numbers respectively are the embracing numbers of those letters in π∗, that is, the k-embracing numbers of the corresponding letters in w. Form the biword

α = ( \begin{matrix}f & | & g \\ f′ & | & g′ \end{matrix} ) .

Then the columns to the left of the bar are excedances, while those to the right are non-excedances. Finally, rearrange the columns of α to form the biword π̃′∗ and read off the permutation π′∗ from the second row. We must show that the excedances in α correspond precisely to the k-excedances in w̃′.

First we note that the column ( \begin{matrix}a \\ b \end{matrix} ) of α corresponds to the column ( \begin{matrix}a \\ b \end{matrix} )

of π̃′∗ and thence either to the column ( \begin{matrix}a \\ b \end{matrix} ) or to the column ( \begin{matrix}a + 1 \\ b \end{matrix} ) of π̃′ according as a ≤ L or a > L.

Case 1: The column ( \begin{matrix}a \\ b \end{matrix} ) of α is an excedance.

Then a is a descent bottom and b is a descent top in π∗. Case 1(i): b ≤ L.

Then a < L. So there is a column ( \begin{matrix}a \\ b \end{matrix} ) in π̃′ corresponding to a column

( \begin{matrix}θ(a) \\ θ(b) \end{matrix} ) in w̃′. Now the proof that ( \begin{matrix}θ(a) \\ θ(b) \end{matrix} ) is an excedance (which in this case is the same as a k-excedance, as θ(b) is small), follows exactly as in Section 4 of [3]. Case 1(ii): b) > L.

If a ≤ ℓ then, as before, the corresponding column in w̃′ is ( \begin{matrix}θ(a) \\ θ(b) \end{matrix} ), which is a k-excedance as θ(a) < θ(b). If a > L then, as a < b, a + 1 ≤ b and so θ(a + 1) ≤ θ(b). Therefore the corresponding column ( \begin{matrix}θ(a + 1) \\ θ(b) \end{matrix} ) of w̃′ is a k-excedance, as θ(b) is large.

Case 2: The column ( \begin{matrix}a \\ b \end{matrix} ) of α is a non-excedance.

Then a is a non descent bottom and b is a non descent top in π∗. Case 2(i): b ≤ L.

Then a ≥ b, so that θ(a) ≥ θ(b). Now if a > L then the column ( \begin{matrix}a + 1 \\ b \end{matrix} )

occurs in π̃′ and the corresponding column in w̃′ is ( \begin{matrix}θ(a + 1) \\ θ(b) \end{matrix} ), which is a

non-k-excedance as θ(b) is small. If a ≤ L then the column ( \begin{matrix}a \\ b \end{matrix} ) occurs in π̃′

and the corresponding column in w̃′ is ( \begin{matrix}θ(a) \\ θ(b) \end{matrix} ), which is a non-k-excedance as θ(b) is small.

Case 2(ii): b > L.

Then also a > L, so the column in w̃′ corresponding to the column ( \begin{matrix}a \\ b \end{matrix} )

of π̃′∗ is ( \begin{matrix}θ(a + 1) \\ θ(b) \end{matrix} ), which we must show is a non k-excedance. Now, let b_{1}

be the smallest letter such that b_{1} > b, θ(b_{1}) = θ(b) and b_{1} occurs to the right of b in the second row of α (that is, in the word g′). We will show that no such letter b_{1} exists. Now such a b_{1} must occur to the left of b in the word w, as large letters in w are coded from right to left. So the k-embracing numbers of b and b_{1} in w satisfy e(b_{1}) ≥ e(b). But as b_{1} occurs to the right of b in g and there is no letter c to the right of b in g in the range b < c < b_{1}, the inversion top numbers of b and b_{1} in g satisfy s(b_{1}) ≤ s(b). By the construction of g we have s(b_{1}) = e(b_{1}) and s(b) = e(b). Hence e(b_{1}) = e(b). Hence there can be no letter between b_{1} and b in w that is greater than b. Hence b_{1} is a descent top, and cannot be a letter of g. Hence we have shown that no letter c > b with θ(c) = θ(b) occurs to the right of b in g. Therefore, if b occurs in the p-th column from the right of g, we have

p ≤ 1 + e(b) + #{ j | j ∈ G′, θ(a_{j}) > θ(b) }
≤ 1 + #{ j | j ∈ G, θ(a_{j}) > θ(b) }.

For to every k-descent of w to the right of b that embraces b there corresponds a non k-descent bottom of w to the right of w that is greater than b, and to every non k-descent top of w that is greater than b there corresponds a non k-descent bottom of w that is greater than b. Thus we have θ(a + 1) > θ(b) as required.

It now follows that the set of k-excedance places in w′ equals the set of excedance places in π′∗, that is, the set of k-excedance places in π′. Similarly, the set of k-excedance tops in w′ equals the set of k-excedance tops in π′. Now, by the previous argument, w′_{E} = θ(f′) and w′_{N} = θ(g′′) (where g′′ is obtained from g′ by stripping off the final letter ∗.) By the proof of Case 2(ii) above, inv_{k} w′_{N} = inv π′∗_{N}, and by a similar argument, Imv_{k} w′_{E} = Imv π′∗_{E}. Thus Ine_{k} w′ = Ine π′∗ = Ine_{k} π′. Hence equation (10) follows.

To complete the proof, we must show that Φ_{k,W} is a bijection. The proof is not essentially different from that presented in [3] for the mapping Φ_{W}, that is, for the case k = 0, but for completeness we set down the proof here.

It clearly suffices to show that Φ_{k,W} is an injection. Suppose that for words w_{1} and w_{2} we have

w_{t} ↦ π_{t} = v_{k}(w_{t}) ↦ π′_{t} = Φ(π_{k}) ↦ w′

for t = 1, 2. Since the map w_{t} ↦ π′_{t} is clearly an injection, it suffices to prove that π′_{1} = π′_{2}. We refer to the sets and words involved in the construction of Φ_{k,W}(w_{t}) by F_{t}, f_{t}, G_{t}, etc., for t = 1, 2. Now F_{t} = E(π′_{t}) = E(w′), = {i_{1}, i_{2}, …, i_{r}}, say, for t = 1, 2, and G_{t} is the complementary set. Hence the words f_{t} and g_{t} can be determined. Further, f′_{t} is an inverse image under θ of the word a_{i_{1}}a_{i_{2}} … a_{i_{r}}, and we have a similar result for g′_{t}. We can calculate the side numbers of π′_{t}, as these equal the side numbers of w′. Now we can recover f′_{t} and g′_{t}, provided that we can determine the relative size of any two letters of these words that have the same image under θ. Suppose that θ(v′_{i}) = θ(v′_{j}) = a for letters v′_{i} and v′_{j} of π′_{t}. We distinguish several cases.

1. Suppose that s(v′_{i}) > s(v′_{j}). Then v′_{i} must occur to the left of v′_{j} in the word π_{t}. Hence, by the way in which the value v(w_{t}) is defined, v′_{i} > v′_{j} or v′_{i} < v′_{j} according as a is large or small.

2. Suppose that s(v′_{i}) = s(v′_{j}).

(a) If v′_{i} ∈ F′_{t} and v′_{j} ∈ G′_{t} then v′_{i} is a descent top in π_{t} and v′_{j} is not a descent top in π_{t}. Hence v′_{i} must occur to the left of v′_{j} in the permutation π_{t}, otherwise s(v′_{j}) = e(v′_{j}) > e(v′_{i}) = s(v′_{i}). Thus v′_{i} > v′_{j} or v′_{i} < v′_{j} as in the previous case.

(b) Let v′_{i}, v_{j} ∈ G′_{t} with i < j. We may assume that θ(v′_{m}) ≠ a for any m between i and j. Then v′_{i} < v′_{j}, for otherwise (i, j) would be an inversion in π′_{t} and we would have s(v′_{i}) ≠ s(v′_{j}).

(c) Let v′_{i}, v′_{j} ∈ F′_{t} with i < j. Then both v′_{i} and v′_{j} are descent tops in π_{t}. Hence a must be a large letter. As in the previous case, we have v′_{i} < v′_{j}.

Hence f′_{t} and g′_{t} are completely determined by w′. Thus π′_{1} = π′_{2} and Φ_{k,W} is an injection.

This completes the proof of Theorem 4.

Example 6 Consider the word w = 3 1 4 1 2 1 3 4 of example 4, with k = 2. Write π∗ = v_{2}(w) = 7 1 9 2 4 3 6 8 5. As in example 5 we have π′∗ = Φ_{2}(π)∗ = 4 8 9 1 7 2 3 6 5. Decoding π then gives w′ = Φ_{2,W}(w) = 2 4 4 1 3 1 1 3. From examples 1, 3 and 4 we have

(des_{2}, mad_{2}, mak_{2}) w = (exc_{2}, inv_{2}, den_{2}) w′
= (4, 21, 14).

References

[1] R . J. Clarke and D. Foata: Eulerian Calculus I: Univariate statistics, Europ. J. Combinatorics 15 (1994), 345–362.

[2] R. J. Clarke and D. Foata: Eulerian Calculus II: An extension of Han’s fundamental transformation, Europ. J. Combinatorics 15 (1994), 345– 362.

[3] R. J. Clarke, E. Steingrímsson and J. Zeng: New Euler-Mahonian statistics on permutations and words, to appear in Adv. Appl. Math.

[4] D. Foata: Unpublished note (1993).

[5] G.-N. Han: The k-extension of a Mahonian statistic, Adv. Appl. Math., 16 (1995), 297–305.

[6] J. Zeng: Enumérations de permutations et J-fractions continues, Europ. J. Combinatorics, 14 (1993), 373–382.