COMBINATORICS OF GENERALIZED q-EULER ... - Semantic Scholar
COMBINATORICS OF GENERALIZED q-EULER NUMBERS TIM HUBER AND AE JA YEE
Abstract. New enumerating functions for the Euler numbers are considered. Several of the relevant generating functions appear in connection to entries in Ramanujan’s Lost Notebook. The results presented here are, in part, a response to a conjecture made by M. E. H. Ismail and C. Zhang about the symmetry of polynomials in Ramanujan’s expansion for a generalization of the Rogers-Ramanujan series. Related generating functions appear in the work of H. Prodinger and L. L. Cristea in their study of geometrically distributed random variables. An elementary combinatorial interpretation for each of these enumerating functions is given in terms of a related set of statistics.
1. Introduction The Euler numbers En are the integers defined by ∞ X En xn = sec x + tan x. n! n=0
(1.1)
In 1879, D. Andr´e [1, 2] gave a combinatorial interpretation for the Euler numbers En . These numbers count the number of permutations π = π1 π2 · · · πn of elements in the set [n] := {1, 2, . . . , n} such that the sign of πi − πi+1 equals (−1)i , 1 ≤ i < n. Such permutations are called alternating or up-down permutations. Alternating permutations have rich combinatorial structure and have been studied extensively over the last century [6, 7, 8, 9, 10, 12, 20, 28]. Particular emphasis has been placed upon the enumeration of alternating permutations by various weights and conditions. In this paper, we undertake a combinatorial analysis of several new q-analogues of the Euler numbers. The resulting expressions provide new enumerations for alternating permutations. The associated generating functions are quotients of basic hypergeometric series and arise in several contexts related to the work of S. Ramanujan [15, 16, 17, 27]. In particular, the generating functions from Section 4 appear in the expansions of Ramanujan’s Hadamard product of the generalized Rogers-Ramanujan series from page 57 of his Lost Notebook [23], [3, Chapter 13]: ! 2 ∞ ∞ X Y qn zn zq 2n−1 P∞ jn = 1+ , (1.2) (q; q) 1 − n j=1 q yj n=0 n=1 where 1 y1 = , (1 − q)ψ 2 (q)
y2 = 0,
P∞ (2n−1)q2n−1 q + q3 n=1 1−q 2n−1 y3 = − , (q; q)3 ψ 2 (q) (1 − q)3 ψ 6 (q)
y4 = y 1 y 3 . (1.3)
1
2
TIM HUBER AND AE JA YEE
The functions ψ(q) and (α; q)n appearing in (1.2) and (1.3) are defined by ψ(q) :=
∞ X
q
n(n+1)/2
n=0
,
(α; q)n =
n−1 Y
1 − αq j ,
|q| < 1.
j=0
In [18], Ismail and Zhang observed that the polynomials appearing in the expansion −2 (1.2), as the coefficients of (q; q)−1 j ψ (q) in yj , are symmetric about the middle coefficient(s). They conjectured that this is the case for each of the polynomials. We prove Ismail and Zhang’s conjecture in Sections 4, 5, 6, and 7 by unraveling the combinatorial significance of these polynomials. The series appearing in this paper arise in an entirely different setting in the work of Prodinger and Cristea [21, 22]. These authors employ generating functions to determine the probability that a random word over the infinite alphabet {1, 2, 3, . . . , } satisfies certain inequality conditions. They assume that, within a word, each letter j occurs with (geometric) probability pq j−1 , independently, for 0 < q < 1 and p = 1 − q. In Sections 2–3, we derive direct combinatorial interpretations for certain generating functions from [21, 22]. Quotients of the series considered in the present paper also have beautiful continued fraction representations [14, 21]. For positive integers r1 , r2 , s1 , s2 and nonnegative integers A, B, C, D with s1 − r1 = 2A − 1 and s2 − r2 = 2C − 1, consider the following q-analogue of tan x: 2 ∞ X (−1)n q An +Bn x2n+1 0, 0, . . . , 0 2 A+B 2 q ,q x ∞ r φs (q; q)2n+1 X f2n+1 (q) x2n+1 x 1 1 0, 0, · · · , q 3 n=0 , = ∞ = X (−1)n q Cn2 +Dn x2n (q; q) 1 − q 0, 0, . . . , 0 2n+1 2 C+D 2 n=0 q ,q x r2 φs2 0, 0, · · · , q (q; q) 2n n=0 (1.4) where X ∞ (a1 ; q)n (a2 ; q)n · · · (ar ; q)n a1 , a2 , . . . , ar q, z = [(−1)n q n(n−1)/2 ]1+s−r z n . r φs b1 , b2 , · · · , bs (q; q)n (b1 ; q)n (b2 ; q)n · · · (bs ; q)n n=0
When (A, B, C, D) = (0, 0, 0, 0), f2n+1 (q) is the q-tangent number T2n+1 (q) of F. H. Jacko −2 son [19]. In [16], Huber proves that the coefficients T2n+1 (q) of (q; q)−1 2n+1 ψ (q) in y2n+1 in (1.2) are f2n+1 (q) for (A, B, C, D) = (1, 1, 1, 0). In this paper, we discuss q-tangent αβ numbers corresponding to (A, B, C, D) given in the following table. Let τ2n+1 represent the probability that a word from {1, 2, 3, · · · } of length 2n + 1 defined in the preceeding paragraph satisfies the inequality conditions αβ. f2n+1 (A, B) (C, D) Probability ≤> ≥< T2n+1 (0, 0) (0, 0) τ2n+1 , τ2n+1 o T2n+1 (1, 1) (1, 0) τ2n+1 >< e T2n+1 (1, 0) (1, 0) τ2n+1 The column on the right contains the numbers considered by Prodinger for which the series (1.4) is an associated generating function [21, Theorem 2.2].
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
3
For each value of A, B, C and D, the quotient (1.4) induces a corresponding recursion relation for the function f2n+1 (q). From these formulas, we obtain the following related polynomials. For a polynomial p(q), let pˆ(q) denote the dual of p(q) (see [21, Remark 3.3]). Several of the dual polynomials occur in connection with probabilities from [21]. (A, B) (0, 1) (2, 1) (2, 0) (1, 0) (1, 1)
(C, D) (0, 1) (2, −1) (2, −2) (1, −1) (1, −1)
f2n+1 Relevant Relations Probability n des des T2n+1 T2n+1 (q) = q T2n+1 (q) Tˆ2n+1 Tˆ2n+1 (q) = T2n+1 (q) des des des Tˆ2n+1 Tˆ2n+1 (q) = q −2n T2n+1 (q) ≤≥ o o −n o ˆ ˆ T2n+1 T2n+1 (q) = q T2n+1 (q) τ2n+1 ≥≤ e e e Tˆ2n+1 Tˆ2n+1 (q) = q −n−1 T2n+1 (q) τ2n+1
In Section 2, a well-known arithmetic interpretation of the classical q-tangent numbers T2n+1 (q) is discussed. We provide an elementary proof of this interpretation that demonstrates fundamental ideas used throughout the paper. The new q-analogue des (q) is also discussed in the same section. In Section 3, we deduce combinatorial T2n+1 interpretations for new q-analogues of the secant numbers appearing in [21, 22] defined, for s2 − r2 = 2C − 1, by ∞ X g2n (q) x2n n=0
(q; q)2n
=
r2 φs2
1 . 0, 0, . . . , 0 2 C+D 2 q ,q x 0, 0, · · · , q
(C, D) g2n Relevant Relations Probability ≤> (0, 0) S2n σ2n e o , S2n (1, 0) qS2n σ2n ≤≥ o o o (1, −1) Sˆ2n Sˆ2n (q) = q 1−n S2n (q) σ2n πj . We denote by inv(π) the number of inversions of the permutation π. The descent set D(π) is defined by {i | πi > πi+1 }, and des(π) denotes the size of D(π). Define the q-binomial coefficient by (q; q)n , if 0 ≤ n ≤ k, n = (q; q)k (q; q)n−k k 0, otherwise. The following lemma is one of several combinatorial interpretations for the q-binomial coefficient. We will refer to this lemma often in the remainder of the paper. For a proof of the lemma, see [26, p. 132]. Lemma 1.1. X
q
inv(π)
π
n = , k
where the sum is over all permutations π with D(π) ⊂ {k}. (k)
A more instructive view of Lemma 1.1 follows by defining Pn , for a given n and k ≤ n, to be the set of all permutations π on [n] such that π1 < π2 < π3 < · · · < πk−1 < πk , Then it follows from Lemma 1.1 that X (k) π∈Pn
q
πk+1 < πk+2 < · · · < πn−1 < πn .
inv(π)
n = . k
We denote by An the set of alternating permutations π1 π2 · · · πn on the set [n] with π1 < π2 > π3 < · · · . For instance, A1 = {1}, A2 = {12}, A3 = {132, 231}, A4 = {1324, 1423, 2314, 2413, 3412} A5 = {13254, 23154, 14253, 24153, 14352, 34152, 24351, 34251, 15243, 25143, 15342, 35142, 25341, 35241, 45132, 45231}. 2. The classical q-tangent numbers Jackson’s q-analogues of the sine and cosine functions [19] are ∞ X n=0
(−1)n x2n+1 (q; q)2n+1 /(1 − q)2n+1
and
∞ X n=0
(−1)n x2n . (q; q)2n /(1 − q)2n
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
5
By considering quotients of these functions, we arrive at a q-analogue of the tangent numbers, T2n+1 (q), defined by ∞ ∞ ∞ X T2n+1 (q)x2n+1 X (−1)n x2n+1 X (−1)n x2n = / . (2.1) (q; q) (q; q) (q; q) 2n+1 2n+1 2n n=0 n=0 n=0 If we replace x by x(1−q) in (2.1) and let q → 1− , the corresponding identity reduces to the relation obtained by equating odd parts on each side of equation (1.1). Multiplying both sides of (2.1) by the denominator on the right side and equating coefficients of x, we obtain n−1 X 2n + 1 n T2n+1 (q) = (−1) + (−1)n−k−1 T2k+1 (q). (2.2) 2k + 1 k=0
The following combinatorial interpretation of the polynomials T2n+1 (q) is well known [5, 13, 25]. We include a proof based upon the recursion (2.2) as an aid to the reader, since later results in the paper have proofs that are similar in nature. Theorem 2.1. For a nonnegative integer n, we have X T2n+1 (q) = q inv(π) . π∈A2n+1
Proof. Let X
f2n+1 (q) =
q inv(π) .
π∈A2n+1
For n = 0, it is clear that f1 (q) = 1 = T1 (q). For any positive integer n, we will prove that f2n+1 (q) satisfies the recurrence (2.2). (k) For a positive integer k ≤ n, let A2n+1 be the set of permutations π on [2n + 1] such that π1 < π2 < π3 < · · · < π2k , π2k+1 < π2k+2 > π2k+3 < · · · < π2n > π2n+1 . From Lemma 1.1 and the definition of f2n+1 (q), we see that X 2n + 1 inv(π) q = f2(n−k)+1 (q). 2k
(2.3)
(k) π∈A2n+1
(k)
For a positive integer k ≤ n, we denote by B2n+1 the set of permutations π on [2n + 1] such that π1 < π2 < π3 < · · · < π2k > π2k+1 < π2k+2 > π2k+3 < · · · < π2n > π2n+1 . (k)
We now compute the generating function for permutations π ∈ B2n+1 X q inv(π) . (k)
π∈B2n+1
6
TIM HUBER AND AE JA YEE (k)
(k)
From the definitions of A2n+1 and B2n+1 , we see that for any k, 1 ≤ k ≤ n, (k)
(k)
(k+1)
A2n+1 = B2n+1 ∪ B2n+1 , (n+1)
where B2n+1 = {π | π1 < π2 < · · · < π2n < π2n+1 }. Thus X X X q inv(π) = q inv(π) − q inv(π) . (k)
(k)
π∈B2n+1
(2.4)
(k+1)
π∈A2n+1
π∈B2n+1
By iterating (2.4), we deduce X
q
inv(π)
=
n X
(−1)k−1
k=1
(1) π∈B2n+1
X
X
q inv(π) + (−1)n
(k) π∈A2n+1
q inv(π) .
(2.5)
(n+1) π∈B2n+1
(1)
Note that B2n+1 = A2n+1 . Therefore, it follows from (2.3) that (2.5) is equivalent to n X 2n + 1 f2n+1 (q) = (−1)k−1 f2(n−k)+1 (q) + (−1)n , 2k k=1
which completes the proof.
The following theorem gives the generating function for alternating permutations π in A2n+1 by weight inv(π) + des(π). Theorem 2.2. Define ∞ X (−1)n q n x2n+1 ∞ des X (q)x2n+1 T2n+1 n=0
(q; q)2n+1
=
n=0 ∞ X n=0
(q; q)2n+1 (−1)n q n x2n (q; q)2n
.
Then des T2n+1 (q) =
X
q inv(π)+des(π) .
π∈A2n+1
Proof. Recalling the definition of T2n+1 (q) given by (2.1) and noting that des(π) = n for any π ∈ A2n+1 , we have X q inv(π)+des(π) = q n T2n+1 (q). π∈A2n+1 des des By comparing the recurrence relations for T2n+1 (q) and T2n+1 (q), we see that T2n+1 (q) = n q T2n+1 (q). des The dual functions corresponding to T2n+1 (q) and T2n+1 (q) are discussed in the following theorem.
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
7
Theorem 2.3. Define 2 ∞ X (−1)n q 2n +n x2n+1
∞ ˆ X T2n+1 (q)x2n+1
(q; q)2n+1
n=0
=
n=0 ∞ X n=0
(q; q)2n+1
,
2
(−1)n q 2n −n x2n (q; q)2n
and 2 ∞ X (−1)n q 2n x2n+1
∞ ˆ des X T2n+1 (q)x2n+1
(q; q)2n+1
n=0
=
n=0 ∞ X n=0
(q; q)2n+1 2
(−1)n q 2n −2n x2n (q; q)2n
.
Then we have Tˆ2n+1 (q) = T2n+1 (q)
and
des des Tˆ2n+1 (q) = q −2n T2n+1 (q).
Proof. From the definition of Tˆ2n+1 (q), it is clear that Tˆ1 (q) = 1 = T1 (q). Note that n 2n2 +n
Tˆ2n+1 (q) = (−1) q
+
n−1 X 2n + 1 k=0
2k + 1
(−1)n−k−1 q 2(n−k)
2 −(n−k)
Tˆ2k+1 (q).
For any positive integer n, we will show that T2n+1 (q) satisfies the same recurrence as Tˆ2n+1 (q). (k) For a nonnegative integer k ≤ n, let A2n+1 be the set of permutations π on [2n + 1] such that π1 < π2 > π3 < · · · < π2k > π2k+1 , π2k+2 > π2k+3 > · · · > π2n > π2n+1 Since there are 2(n − k)2 − (n − k) inversions in a permutation π2k+2 π2k+3 · · · π2n+1 such that π2k+2 > π2k+3 > · · · > π2n > π2n+1 , Lemma 1.1 and the definition of T2n+1 (q) imply X 2n + 1 2(n−k)2 −(n−k) inv(π) q = q T2k+1 (q). (2.6) 2k + 1 (k)
π∈A2n+1 (k)
For a positive integer k ≤ n, let B2n+1 denote the set of permutations π on [2n + 1] satisfying π1 < π2 > π3 < · · · < π2k > π2k+1 > π2k+2 > π2k+3 > · · · > π2n > π2n+1 .
8
TIM HUBER AND AE JA YEE (k)
We now compute the generating function for permutations π ∈ B2n+1 X q inv(π) . (k)
π∈B2n+1 (k)
(k)
From the definitions of A2n+1 and B2n+1 , we see that for any k, 0 ≤ k < n, (k)
(k)
(k+1)
A2n+1 = B2n+1 ∪ B2n+1 , (0)
where B2n+1 = {π | π1 > π2 > · · · > π2n > π2n+1 }. Thus, X X X q inv(π) = q inv(π) − q inv(π) . (k+1)
(k)
π∈B2n+1
(2.7)
(k)
π∈A2n+1
π∈B2n+1
By iterating (2.7), we deduce X
q
inv(π)
=
n−1 X k=0
(n)
π∈B2n+1
X
(−1)n−k−1
X
q inv(π) + (−1)n
q inv(π) ,
(0)
(k)
π∈B2n+1
π∈A2n+1
which is equivalent to T2n+1 (q) =
n−1 X 2n + 1 2k + 1
k=0
(−1)n−k−1 q 2(n−k)
2 −(n−k)
2 +n
T2k+1 (q) + (−1)n q 2n
by (2.6). Therefore, Tˆ2n+1 (q) = T2n+1 (q). des des It follows from the recurrences for Tˆ2n+1 and Tˆ2n+1 that Tˆ2n+1 (q) = q n Tˆ2n+1 (q). −2n des des This identity is equivalent to Tˆ2n+1 (q) = q T2n+1 (q) since Tˆ2n+1 (q) = T2n+1 (q) = q −n T des (q). By equating coefficients of x2n+1 in the generating functions for T2n+1 (q) and Tˆ2n+1 (q) we obtain the following result. Corollary 2.4. For any nonnegative integer n, X n n X 2j 2 +j 2n + 1 2j 2 −j 2n + 1 q = q . 2j + 1 2j j=0
j=0
3. New q-Secant Numbers The classical q-secant numbers enumerate alternating permutations on the set [2n] for n ≥ 1 by the number of inversions. Theorem 3.1. Define ∞ X S2n (q)x2n n=0
(q; q)2n
=
∞ X (−1)n x2n n=0
(q; q)2n
Then S2n (q) =
X π∈A2n
q inv(π) .
!−1 .
(3.1)
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
9
For a proof of Theorem 3.1, see [4, 24, 25]. The following theorem provides a combinatorial interpretation for a new class of secant numbers. Theorem 3.2. Define !−1
∞ o X (q)x2n S2n = (q; q)2n n=0
2 ∞ X (−1)n q n +1 x2n
∞ ˆo X S2n (q)x2n = (q; q) 2n n=0
∞ X (−1)n q n(n−1) x2n
n=0
(q; q)2n
(q; q)2n
n=0
, !−1 ,
Then, for n ≥ 1, X
o (q) = S2n
q inv(π)+des(πo ) ,
π∈A2n o o Sˆ2n (q) = q 1−n S2n (q).
Proof. Let X
g2n (q) =
q inv(π)+des(πo ) .
π∈A2n o We first prove that g2n (q) = S2n (q). From the definitions of inv(π) and des(πo ), it is clear that
g2 (q) = 1 = S2o (q). o (q) satisfy S0o (q) = q −1 and The polynomials S2n
o S2n (q)
=
n−1 X 2n k=0
2k
2
o (−1)n−k−1 q (n−k) S2k (q),
for n ≥ 1.
(3.2)
We define g0 (q) = q −1 . We will show that the polynomials g2n (q) satisfy (3.2) for n > 1. (k)
For a positive integer k ≤ n, let A2n be the set of permutations π on [2n] such that π1 < π2 > π3 < · · · > π2k−1 < π2k , π2k+2 > π2k+4 > · · · > π2n > π2n−1 > π2n−3 > · · · > π2k+1 . From Lemma 1.1 and the definition of g2n , we see that X 2n (n−k)2 inv(π)+des(π1 π3 ···π2k−1 )+(n−k) q = q g2k (q). 2k (k) π∈A2n
(3.3)
10
TIM HUBER AND AE JA YEE (k)
For a positive integer k < n, we decompose A2n into disjoint subsets as follows: (k)
A2n ={π | π2k > π2k+2 } ∪ {π | π2k < π2k+2 } ={π | π2k > π2k+2 > π2k−1 } ∪ {π | π2k > π2k−1 > π2k+2 } ∪ {π | π2k+2 > π2k > π2k−1 } ={π | π2k > π2k+2 > π2k+1 > π2k−1 } ∪ {π | π2k > π2k+2 > π2k−1 > π2k+1 } ∪ {π | π2k > π2k−1 > π2k+2 > π2k+1 } ∪ {π | π2k+2 > π2k > π2k−1 > π2k+1 } ∪ {π | π2k+2 > π2k > π2k+1 > π2k−1 } ∪ {π | π2k+2 > π2k+1 > π2k > π2k−1 } ={π | π2k > π2k+2 > π2k+1 > π2k−1 } ∪ {π | π2k > π2k−1 > π2k+1 } ∪ {π | π2k+2 > π2k > π2k+1 > π2k−1 } ∪ {π | π2k+2 > π2k+1 > π2k > π2k−1 } (k)
(k)
(k)
(k)
=:B2n ∪ C2n ∪ D2n ∪ E2n .
(3.4)
(k)
Note that B2n is the set of alternating permutations π on [2n] such that π1 < π2 > π3 < · · · > π2k−1 < π2k , π2k > π2k+2 > · · · > π2n > π2n−1 > π2n−3 > · · · > π2k+1 > π2k−1 , (k)
(k−1)
(n)
from which it is clear that B2n is a subset of A2n . We define B2n = A2n . For a (k) permutation π ∈ B2n with 1 < k ≤ n, if π2k−3 > π2k−1 , then π2k−2 > π2k−3 > π2k−1 , (k−1)
which shows that such π satisfy the conditions of C2n
. Thus
(k)
B2n ={π | π2k−3 > π2k−1 } ∪ {π | π2k−3 < π2k−1 } (k−1)
=C2n
∪ {π | π2k−3 < π2k−1 }.
(3.5) (k)
We now compute the generating function for permutations π ∈ B2n X q inv(π)+des(π1 π3 ···π2k−1 )+(n−k) . (k)
π∈B2n (k)
Let π ∈ B2n . If π2k−3 > π2k−1 , then inv(π) + des(π1 π3 · · · π2k−1 ) + (n − k) = inv(π) + des(π1 π3 · · · π2k−3 ) + (n − k + 1). However, if π2k−3 < π2k−1 , then inv(π) + des(π1 π3 · · · π2k−1 ) + (n − k) = inv(π) + des(π1 π3 · · · π2k−3 ) + (n − k + 1) − 1. In this case, we look for a permutation σ such that inv(π) + des(π1 π3 · · · π2k−1 ) = inv(σ) + des(σ1 σ3 · · · σ2k−3 ) + 1. Let m be defined by π2k+m = max{π2k+j | π2k+j < π2k−2 , j ≥ −1}. There exists such an m since π is an alternating permutation, so that π2k−1 < π2k−2 . It follows that π2k−1 ≤ π2k+m < π2k−2 . We switch π2k−2 and π2k+m , and denote
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
11
the resulting partition by π ¯ . Switching π2k−2 with π2k+m results in a decrease of the inversion number, namely inv(π) = inv(¯ π ) + 1. Moreover, since π ¯2i+1 = π2i+1 for i < k and π2k−3 < π2k−1 , des(π1 π3 · · · π2k−1 ) = des(¯ π1 π ¯3 · · · π ¯2k−3 ). Thus inv(π) + des(π1 π3 · · · π2k−1 ) = inv(¯ π ) + des(¯ π1 π ¯3 · · · π ¯2k−3 ) + 1. If π2k−2 were switched with π2k+m for m ≥ 0, then π2k−1 < π2k+m and π2k ≥ π2k+m . Hence, from the definition of π ¯ , we see that π ¯1 < π ¯2 > π ¯3 < · · · < π ¯2k−2 , π ¯2k > π ¯2k+2 > · · · > π ¯2n > π ¯2n+1 > π ¯2n−1 > · · · > π ¯2k+1 > π ¯2k−1 , π ¯2k > π ¯2k−2 > π ¯2k−1 > π ¯2k−3 , (k−1)
which shows π ¯ ∈ D2n+1 . If π2k−2 and π2k−1 were switched, namely m = −1, then π2k−2 < πi for i ≥ 2k. Hence, from the definition of π ¯ , we see that π ¯1 < π ¯2 > π ¯3 < · · · < π ¯2k−2 , π ¯2k > π ¯2k+2 > · · · > π ¯2n > π ¯2n+1 > π ¯2n−1 > · · · > π ¯2k+1 > π ¯2k−1 , π ¯2k > π ¯2k−1 > π ¯2k−2 > π ¯2k−3 , (k−1)
which shows π ¯ ∈ E2n+1 . Thus, by (3.4) and (3.5), for any k, 1 < k ≤ n, (k)
(k)
{π | π ∈ B2n , π2k−3 > π2k−1 } ∪ {¯ π | π ∈ B2n , π2k−3 < π2k−1 } (k−1)
= C2n
(k−1)
= A2n
(k−1)
∪ D2n
(k−1)
\ B2n
(k−1)
∪ E2n
.
Therefore, X q inv(π)+des(π1 π3 ···π2k−1 )+(n−k) (k)
π∈B2n
=
X (k) π∈B2n π2k−3 >π2k−1
=
X
X (k−1)
π∈A2n
q inv(π)+des(π1 π3 ···π2k−3 )+(n−k)
(k) π∈B2n π2k−3 π2k−1
=
X
q inv(π)+des(π1 π3 ···π2k−3 )+(n−k+1) +
q inv(π)+des(π1 π3 ···π2k−3 )+(n−k+1) −
q inv(¯π)+des(¯π1 π¯3 ···¯π2k−3 )+(n−k+1)
(k) π∈B2n π2k−3 1, we deduce n−1 X X X inv(π)+des(π1 π3 ···π2k−1 )+(n−k) q inv(π)+des(πo ) = (−1)n−k−1 q k=1
(n)
π∈B2n
(k)
π∈A2n
+ (−1)n−1
X
q inv(π)+n−1 ,
(1)
π∈B2n
which is equivalent to g2n (q) =
n−1 X 2n k=1
=
2k
n−1 X 2n k=0
2k
2
2 −1
(−1)n−k−1 q (n−k) g2k (q) + (−1)n−1 q n 2
(−1)n−k−1 q (n−k) g2k (q),
where the second equality holds since g0 (q) = q −1 . o o o Using the recurrences satisfied by S2n (q) and Sˆ2n (q), we can easily prove that Sˆ2n (q) = 1−n o q S2n (q) for n ≥ 1. We omit the details. For a permutation π = π1 π2 · · · , we define πe = π0 π2 π4 · · · , where π0 = ∞. Theorem 3.3. Define e S2n (q) =
X
q inv(π)+des(πe ) .
(3.7)
π∈A2n
Then we have e o S2n (q) = qS2n (q).
Proof. For a permutation π ∈ A2n , define the map σ by σ(π) = (2n + 1 − π2n )(2n + 1 − π2n−1 ) · · · (2n + 1 − π2 )(2n + 1 − π1 ). Then σ(π) is an alternating permutation in A2n with inv(π) = inv(σ(π))
and
des(πo ) + 1 = des(σ(π)e ),
since σ(π)e = (∞)(2n + 1 − π2n−1 )(2n + 1 − π2n−3 ) · · · (2n + 1 − π3 )(2n + 1 − π1 ). e o Therefore, it follows that S2n (q) = qS2n (q). It follows from Theorems 3.2 and 3.3 that !−1 2 ∞ ∞ e X X (−1)n q n x2n S2n (q)x2n . = (q; q) (q; q) 2n 2n n=0 n=0 o o Theorem 3.4. The polynomials S2n (q) and Sˆ2n (q) are symmetric about the middle coefficient. o Proof. It suffices to show that S2n (q) is symmetric. First note that the alternating permutation with the least weight is
π = 1 3 2 5 4 7 · · · (2n − 1) (2n − 2) 2n
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
13
with inv(π) = n − 1 and des(πo ) = 0; while the alternating permutation with the largest weight is π = (2n − 1) 2n (2n − 3) (2n − 2) · · · 1 2 with inv(π) = 2n2 − 2n and des(πo ) = n − 1. Thus 2 −n−1
o S2n (q) = q n−1 + · · · + q 2n
.
o S2k (q) is symmetric about q k It is clear that S2o (q) = 1 is symmetric. Suppose x k < n. Since the q-binomial coefficient is symmetric, y 2n (n−k)2 o q S2k (q) 2k
2 −1
for
is symmetric. The exponent of the middle term is (n − k)2 + (2nk − 2k 2 ) + k 2 − 1 = n2 − 1. 2 −1
o Therefore, S2n (q) is symmetric about q n
.
Note. Is there a combinatorial proof of Theorem 3.4 analogous to that of Theorem 5.3? Theorem 3.5. Let ∞ ˆ X S2n (q)x2n n=0
(q; q)2n
=
∞ X (−1)n q n(2n−1) x2n
(q; q)2n
n=0
!−1 .
Then Sˆ2n (q) =
X
q inv(π) ,
π
where the sum is over all alternating permutations π on [2n] with π1 > π2 < π3 > π4 < · · · < π2n−1 > π2n . Proof. We first note that the polynomials Sˆ2n (q) satisfy Sˆ0 (q) = 1 and n−1 X 2n ˆ S2n (q) = (−1)n−k−1 q (n−k)(2(n−k)−1) Sˆ2k (q), for n ≥ 1. 2k k=0
For any alternating permutation π ∈ A2n , define π ¯ by π ¯ = π2n π2n−1 π2n−2 · · · π2 π1 . Then π ¯ is clearly a down-up permutation. Furthermore inv(π) + inv(¯ π ) = n(2n − 1). Thus X π∈A2n
q inv(¯π) =
X
q n(2n−1)−inv(π)
π∈A2n
= q n(2n−1) S2n (q −1 ).
(3.8)
14
TIM HUBER AND AE JA YEE
The theorem is equivalent to Sˆ2n (q) = q n(2n−1) S2n (q −1 ). We now show that q n(2n−1) S2n (q −1 ) satisfies the same recursion as Sˆ2n (q). Substitute q k(2k−1) S2k (q −1 ) for Sˆ2k in (3.8). Then n−1 X 2n (−1)n−k−1 q (n−k)(2(n−k)−1) q k(2k−1) S2k (q −1 ) 2k q k=0 n−1 X 2 2n = (−1)n−k−1 q 2n −n S2k (q −1 ) 2k q−1 k=0 n(2n−1)
=q
S2n (q −1 ),
where the last equality follows from the recursion formula for S2n (q).
Note: The polynomials S2n (q) and Sˆ2n (q) appearing in Theorems 3.1 and 3.5 are the only polynomials considered in this paper that are not symmetric about the middle coefficient(s). 4. New q-tangent numbers associated with odd indices o Define T2n+1 (q) by 2 ∞ X (−1)n q n +n x2n+1
∞ o X (q)x2n+1 T2n+1
(q; q)2n+1
n=0
=
n=0 ∞ X n=0
Then, by the definition of
o S2n (q)
∞ o X (q)x2n+1 T2n+1 n=0
(q; q)2n+1
(q; q)2n+1 2
(−1)n q n x2n (q; q)2n
.
in Theorem 3.2,
2 ∞ ∞ o X (−1)n q n +n x2n+1 X qS2n (q)x2n = . (q; q) (q; q) 2n+1 2n n=0 n=0
Thus o T2n+1 (q)
=
n X 2n + 1 k=0
2k
(−1)n−k q (n−k)
2 +(n−k)+1
o S2k (q).
(4.1)
−2 o In [16], Huber proves that the coefficients (q; q)−1 2n+1 ψ (q) in y2n+1 in (1.2) are T2n+1 (q), whose combinatorial interpretation is given the following theorem. To define the corresponding statistic we denote, for a given permutation π = π1 π2 · · · π2n+1 , the permutation πo = π1 π3 π5 · · · π2n−1 π2n+1 .
Theorem 4.1. For each nonnegative integer n, we have X o T2n+1 (q) = q inv(π)+des(πo ) . π∈A2n+1
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
15
Proof. Let X
f2n+1 (q) =
q inv(π)+des(πo ) .
π∈A2n+1
For n = 0, it is clear that f1 (q) = 1 = T1o (q). For any positive integer n, we will show that f2n+1 (q) satisfies the equation (4.1). (k) For a positive integer k ≤ n, let A2n+1 be the set of permutations π on [2n + 1] such that π1 < π2 > π3 < · · · > π2k−1 < π2k , π2k+2 > π2k+4 > · · · > π2n > π2n+1 > π2n−1 > · · · > π2k+1 .
(4.2)
From Lemma 1.1 and Theorem 3.2, we see that for k ≤ n, X 2n + 1 (n−k)2 o inv(π)+des(π1 π3 ···π2k−1 ) q = S2k (q). q 2k
(4.3)
(k)
π∈A2n+1 (n)
We decompose A2n+1 into disjoint subsets as follows: (n)
A2n+1 ={π | π2n > π2n+1 } ∪ {π | π2n < π2n+1 } ={π | π2n > π2n+1 > π2n−1 } ∪ {π | π2n > π2n−1 > π2n+1 } ∪ {π | π2n < π2n+1 } (n)
(n)
(n)
(n)
(n)
=:B2n+1 ∪ D2n+1 ∪ E2n+1 .
(4.4) (k)
Note that A2n+1 = B2n+1 ∪ D2n+1 . For k < n, we decompose A2n+1 into disjoint subsets as follows: (k)
A2n+1 ={π | π2k > π2k+1 } ∪ {π | π2k < π2k+1 } ={π | π2k > π2k+1 > π2k−1 } ∪ {π | π2k > π2k−1 > π2k+1 } ∪ {π | π2k < π2k+1 } ={π | π2k > π2k+2 > π2k+1 > π2k−1 } ∪ {π | π2k+2 > π2k > π2k+1 > π2k−1 } ∪ {π | π2k > π2k−1 > π2k+1 } ∪ {π | π2k < π2k+1 } (k)
(k)
(k)
(k)
=:B2n+1 ∪ C2n+1 ∪ D2n+1 ∪ E2n+1 .
(4.5)
(k)
(k)
Note that, since the permutations in B2n+1 satisfy the conditions of (4.2), B2n+1 is the set of alternating permutations π on [2n + 1] such that π1 < π2 > π3 < · · · < π2k , π2k > π2k+2 > · · · > π2n > π2n+1 > π2n−1 > · · · > π2k−1 . Furthermore, for any positive k ≤ n, (k+1)
(k)
(k)
(k)
B2n+1 = B2n+1 ∪ C2n+1 ∪ D2n+1 ,
16
TIM HUBER AND AE JA YEE (n)
where B (n+1) = A2n+1 and C2n+1 = ∅. For k ≥ 1, we now compute the generating (k+1) function for permutations π ∈ B2n+1 X q inv(π)+des(π1 π3 ···π2k+1 ) . (k+1)
π∈B2n+1 (k+1)
(k)
Let π ∈ B2n+1 . If π2k−1 > π2k+1 , namely π ∈ D2n+1 , then inv(π) + des(π1 π3 · · · π2k+1 ) = inv(π) + des(π1 π3 · · · π2k−1 ) + 1. (k)
(k)
However, if π2k−1 < π2k+1 , namely π ∈ B2n+1 ∪ C2n+1 , then inv(π) + des(π1 π3 · · · π2k+1 ) = inv(π) + des(π1 π3 · · · π2k−1 ). In this case, we look for a permutation σ such that inv(π) + des(π1 π3 · · · π2k+1 ) = inv(σ) + des(σ1 σ3 · · · σ2k−1 ) + 1. Let the positive integer m be defined by π2k+m = max{π2k+j | π2k+j < π2k , 1 ≤ j ≤ 2(n − k) + 1}. Since π2k+1 < π2k , there exists such an m. We switch π2k+m with π2k and denote the resulting partition by π ¯ . Switching π2k with π2k+m results in a decrease of the inversion number, namely inv(π) = inv(¯ π ) + 1. Moreover, if π2k were switched with π2k+m for m > 1, it is trivial that des(π1 π3 · · · π2k+1 ) = des(¯ π1 π ¯3 · · · π ¯2k−1 ). If π2k were switched with π2k+1 , since π2k−1 < π2k+1 , des(π1 π3 · · · π2k+1 ) = des(¯ π1 π ¯3 · · · π ¯2k−1 ). Thus, in either case, inv(π) + des(π1 π3 · · · π2k+1 ) = inv(¯ π ) + des(¯ π1 π ¯3 · · · π ¯2k−1 ) + 1. (k)
From the definition of π ¯ , if π2k were switched with π2k+1 , namely π ∈ C2n+1 , then π ¯1 < π ¯2 > π ¯3 < · · · < π ¯2k−2 > π ¯2k−1 < π ¯2k , π ¯2k+2 > π ¯2k+4 > · · · > π ¯2n > π ¯2n+1 > π ¯2n−1 > · · · > π ¯2k+1 , π ¯2k < π ¯2k+1 , (k)
which shows that π ¯ ∈ E2n+1 . If π2k were switched with π2k+2 , then π ¯1 < π ¯2 > π ¯3 < · · · < π ¯2k−2 > π ¯2k−1 < π ¯2k , π ¯2k+2 > π ¯2k+4 > · · · > π ¯2n > π ¯2n+1 > π ¯2n−1 > · · · > π ¯2k+1 , π ¯2k+2 > π ¯2k > π ¯2k+1 > π ¯2k−1 ,
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
17
from the maximality of π2k+m ; if π2k were switched with π2k+m for m > 2, then π ¯1 < π ¯2 > π ¯3 < · · · < π ¯2k−2 > π ¯2k−1 < π ¯2k , π ¯2k+2 > π ¯2k+4 > · · · > π ¯2n > π ¯2n+1 > π ¯2n−1 > · · · > π ¯2k+1 , π ¯2k+2 > π ¯2k > π ¯2k+1 > π ¯2k−1 , since π2k+2 > π2k+m . Thus, if π2k were switched with π2k+m for m > 1, namely (k) (k) π ∈ B2n+1 , then π ¯ ∈ C2n+1 . Thus, for any k, 1 ≤ k ≤ n, (k+1)
(k+1)
{π | π ∈ B2n+1 , π2k−1 > π2k+1 } ∪ {¯ π | π ∈ B2n+1 , π2k−1 < π2k+1 } (k)
(k)
(k)
{¯ π | π ∈ B2n+1 } ∪ {¯ π | π ∈ C2n+1 } ∪ {π | π ∈ D2n+1 } (k)
(k)
(k)
= C2n+1 ∪ D2n+1 ∪ E2n+1 (k)
(k)
= A2n+1 \ B2n+1 ,
(4.6)
where the last equality of (4.6) follows from (4.5). Therefore, X q inv(π)+des(π1 π3 ···π2k+1 ) (k+1)
π∈B2n+1
=
X
X
q inv(π)+des(π1 π3 ···π2k−1 )+1
(k) π∈B2n+1 π2k−1 π2k+1
=
X
q inv(π)+des(π1 π3 ···π2k−1 ) +
q inv(π)+des(π1 π3 ···π2k−1 )+1 +
X
q inv(¯π)+des(¯π1 π¯3 ···¯π2k−1 )+1
(k) π∈B2n+1 π2k−1 π2k+1
X X inv(π)+des(π1 π3 ···π2k−1 ) = q q − q inv(π)+des(π1 π3 ···π2k−1 ) . (k) π∈A2n+1
(4.7)
(k) π∈B2n+1
By iterating (4.7), we deduce X q inv(π)+des(πo ) (n+1)
π∈B2n+1
=
n X
X X inv(π)+des(π1 π3 ···π2k−1 )+(n−k)+1 n (−1)n−k q q inv(π)+n , + (−1)
k=1
(k)
(1)
π∈A2n+1
π∈B2n+1
which is equivalent to f2n+1 (q) =
n X 2n + 1 k=0
2k (n+1)
(−1)n−k q (n−k)
(1)
2 +(n−k)+1
by (4.3) and the definitions of B2n+1 , B2n+1 , and S0o (q) = q −1 .
o S2k (q).
18
TIM HUBER AND AE JA YEE o Define Tˆ2n+1 (q) by 2 ∞ X (−1)n q n x2n+1
∞ ˆo X T2n+1 (q)x2n+1 n=0
(q; q)2n+1
=
n=0 ∞ X n=0
(q; q)2n+1 2
(−1)n q n −n x2n (q; q)2n
.
Theorem 4.2. For a nonnegative integer n, we have o o Tˆ2n+1 (q) = q −n T2n+1 (q). o o Proof. The theorem follows from the recurrences satisfied by T2n+1 (q) and Tˆ2n+1 (q).
5. New q-tangent numbers associated with even indices e Let T2n+1 (q) be the polynomial satisfying 2 ∞ X (−1)n q n x2n+1
∞ e X (q)x2n+1 T2n+1 n=0
(q; q)2n+1
=
n=0 ∞ X n=0
(q; q)2n+1 2
(−1)n q n x2n (q; q)2n
.
e (q), it follows that From the generating function for T2n+1 n−1 X 2 2n + 1 e n n2 e T2n+1 (q) = (−1) q + (−1)n−k−1 q (n−k) T2k+1 (q). 2k + 1
(5.1)
k=0
Recall πe for a permutation π defined in Section 3, namely πe = π0 π2 π4 · · · with π0 = ∞. Theorem 5.1. For a nonnegative integer n, X e T2n+1 (q) = q inv(π)+des(πe ) . π∈A2n+1
Proof. Let f2n+1 (q) =
X
q inv(π)+des(πe ) .
π∈A2n+1
For n = 0, it is clear that f1 (q) = 1 = T1e (q). For any positive integer n, we will show that f2n+1 (q) satisfies the recurrence (5.1). (k) For a nonnegative integer k ≤ n, let A2n+1 be the set of permutations π on [2n + 1] such that π1 < π2 > π3 < · · · < π2k > π2k+1 , π2k+2 > π2k+4 > · · · > π2n > π2n+1 > π2n−1 > · · · > π2k+3 .
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
From Lemma 1.1 and the definition of f2n+1 , we see that X 2n + 1 (n−k)2 inv(π)+des(π0 π2 π4 ···π2k ) q = q f2k+1 (q). 2k + 1
19
(5.2)
(k) π∈A2n+1
(k)
For k < n, we decompose A2n+1 into disjoint subsets as follows: (k)
A2n+1 ={π | π2k+1 < π2k+2 } ∪ {π | π2k+1 > π2k+2 } ={π | π2k+1 < π2k+2 < π2k } ∪ {π | π2k+1 < π2k < π2k+2 } ∪ {π | π2k+1 > π2k+2 } ={π | π2k+1 < π2k+3 < π2k+2 < π2k } ∪ {π | π2k+3 < π2k+1 < π2k+2 < π2k } ∪ {π | π2k+1 < π2k < π2k+2 } ∪ {π | π2k+1 > π2k+2 } (k)
(k)
(k)
(k)
=:B2n+1 ∪ C2n+1 ∪ D2n+1 ∪ E2n+1 .
(5.3)
(k)
Note that B2n+1 is the set of alternating permutations π on [2n + 1] such that π1 < π2 > π3 < · · · < π2k > π2k+1 , π2k > π2k+2 > · · · > π2n > π2n+1 > π2n−1 > · · · > π2k+3 > π2k+1 . (n)
(k)
We define B2n+1 = A2n+1 . For a permutation π ∈ B2n+1 with 1 ≤ k ≤ n, if π2k−2 < π2k , then π2k−1 < π2k−2 < π2k , (k−1)
which shows that such π satisfies the conditions of D2n+1 . Thus (k)
B2n+1 ={π | π2k−2 < π2k } ∪ {π | π2k−2 > π2k } (k−1)
=D2n+1 ∪ {π | π2k−2 > π2k }. (k)
We now compute the generating function for permutations π ∈ B2n+1 X q inv(π)+des(π0 π2 π4 ···π2k ) . (k)
π∈B2n+1 (k)
Let π ∈ B2n+1 . If π2k−2 < π2k , then inv(π) + des(π0 π2 π4 · · · π2k ) = inv(π) + des(π0 π2 π4 · · · π2k−2 ). However, if π2k−2 > π2k , then inv(π) + des(π0 π2 π4 · · · π2k ) = inv(π) + des(π0 π2 π4 · · · π2k−2 ) + 1. In this case, we look for a permutation σ such that inv(π) + des(π0 π2 π4 · · · π2k ) = inv(σ) + des(σ0 σ2 σ4 · · · σ2k−2 ). Let m be defined by π2k+m = min{π2k+j | π2k+j > π2k−1 , 0 ≤ j ≤ 2n − 2k + 1}.
20
TIM HUBER AND AE JA YEE
There exists such an m since π is an alternating permutation, namely π2k−1 < π2k . So, π2k+m ≤ π2k < π2k−2 . We switch π2k+m with π2k−1 and denote the resulting partition by π ¯ . Switching π2k−1 with π2k+m results in increasing of an inversion, namely inv(π) + 1 = inv(¯ π ). Moreover, since π2i = π ¯2i for i < k, des(π0 π2 π4 · · · π2k−2 ) = des(¯ π0 π ¯2 π ¯4 · · · π ¯2k−2 ). Thus inv(π) + des(π2 π4 · · · π2k ) = inv(π) + des(π0 π2 π4 · · · π2k−2 ) + 1 = inv(¯ π ) + des(¯ π0 π ¯2 π ¯4 · · · π ¯2k−2 ). If π2k−1 were switched with π2k+m for m > 0, then π2k+j < π2k . Hence, from the definition of π ¯ , we see that π ¯1 < π ¯2 > π ¯3 < · · · < π ¯2k−2 > π ¯2k−1 , π ¯2k > π ¯2k+2 > · · · > π ¯2n > π ¯2n+1 > π ¯2n−1 > · · · > π ¯2k+1 , π ¯2k−2 > π ¯2k > π ¯2k−1 > π ¯2k+1 , (k−1)
which shows, from (5.3), that π ¯ ∈ C2n+1 . If π2k−1 and π2k were switched, namely m = 0, then π2k−1 > πi for i > 2k. Hence, from the definition of π ¯ , we see that π ¯1 < π ¯2 > π ¯3 < · · · < π ¯2k−2 > π ¯2k−1 , π ¯2k > π ¯2k+2 > · · · > π ¯2n > π ¯2n+1 > π ¯2n−1 > · · · > π ¯2k+1 , π ¯2k−2 > π ¯2k−1 > π ¯2k > π ¯2k+1 . (k−1)
(5.4)
(k−1)
Note that, since E2n+1 ⊆ A2n+1 , we see that (k−1)
E2n+1 = {π | π2k−2 > π2k−1 > π2k > π2k+1 }. (k−1)
Therefore, (5.4) implies π ¯ ∈ E2n+1 . Hence, by (5.3), for any k, 1 ≤ k ≤ n (k)
(k)
{π | π ∈ B2n+1 , π2k−2 < π2k } ∪ {¯ π | π ∈ B2n+1 , π2k−2 > π2k } (k−1)
(k−1)
(k−1)
= C2n+1 ∪ D2n+1 ∪ E2n+1 (k−1)
(k−1)
= A2n+1 \ B2n+1 .
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
Therefore, X
21
q inv(π)+des(π0 π2 π4 ···π2k )
(k)
π∈B2n+1
X
=
(k) π∈B2n+1 π2k−2 π2k
X
q inv(π)+des(π0 π2 π4 ···π2k−2 ) +
(k) π∈B2n+1 π2k−2 π2k
q inv(π)+des(π0 π2 π4 ···π2k−2 ) −
(k−1) π∈A2n+1
X
q inv(π)+des(π0 π2 π4 ···π2k−2 ) .
(5.5)
(k−1) π∈B2n+1
By iterating (5.5), we deduce X q inv(π)+des(πe ) (n)
π∈B2n+1
=
n−1 X
X X inv(π)+des(π0 π2 π4 ···π2k ) n (−1)n−k−1 q q inv(π) , + (−1)
k=0
(k)
(0)
π∈A2n+1
π∈B2n+1
which is equivalent to f2n+1 (q) =
n−1 X 2n + 1 k=0
2k + 1
2
2
(−1)n−k−1 q (n−k) f2k+1 (q) + (−1)n q n .
Theorem 5.2. Define 2 ∞ X (−1)n q n +n x2n+1
∞ ˆe X T2n+1 (q)x2n+1 n=0
(q; q)2n+1
=
n=0 ∞ X n=0
(q; q)2n+1 2
(−1)n q n −n x2n (q; q)2n
.
Then, for n ≥ 1, e e (q) = q −n−1 T2n+1 (q). Tˆ2n+1
Proof. Note that 2 e Tˆ2n+1 (q) = (−1)n q n +n +
n−1 X 2n + 1 k=0
2k + 1
(−1)n−k−1 q (n−k)
2 −(n−k)
e Tˆ2k+1 (q).
(5.6)
22
TIM HUBER AND AE JA YEE
To prove Theorem 5.2, multiply both sides of (5.6) by q n+1 to obtain, for n ≥ 1, 2 2n + 1 n+1 ˆ e n (n+1)2 q T2n+1 (q) = (−1) q + (−1)n−1 q n +1 (5.7) 1 n−1 X 2 2n + 1 e (q). + (−1)n−k−1 q (n−k) +k+1 Tˆ2k+1 2k + 1 k=1
Note that n (n+1)2
(−1) q
2 2 2n + 1 + (−1)n−1 q n +1 = (−1)n−1 q n +1 (1 + q + · · · + q 2n−1 ) 1 2 2n + 1 n n2 = (−1) q + (−1)n−1 q n . 1
(5.8)
e (q) is identical to the Inserting (5.8) into (5.7), we see that the recursion (5.1) for T2n+1 n+1 ˆ e recursion for q T2n+1 (q) in (5.7). o In [18], Ismail and Zhang conjectured that the polynomials T2n+1 (q) are symmetric about the middle coefficient(s). We prove their conjecture in the following theorem. o e Theorem 5.3. The polynomials T2n+1 (q), T2n+1 (q), and T2n+1 (q) are symmetric about the middle coefficient(s).
Proof. For each alternating permutation π = π1 π2 . . . π2n+1 , the permutation π = π2n+1 π2n . . . π1 is also an alternating permutation. Recall that des(πe ) = π0 π2 · · · π2n and des(π e ) = π0 π2n · · · π2 . From the definition of π, it follows that n(2n + 1) = inv(π) + inv(π), 2n(n + 1) = inv(π) + des(πo ) + inv(π) + des(π o ), 2n(n + 1) + 1 = inv(π) + des(πe ) + inv(π) + des(π e ). Therefore, the inversion map π → π is a bijection between {π | inv(π) = k}
and
{π | inv(π) = n(2n + 1) − k},
{π | inv(π) + des(πo ) = k}
and
{π | inv(π) + des(πo ) = 2n(n + 1) − k},
{π | inv(π) + des(πe ) = k}
and
{π | inv(π) + des(πe ) = 2n(n + 1) + 1 − k}.
If k = n(2n + 1)/2, then {π | inv(π) = k} = {π | inv(π) = n(2n + 1) − k}. Otherwise, the two sets are disjoint. It follows that, if n is even, the coefficients of T2n+1 are symmetric about the term |{π | inv(π) = n(2n + 1)/2}| · q n(2n+1)/2 . If n is odd, the coefficients are symmetric about the terms corresponding to q bn(2n+1)/2c o e and q bn(2n+1)/2c+1 . The coefficients of T2n+1 (q) and T2n+1 (q) can similarly be seen to be symmetric about the middle term(s). Definition. A polynomial p : C → C of degree n is said to be reciprocal if 1 n p(z) = ±z p . z
(5.9)
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
23
The following corollary follows from Theorems 2.3, 4.2, 5.2, and 5.3. e o des (q) are reciprocal. More (q), and Tˆ2n+1 (q), Tˆ2n+1 Corollary 5.4. The polynomials Tˆ2n+1 des o e precisely, if f (q) ∈ {Tˆ2n+1 (q), Tˆ2n+1 (q), Tˆ2n+1 (q) | n ≥ 0}, and f has degree n, then
f (1/q) = f (q)/q n . 6. Higher Order q-Euler Numbers The tangent numbers of order k are defined by the Taylor series coefficients in the expansion of tank z about z = 0. Since d tan2 z = dz we see that for n ≥ 1, the numbers d2n 2 tan z = dz 2n z=0
d2 tan z, dz 2 d2n+1 tan z dz 2n+1 z=0
(6.1)
each enumerate the alternating permutations on [2n + 1]. Equivalently, the first and second order tangent numbers are identical. The q-extensions of second order q-tangent numbers, in contrast, generate polynomials distinct from those of first order. In the following theorem, we offer a combinatorial interpretation for the second order q-tangent numbers arising in the previous sections. Before we state the theorem, we define the permutation statistics α, β, and γ on A2n+1 by α(π) = inv(π) + max(π) − 2n − 1, 3 + sign(πmax(π)−1 − πmax(π)+1 ) , 2 γ(π) = inv(π) + des(πe ) + max(π) − 2n − 2,
β(π) = inv(π) + des(πo ) + max(π) − 2n −
where max(π) denotes the index of 2n + 1 in π. Throughout the section, we denote πij = πi πi+1 · · · πj for a given permutation π. o e Theorem 6.1. Let T2n+1 (q), T2n+1 (q), T2n+1 (q) denote the q-analogues of the tangent (2) (2) (2) numbers defined by Theorems 2.1, 4.1, and 5.1, respectively. Define T2n (q), T o 2n (q), T e 2n (q) by !2 ∞ (2) X T2n+1 (q)z 2n+1 X T2n (q)z 2n = , (q; q)2n (q; q)2n+1 n=0 n=0 !2 ∞ o 2n+1 o (2) 2n X X T (q)z T 2n (q)z 2n+1 = , (q; q)2n (q; q)2n+1 n=0 n=0 !2 ∞ (2) X X T e (q)z 2n+1 T e 2n (q)z 2n 2n = . (q; q) (q; q) 2n 2n+1 n=0 n=0
24
TIM HUBER AND AE JA YEE
Then (2)
T2n (q) =
X
q α(π) ,
X
(2)
T o 2n (q) =
π∈A2n+1
q β(π) ,
(2)
T e 2n (q) =
π∈A2n+1
X
q γ(π) .
π∈A2n+1
Proof. For a permutation π in A2n+1 , let π2k = 2n + 1 for some k, 1 ≤ k ≤ n. Then inv(π) =|{(i, j) | i < 2k < j and πi > πj }| + |{(i, j) | i < j < 2k and πi > πj }| + |{(i, j) | 2k < i < j and πi > πj }| + |{(2k, j) | 2k < j}| 2n+1 =|{(i, j) | i < 2k < j and πi > πj }| + inv(π12k−1 ) + inv(π2k+1 ) + 2(n − k) + 1.
Thus, 2n+1 α(π) = |{(i, j) | i < 2k < j and πi > πj }| + inv(π12k−1 ) + inv(π2k+1 ). (2)
We now show that T2n (q) is the generating function for permutations in A2n+1 with (2) weight α. The polynomials T2n (q) satisfy n X 2n (2) T (q)T2(n−k)+1 (q). (6.2) T2n (q) = 2k − 1 q 2k−1 k=1
2n in the summand on the right hand By Lemma 1.1, the q-binomial coefficient 2k − 1 q
2n+1 side of (6.2) counts the inversions between the two sub-permutations π12k−1 and π2k+1 , namely
|{(i, j) | i < 2k < j and πi > πj }|; 2n+1 T2k−1 (q) and T2(n−k)+1 (q) count the inversions of π12k−1 and π2k+1 , respectively. There(2) fore, T2n (q) is the generating function for permutations π in A2n+1 with weight α(π). (2) (2) The arithmetic interpretations for the polynomials T o 2n (q) and T e 2n (q) are similarly derived. We omit the details.
In the next theorem, we present a corresponding interpretation for the second order q-secant numbers studied in Section 3. The proof for each interpretation is similar to that of Theorem 6.1. o Theorem 6.2. Let S2n (q) and S2n (q) denote the q-analogues of the secant numbers (2) (2) defined by Theorems 3.1 and 3.2, respectively. Define S2n (q) and S o 2n (q) by !2 ∞ (2) 2n 2n X X S2n (q)z S2n (q)z = , (q; q)2n (q; q)2n n=0 n=0 !2 ∞ (2) X X S o (q)z 2n S o 2n (q)z 2n 2n = . (q; q) (q; q) 2n 2n n=0 n=0
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
25
Then (2)
S2n (q) =
X
q δ(π) ,
π∈A2n+1 (2)
S o 2n (q) =
X
q υ(π) =
π∈A2n+1
X
q ω(π) ,
π∈A2n+1
where 2n + 1 − min(π) δ(π) = inv(π) − 2 · + − min(π) + 1, 2 min(π) + 3 2n+1 υ(π) = δ(π) + des(πo ) − 2 · des((πmin(π)+1 )o ) + n − 2 min(π) + 2 + sign(πmax(π)−1 − πmax(π)+1 ) 2n+1 ω(π) = δ(π) + des(πe ) − 2 · des((πmin(π)+1 )e ) + n − 2 + sign(π2n+1 − 1) + sign(π1 − 1) − 2. 2n+1 inv(πmin(π)+1 )
Proof. For a permutation π in A2n+1 , let π2k+1 = 1 for some k, 0 ≤ k ≤ n. Then inv(π) =|{(i, j) | i < 2k + 1 < j and πi > πj }| + |{(i, j) | i < j < 2k + 1 and πi > πj }| + |{(i, j) | 2k + 1 < i < j and πi > πj }| + |{(i, 2k + 1) | i < 2k + 1}| 2n+1 =|{(i, j) | i < 2k + 1 < j and πi > πj }| + inv(π12k ) + inv(π2k+2 ) + 2k.
Thus, δ(π) = |{(i, j) | i < 2k + 1 < j and πi > πj }| +
inv(π12k )
−
2n+1 inv(π2k+2 )
2(n − k) + 2
2n+1 ), = |{(i, j) | i < 2k + 1 < j and πi > πj }| + inv(π12k ) + inv(π2k+2
(6.3)
2n+1 where π2k+2 = π2n+1 π2n · · · π2k+3 π2k+2 . (2) We now show that S2n (q) is the generating function of permutations in A2n+1 with (2) weight δ. The polynomials S2n (q) satisfy n X 2n (2) S2n (q) = S (q)S2(n−k) (q). (6.4) 2k q 2k k=0 2n By Lemma 1.1, the q-binomial coefficient in the summand on the right hand 2k q 2n+1 side of (6.4) counts the inversions between the two sub-permutations π12k and π2k+2 , namely
|{(i, j) | i < 2k + 1 < j and πi > πj }|; 2n+1 S2k (q) and S2(n−k) (q) count the inversions of π12k and π2k+1 , respectively, since π12k ∈ A2k (2) 2n+1 and π2k+1 ∈ A2(n−k) . Therefore, S2n (q) is the generating function for permutations π in A2n+1 with weight δ(π).
26
TIM HUBER AND AE JA YEE (2)
To obtain the claimed arithmetic interpretation for the polynomials S o 2n (q), note that, with the convention that des(∅) = 0, where ∅ denotes the empty permutation, des(π2n+1 π2n−1 . . . πmin(π)+2 ) + sign(π2n+1 − 1) − 1 =
2n − min(π) − 1 − des(πmin(π)+2 πmin(π)+4 · · · π2n+1 ). 2
Hence, by (6.3), we find that 2n+1 υ(π) =δ(π) + des(πo ) − 2 · des((πmin(π)+2 )o ) + n −
min(π) + 3 2
2n − min(π) − 1 2 = δ(π) + des(π1 π3 · · · πmin(π)−2 ) + des(π2n+1 π2n−1 · · · πmin(π)+2 ) + sign(π2n+1 − 1) − 1
=δ(π) + des(π1 π3 · · · πmin(π)−2 ) − des(πmin(π)+2 πmin(π)+3 · · · π2n+1 ) + min(π)−1
=|{(i, j) | i < min(π) < j and πi > πj }| + inv(π1
min(π)−1
) + des((π1
)o )
2n+1 2n+1 + inv(πmin(π)+1 ) + des((πmin(π)+1 )o ) + sign(π2n+1 − 1) − 1, 2n+1 = π2n+1 π2n · · · π2k+3 πmin(π)+1 . Therefore, since S0o (q) = q −1 , by Lemma where πmin(π)+1 (2)
1.1, Theorem 3.2, and the definition of S o 2n (q), we see that X (2) S o 2n (q) = q υ(π) . π∈A2n+1 e (q) defined by (3.7) and using the fact By squaring the generating function for S2n o e that S2n (q) = S2n (q), we readily observe that the weight ω(π) corresponds to the same enumeration for A2n+1 as υ(π). We omit the details.
7. Concluding Remarks Define the Bell polynomials Bn,k (x1 , x2 , · · · , xn−k+1 ) via the generating function [11] ! ∞ ∞ X n X X xm z m uk z n exp u =1+ Bn,k (x1 , x2 , xn−k+1 ) . m! n! m=1 n=1 k=1 Then, by Fa´a Di Bruno’s formula [11, p. 137] and Theorem 3.1, (q; q)2n S2n (q) = (2n)!
1+
2n X
! (−1)v v!B2n,v (α1 , . . . , α2n−v+1 ) ,
(7.1)
v=1
where ( (−1)k/2 k!/(q; q)k , if k is even, αk = 0, if k is odd. From (2.1) and (3.1), we obtain T2n+1 (q) =
n X 2n + 1 j=0
2j
(−1)n−j S2j (q).
(7.2)
Closed formulas and relations for the other q-Euler numbers can be similarly derived.
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
27
o (q) are the As mentioned in the introduction, the generalized tangent numbers T2n+1 −2 polynomials arising in the coefficient of ψ (q) in (q; q)2n+1 y2n+1 of (1.2). Constant (2) multiples of the second order extensions T o 2n+1 (q) appear in (1.2) as the coefficient of ψ −4 (q) in the corresponding expansion of (q; q)2n y2n for n ≥ 2. Ismail and Zhang [18, Theorem 4.1] prove that each yj can be expressed as a polynomial in theta functions over the field of rational functions in q. The authors of [18] suggest that polynomials appearing in the numerators of these expansions, denoted by Dr,s,t (q), have interesting o (q) addresses the combinatorics of the combinatorial properties. Our study of T2n+1 polynomials Dr,0,1 (q) and Dr,0,2 (q). Recursion formulas for yj appearing in [15, 16] show that the polynomials Dr,s,t (q) arise as linear combinations of finite products Q mj o j (T2nj +1 (q)) . Therefore, the current paper shows that the polynomials Dr,s,t (q) have integer coefficients1. Ismail and Zhang conjecture [18, Conjecture 4.3] that the polynomials Dr,s,t (q) are symmetric about the middle coefficient(s). The symmetry o of the polynomials T2n (q) established in the current paper shows that, to prove the +1 general Ismail-Zhang conjecture, it suffices the form of the representation Q oto determine mj (q)) . From such formulas, arithmetic for Dr,s,t (q) in terms of the products j (T2n j +1 interpretations for the polynomials Dr,s,t (q) for all r, s, t ∈ N may be deduced.
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]
D. Andr´e. D´eveloppement de sec x et de tan x. C. R. Acad. Sci. Paris, 88:965–967, 1879. D. Andr´e. M´emoire sur les permutations altern´ees. J. Math. Pur. Appl., 6:167–184, 1881. G. E. Andrews and B. C. Berndt. Ramanujan’s lost notebook. Part I. Springer, New York, 2005. G. E. Andrews and D. Foata. Congruences for the q-secant numbers. European J. Combin., 1(4):283–287, 1980. G. E. Andrews and I. Gessel. Divisibility properties of the q-tangent numbers. Proc. Amer. Math. Soc., 68(3):380–384, 1978. L. Carlitz. Enumeration of up-down permutations by number of rises. Pacific J. Math., 45:49–58, 1973. L. Carlitz. Enumeration of up-down sequences. Discrete Math., 4:273–286, 1973. L. Carlitz. Permutations, sequences and special functions. SIAM Rev., 17:298–322, 1975. L. Carlitz and R. Scoville. Enumeration of up-down permutations by upper records. Monatsh. Math., 79:3–12, 1975. R. Chapman and L. K. Williams. A conjecture of Stanley on alternating permutations. Electron. J. Combin., 14(1):Note 16, 7 pp. (electronic), 2007. L. Comtet. Advanced combinatorics. D. Reidel ublishing Co., Dordrecht, 1974. The art of finite and infinite expansions. Revised and enlarged edition. S. Dulucq and R. Simion. Combinatorial statistics on alternating permutations. J. Algebraic Combin., 8(2):169–191, 1998. D. Foata. Further divisibility properties of the q-tangent numbers. Proc. Amer. Math. Soc., 81(1):143–148, 1981. M. Fulmek. A continued fraction expansion for a q-tangent function. S´em. Lothar. Combin., 45:Art. B45b, 5 pp. (electronic), 2000/01. T. Huber. Zeros of generalized Rogers-Ramanujan series and topics from Ramanujan’s theory of elliptic functions. Ph.D. Thesis, University of Illinois at Urbana-Champaign, 2007. T. Huber. Zeros of generalized Rogers-Ramanujan series: asymptotic and combinatorial properties. Preprint, 2008.
1The
actual polynomials Dr,s,t (q) are not correctly given in [18]. See [16, Remark 4.2].
28
TIM HUBER AND AE JA YEE
[17] T. Huber. Hadamard products for generalized Rogers-Ramanujan series. J. Approx. Th., To appear. [18] M. E. H. Ismail and C. Zhang. Zeros of entire functions and a problem of Ramanujan. Adv. Math., 209(1):363–380, 2007. [19] F. H. Jackson. A basic sine and cosine with symbolic solutions of certain differential equations. Proc. Edinburgh Math. Soc., 22:28–39, 1904. [20] C. Poupard. Using the Entringer numbers to count the alternating permutations according a new parameter. Ann. I.S.U.P., 44(2-3):73–86 (2001), 2000. [21] H. Prodinger. Combinatorics of geometrically distributed random variables: new q-tangent and q-secant numbers. Int. J. Math. Math. Sci., 24(12):825–838, 2000. [22] H. Prodinger and L. L. Cristea. q-enumeration of up-down words by number of rises. Preprint, 2008. [23] S. Ramanujan. The lost notebook and other unpublished papers. Springer-Verlag, Berlin, 1988. With an introduction by George E. Andrews. [24] D. Rawlings. Generalized Worpitzky identities with applications to permutation enumeration. European J. Combin., 2(1):67–78, 1981. [25] R. P. Stanley. Binomial posets, M¨ obius inversion, and permutation enumeration. J. Combinatorial Theory Ser. A, 20(3):336–356, 1976. [26] R. P. Stanley. Enumerative combinatorics. Vol. 1, volume 49 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1997. With a foreword by Gian-Carlo Rota, Corrected reprint of the 1986 original. [27] R. P. Stanley. Alternating permutations and symmetric functions. J. Combin. Theory Ser. A, 114(3):436–460, 2007. [28] V. Strehl. Alternating permutations and modified Ghandi polynomials. Discrete Math., 28(1):89– 100, 1979.
Abstract. New enumerating functions for the Euler numbers are considered. Several of the relevant generating functions appear in connection to entries in Ramanujan’s Lost Notebook. The results presented here are, in part, a response to a conjecture made by M. E. H. Ismail and C. Zhang about the symmetry of polynomials in Ramanujan’s expansion for a generalization of the Rogers-Ramanujan series. Related generating functions appear in the work of H. Prodinger and L. L. Cristea in their study of geometrically distributed random variables. An elementary combinatorial interpretation for each of these enumerating functions is given in terms of a related set of statistics.
1. Introduction The Euler numbers En are the integers defined by ∞ X En xn = sec x + tan x. n! n=0
(1.1)
In 1879, D. Andr´e [1, 2] gave a combinatorial interpretation for the Euler numbers En . These numbers count the number of permutations π = π1 π2 · · · πn of elements in the set [n] := {1, 2, . . . , n} such that the sign of πi − πi+1 equals (−1)i , 1 ≤ i < n. Such permutations are called alternating or up-down permutations. Alternating permutations have rich combinatorial structure and have been studied extensively over the last century [6, 7, 8, 9, 10, 12, 20, 28]. Particular emphasis has been placed upon the enumeration of alternating permutations by various weights and conditions. In this paper, we undertake a combinatorial analysis of several new q-analogues of the Euler numbers. The resulting expressions provide new enumerations for alternating permutations. The associated generating functions are quotients of basic hypergeometric series and arise in several contexts related to the work of S. Ramanujan [15, 16, 17, 27]. In particular, the generating functions from Section 4 appear in the expansions of Ramanujan’s Hadamard product of the generalized Rogers-Ramanujan series from page 57 of his Lost Notebook [23], [3, Chapter 13]: ! 2 ∞ ∞ X Y qn zn zq 2n−1 P∞ jn = 1+ , (1.2) (q; q) 1 − n j=1 q yj n=0 n=1 where 1 y1 = , (1 − q)ψ 2 (q)
y2 = 0,
P∞ (2n−1)q2n−1 q + q3 n=1 1−q 2n−1 y3 = − , (q; q)3 ψ 2 (q) (1 − q)3 ψ 6 (q)
y4 = y 1 y 3 . (1.3)
1
2
TIM HUBER AND AE JA YEE
The functions ψ(q) and (α; q)n appearing in (1.2) and (1.3) are defined by ψ(q) :=
∞ X
q
n(n+1)/2
n=0
,
(α; q)n =
n−1 Y
1 − αq j ,
|q| < 1.
j=0
In [18], Ismail and Zhang observed that the polynomials appearing in the expansion −2 (1.2), as the coefficients of (q; q)−1 j ψ (q) in yj , are symmetric about the middle coefficient(s). They conjectured that this is the case for each of the polynomials. We prove Ismail and Zhang’s conjecture in Sections 4, 5, 6, and 7 by unraveling the combinatorial significance of these polynomials. The series appearing in this paper arise in an entirely different setting in the work of Prodinger and Cristea [21, 22]. These authors employ generating functions to determine the probability that a random word over the infinite alphabet {1, 2, 3, . . . , } satisfies certain inequality conditions. They assume that, within a word, each letter j occurs with (geometric) probability pq j−1 , independently, for 0 < q < 1 and p = 1 − q. In Sections 2–3, we derive direct combinatorial interpretations for certain generating functions from [21, 22]. Quotients of the series considered in the present paper also have beautiful continued fraction representations [14, 21]. For positive integers r1 , r2 , s1 , s2 and nonnegative integers A, B, C, D with s1 − r1 = 2A − 1 and s2 − r2 = 2C − 1, consider the following q-analogue of tan x: 2 ∞ X (−1)n q An +Bn x2n+1 0, 0, . . . , 0 2 A+B 2 q ,q x ∞ r φs (q; q)2n+1 X f2n+1 (q) x2n+1 x 1 1 0, 0, · · · , q 3 n=0 , = ∞ = X (−1)n q Cn2 +Dn x2n (q; q) 1 − q 0, 0, . . . , 0 2n+1 2 C+D 2 n=0 q ,q x r2 φs2 0, 0, · · · , q (q; q) 2n n=0 (1.4) where X ∞ (a1 ; q)n (a2 ; q)n · · · (ar ; q)n a1 , a2 , . . . , ar q, z = [(−1)n q n(n−1)/2 ]1+s−r z n . r φs b1 , b2 , · · · , bs (q; q)n (b1 ; q)n (b2 ; q)n · · · (bs ; q)n n=0
When (A, B, C, D) = (0, 0, 0, 0), f2n+1 (q) is the q-tangent number T2n+1 (q) of F. H. Jacko −2 son [19]. In [16], Huber proves that the coefficients T2n+1 (q) of (q; q)−1 2n+1 ψ (q) in y2n+1 in (1.2) are f2n+1 (q) for (A, B, C, D) = (1, 1, 1, 0). In this paper, we discuss q-tangent αβ numbers corresponding to (A, B, C, D) given in the following table. Let τ2n+1 represent the probability that a word from {1, 2, 3, · · · } of length 2n + 1 defined in the preceeding paragraph satisfies the inequality conditions αβ. f2n+1 (A, B) (C, D) Probability ≤> ≥< T2n+1 (0, 0) (0, 0) τ2n+1 , τ2n+1 o T2n+1 (1, 1) (1, 0) τ2n+1 >< e T2n+1 (1, 0) (1, 0) τ2n+1 The column on the right contains the numbers considered by Prodinger for which the series (1.4) is an associated generating function [21, Theorem 2.2].
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
3
For each value of A, B, C and D, the quotient (1.4) induces a corresponding recursion relation for the function f2n+1 (q). From these formulas, we obtain the following related polynomials. For a polynomial p(q), let pˆ(q) denote the dual of p(q) (see [21, Remark 3.3]). Several of the dual polynomials occur in connection with probabilities from [21]. (A, B) (0, 1) (2, 1) (2, 0) (1, 0) (1, 1)
(C, D) (0, 1) (2, −1) (2, −2) (1, −1) (1, −1)
f2n+1 Relevant Relations Probability n des des T2n+1 T2n+1 (q) = q T2n+1 (q) Tˆ2n+1 Tˆ2n+1 (q) = T2n+1 (q) des des des Tˆ2n+1 Tˆ2n+1 (q) = q −2n T2n+1 (q) ≤≥ o o −n o ˆ ˆ T2n+1 T2n+1 (q) = q T2n+1 (q) τ2n+1 ≥≤ e e e Tˆ2n+1 Tˆ2n+1 (q) = q −n−1 T2n+1 (q) τ2n+1
In Section 2, a well-known arithmetic interpretation of the classical q-tangent numbers T2n+1 (q) is discussed. We provide an elementary proof of this interpretation that demonstrates fundamental ideas used throughout the paper. The new q-analogue des (q) is also discussed in the same section. In Section 3, we deduce combinatorial T2n+1 interpretations for new q-analogues of the secant numbers appearing in [21, 22] defined, for s2 − r2 = 2C − 1, by ∞ X g2n (q) x2n n=0
(q; q)2n
=
r2 φs2
1 . 0, 0, . . . , 0 2 C+D 2 q ,q x 0, 0, · · · , q
(C, D) g2n Relevant Relations Probability ≤> (0, 0) S2n σ2n e o , S2n (1, 0) qS2n σ2n ≤≥ o o o (1, −1) Sˆ2n Sˆ2n (q) = q 1−n S2n (q) σ2n πj . We denote by inv(π) the number of inversions of the permutation π. The descent set D(π) is defined by {i | πi > πi+1 }, and des(π) denotes the size of D(π). Define the q-binomial coefficient by (q; q)n , if 0 ≤ n ≤ k, n = (q; q)k (q; q)n−k k 0, otherwise. The following lemma is one of several combinatorial interpretations for the q-binomial coefficient. We will refer to this lemma often in the remainder of the paper. For a proof of the lemma, see [26, p. 132]. Lemma 1.1. X
q
inv(π)
π
n = , k
where the sum is over all permutations π with D(π) ⊂ {k}. (k)
A more instructive view of Lemma 1.1 follows by defining Pn , for a given n and k ≤ n, to be the set of all permutations π on [n] such that π1 < π2 < π3 < · · · < πk−1 < πk , Then it follows from Lemma 1.1 that X (k) π∈Pn
q
πk+1 < πk+2 < · · · < πn−1 < πn .
inv(π)
n = . k
We denote by An the set of alternating permutations π1 π2 · · · πn on the set [n] with π1 < π2 > π3 < · · · . For instance, A1 = {1}, A2 = {12}, A3 = {132, 231}, A4 = {1324, 1423, 2314, 2413, 3412} A5 = {13254, 23154, 14253, 24153, 14352, 34152, 24351, 34251, 15243, 25143, 15342, 35142, 25341, 35241, 45132, 45231}. 2. The classical q-tangent numbers Jackson’s q-analogues of the sine and cosine functions [19] are ∞ X n=0
(−1)n x2n+1 (q; q)2n+1 /(1 − q)2n+1
and
∞ X n=0
(−1)n x2n . (q; q)2n /(1 − q)2n
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
5
By considering quotients of these functions, we arrive at a q-analogue of the tangent numbers, T2n+1 (q), defined by ∞ ∞ ∞ X T2n+1 (q)x2n+1 X (−1)n x2n+1 X (−1)n x2n = / . (2.1) (q; q) (q; q) (q; q) 2n+1 2n+1 2n n=0 n=0 n=0 If we replace x by x(1−q) in (2.1) and let q → 1− , the corresponding identity reduces to the relation obtained by equating odd parts on each side of equation (1.1). Multiplying both sides of (2.1) by the denominator on the right side and equating coefficients of x, we obtain n−1 X 2n + 1 n T2n+1 (q) = (−1) + (−1)n−k−1 T2k+1 (q). (2.2) 2k + 1 k=0
The following combinatorial interpretation of the polynomials T2n+1 (q) is well known [5, 13, 25]. We include a proof based upon the recursion (2.2) as an aid to the reader, since later results in the paper have proofs that are similar in nature. Theorem 2.1. For a nonnegative integer n, we have X T2n+1 (q) = q inv(π) . π∈A2n+1
Proof. Let X
f2n+1 (q) =
q inv(π) .
π∈A2n+1
For n = 0, it is clear that f1 (q) = 1 = T1 (q). For any positive integer n, we will prove that f2n+1 (q) satisfies the recurrence (2.2). (k) For a positive integer k ≤ n, let A2n+1 be the set of permutations π on [2n + 1] such that π1 < π2 < π3 < · · · < π2k , π2k+1 < π2k+2 > π2k+3 < · · · < π2n > π2n+1 . From Lemma 1.1 and the definition of f2n+1 (q), we see that X 2n + 1 inv(π) q = f2(n−k)+1 (q). 2k
(2.3)
(k) π∈A2n+1
(k)
For a positive integer k ≤ n, we denote by B2n+1 the set of permutations π on [2n + 1] such that π1 < π2 < π3 < · · · < π2k > π2k+1 < π2k+2 > π2k+3 < · · · < π2n > π2n+1 . (k)
We now compute the generating function for permutations π ∈ B2n+1 X q inv(π) . (k)
π∈B2n+1
6
TIM HUBER AND AE JA YEE (k)
(k)
From the definitions of A2n+1 and B2n+1 , we see that for any k, 1 ≤ k ≤ n, (k)
(k)
(k+1)
A2n+1 = B2n+1 ∪ B2n+1 , (n+1)
where B2n+1 = {π | π1 < π2 < · · · < π2n < π2n+1 }. Thus X X X q inv(π) = q inv(π) − q inv(π) . (k)
(k)
π∈B2n+1
(2.4)
(k+1)
π∈A2n+1
π∈B2n+1
By iterating (2.4), we deduce X
q
inv(π)
=
n X
(−1)k−1
k=1
(1) π∈B2n+1
X
X
q inv(π) + (−1)n
(k) π∈A2n+1
q inv(π) .
(2.5)
(n+1) π∈B2n+1
(1)
Note that B2n+1 = A2n+1 . Therefore, it follows from (2.3) that (2.5) is equivalent to n X 2n + 1 f2n+1 (q) = (−1)k−1 f2(n−k)+1 (q) + (−1)n , 2k k=1
which completes the proof.
The following theorem gives the generating function for alternating permutations π in A2n+1 by weight inv(π) + des(π). Theorem 2.2. Define ∞ X (−1)n q n x2n+1 ∞ des X (q)x2n+1 T2n+1 n=0
(q; q)2n+1
=
n=0 ∞ X n=0
(q; q)2n+1 (−1)n q n x2n (q; q)2n
.
Then des T2n+1 (q) =
X
q inv(π)+des(π) .
π∈A2n+1
Proof. Recalling the definition of T2n+1 (q) given by (2.1) and noting that des(π) = n for any π ∈ A2n+1 , we have X q inv(π)+des(π) = q n T2n+1 (q). π∈A2n+1 des des By comparing the recurrence relations for T2n+1 (q) and T2n+1 (q), we see that T2n+1 (q) = n q T2n+1 (q). des The dual functions corresponding to T2n+1 (q) and T2n+1 (q) are discussed in the following theorem.
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
7
Theorem 2.3. Define 2 ∞ X (−1)n q 2n +n x2n+1
∞ ˆ X T2n+1 (q)x2n+1
(q; q)2n+1
n=0
=
n=0 ∞ X n=0
(q; q)2n+1
,
2
(−1)n q 2n −n x2n (q; q)2n
and 2 ∞ X (−1)n q 2n x2n+1
∞ ˆ des X T2n+1 (q)x2n+1
(q; q)2n+1
n=0
=
n=0 ∞ X n=0
(q; q)2n+1 2
(−1)n q 2n −2n x2n (q; q)2n
.
Then we have Tˆ2n+1 (q) = T2n+1 (q)
and
des des Tˆ2n+1 (q) = q −2n T2n+1 (q).
Proof. From the definition of Tˆ2n+1 (q), it is clear that Tˆ1 (q) = 1 = T1 (q). Note that n 2n2 +n
Tˆ2n+1 (q) = (−1) q
+
n−1 X 2n + 1 k=0
2k + 1
(−1)n−k−1 q 2(n−k)
2 −(n−k)
Tˆ2k+1 (q).
For any positive integer n, we will show that T2n+1 (q) satisfies the same recurrence as Tˆ2n+1 (q). (k) For a nonnegative integer k ≤ n, let A2n+1 be the set of permutations π on [2n + 1] such that π1 < π2 > π3 < · · · < π2k > π2k+1 , π2k+2 > π2k+3 > · · · > π2n > π2n+1 Since there are 2(n − k)2 − (n − k) inversions in a permutation π2k+2 π2k+3 · · · π2n+1 such that π2k+2 > π2k+3 > · · · > π2n > π2n+1 , Lemma 1.1 and the definition of T2n+1 (q) imply X 2n + 1 2(n−k)2 −(n−k) inv(π) q = q T2k+1 (q). (2.6) 2k + 1 (k)
π∈A2n+1 (k)
For a positive integer k ≤ n, let B2n+1 denote the set of permutations π on [2n + 1] satisfying π1 < π2 > π3 < · · · < π2k > π2k+1 > π2k+2 > π2k+3 > · · · > π2n > π2n+1 .
8
TIM HUBER AND AE JA YEE (k)
We now compute the generating function for permutations π ∈ B2n+1 X q inv(π) . (k)
π∈B2n+1 (k)
(k)
From the definitions of A2n+1 and B2n+1 , we see that for any k, 0 ≤ k < n, (k)
(k)
(k+1)
A2n+1 = B2n+1 ∪ B2n+1 , (0)
where B2n+1 = {π | π1 > π2 > · · · > π2n > π2n+1 }. Thus, X X X q inv(π) = q inv(π) − q inv(π) . (k+1)
(k)
π∈B2n+1
(2.7)
(k)
π∈A2n+1
π∈B2n+1
By iterating (2.7), we deduce X
q
inv(π)
=
n−1 X k=0
(n)
π∈B2n+1
X
(−1)n−k−1
X
q inv(π) + (−1)n
q inv(π) ,
(0)
(k)
π∈B2n+1
π∈A2n+1
which is equivalent to T2n+1 (q) =
n−1 X 2n + 1 2k + 1
k=0
(−1)n−k−1 q 2(n−k)
2 −(n−k)
2 +n
T2k+1 (q) + (−1)n q 2n
by (2.6). Therefore, Tˆ2n+1 (q) = T2n+1 (q). des des It follows from the recurrences for Tˆ2n+1 and Tˆ2n+1 that Tˆ2n+1 (q) = q n Tˆ2n+1 (q). −2n des des This identity is equivalent to Tˆ2n+1 (q) = q T2n+1 (q) since Tˆ2n+1 (q) = T2n+1 (q) = q −n T des (q). By equating coefficients of x2n+1 in the generating functions for T2n+1 (q) and Tˆ2n+1 (q) we obtain the following result. Corollary 2.4. For any nonnegative integer n, X n n X 2j 2 +j 2n + 1 2j 2 −j 2n + 1 q = q . 2j + 1 2j j=0
j=0
3. New q-Secant Numbers The classical q-secant numbers enumerate alternating permutations on the set [2n] for n ≥ 1 by the number of inversions. Theorem 3.1. Define ∞ X S2n (q)x2n n=0
(q; q)2n
=
∞ X (−1)n x2n n=0
(q; q)2n
Then S2n (q) =
X π∈A2n
q inv(π) .
!−1 .
(3.1)
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
9
For a proof of Theorem 3.1, see [4, 24, 25]. The following theorem provides a combinatorial interpretation for a new class of secant numbers. Theorem 3.2. Define !−1
∞ o X (q)x2n S2n = (q; q)2n n=0
2 ∞ X (−1)n q n +1 x2n
∞ ˆo X S2n (q)x2n = (q; q) 2n n=0
∞ X (−1)n q n(n−1) x2n
n=0
(q; q)2n
(q; q)2n
n=0
, !−1 ,
Then, for n ≥ 1, X
o (q) = S2n
q inv(π)+des(πo ) ,
π∈A2n o o Sˆ2n (q) = q 1−n S2n (q).
Proof. Let X
g2n (q) =
q inv(π)+des(πo ) .
π∈A2n o We first prove that g2n (q) = S2n (q). From the definitions of inv(π) and des(πo ), it is clear that
g2 (q) = 1 = S2o (q). o (q) satisfy S0o (q) = q −1 and The polynomials S2n
o S2n (q)
=
n−1 X 2n k=0
2k
2
o (−1)n−k−1 q (n−k) S2k (q),
for n ≥ 1.
(3.2)
We define g0 (q) = q −1 . We will show that the polynomials g2n (q) satisfy (3.2) for n > 1. (k)
For a positive integer k ≤ n, let A2n be the set of permutations π on [2n] such that π1 < π2 > π3 < · · · > π2k−1 < π2k , π2k+2 > π2k+4 > · · · > π2n > π2n−1 > π2n−3 > · · · > π2k+1 . From Lemma 1.1 and the definition of g2n , we see that X 2n (n−k)2 inv(π)+des(π1 π3 ···π2k−1 )+(n−k) q = q g2k (q). 2k (k) π∈A2n
(3.3)
10
TIM HUBER AND AE JA YEE (k)
For a positive integer k < n, we decompose A2n into disjoint subsets as follows: (k)
A2n ={π | π2k > π2k+2 } ∪ {π | π2k < π2k+2 } ={π | π2k > π2k+2 > π2k−1 } ∪ {π | π2k > π2k−1 > π2k+2 } ∪ {π | π2k+2 > π2k > π2k−1 } ={π | π2k > π2k+2 > π2k+1 > π2k−1 } ∪ {π | π2k > π2k+2 > π2k−1 > π2k+1 } ∪ {π | π2k > π2k−1 > π2k+2 > π2k+1 } ∪ {π | π2k+2 > π2k > π2k−1 > π2k+1 } ∪ {π | π2k+2 > π2k > π2k+1 > π2k−1 } ∪ {π | π2k+2 > π2k+1 > π2k > π2k−1 } ={π | π2k > π2k+2 > π2k+1 > π2k−1 } ∪ {π | π2k > π2k−1 > π2k+1 } ∪ {π | π2k+2 > π2k > π2k+1 > π2k−1 } ∪ {π | π2k+2 > π2k+1 > π2k > π2k−1 } (k)
(k)
(k)
(k)
=:B2n ∪ C2n ∪ D2n ∪ E2n .
(3.4)
(k)
Note that B2n is the set of alternating permutations π on [2n] such that π1 < π2 > π3 < · · · > π2k−1 < π2k , π2k > π2k+2 > · · · > π2n > π2n−1 > π2n−3 > · · · > π2k+1 > π2k−1 , (k)
(k−1)
(n)
from which it is clear that B2n is a subset of A2n . We define B2n = A2n . For a (k) permutation π ∈ B2n with 1 < k ≤ n, if π2k−3 > π2k−1 , then π2k−2 > π2k−3 > π2k−1 , (k−1)
which shows that such π satisfy the conditions of C2n
. Thus
(k)
B2n ={π | π2k−3 > π2k−1 } ∪ {π | π2k−3 < π2k−1 } (k−1)
=C2n
∪ {π | π2k−3 < π2k−1 }.
(3.5) (k)
We now compute the generating function for permutations π ∈ B2n X q inv(π)+des(π1 π3 ···π2k−1 )+(n−k) . (k)
π∈B2n (k)
Let π ∈ B2n . If π2k−3 > π2k−1 , then inv(π) + des(π1 π3 · · · π2k−1 ) + (n − k) = inv(π) + des(π1 π3 · · · π2k−3 ) + (n − k + 1). However, if π2k−3 < π2k−1 , then inv(π) + des(π1 π3 · · · π2k−1 ) + (n − k) = inv(π) + des(π1 π3 · · · π2k−3 ) + (n − k + 1) − 1. In this case, we look for a permutation σ such that inv(π) + des(π1 π3 · · · π2k−1 ) = inv(σ) + des(σ1 σ3 · · · σ2k−3 ) + 1. Let m be defined by π2k+m = max{π2k+j | π2k+j < π2k−2 , j ≥ −1}. There exists such an m since π is an alternating permutation, so that π2k−1 < π2k−2 . It follows that π2k−1 ≤ π2k+m < π2k−2 . We switch π2k−2 and π2k+m , and denote
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
11
the resulting partition by π ¯ . Switching π2k−2 with π2k+m results in a decrease of the inversion number, namely inv(π) = inv(¯ π ) + 1. Moreover, since π ¯2i+1 = π2i+1 for i < k and π2k−3 < π2k−1 , des(π1 π3 · · · π2k−1 ) = des(¯ π1 π ¯3 · · · π ¯2k−3 ). Thus inv(π) + des(π1 π3 · · · π2k−1 ) = inv(¯ π ) + des(¯ π1 π ¯3 · · · π ¯2k−3 ) + 1. If π2k−2 were switched with π2k+m for m ≥ 0, then π2k−1 < π2k+m and π2k ≥ π2k+m . Hence, from the definition of π ¯ , we see that π ¯1 < π ¯2 > π ¯3 < · · · < π ¯2k−2 , π ¯2k > π ¯2k+2 > · · · > π ¯2n > π ¯2n+1 > π ¯2n−1 > · · · > π ¯2k+1 > π ¯2k−1 , π ¯2k > π ¯2k−2 > π ¯2k−1 > π ¯2k−3 , (k−1)
which shows π ¯ ∈ D2n+1 . If π2k−2 and π2k−1 were switched, namely m = −1, then π2k−2 < πi for i ≥ 2k. Hence, from the definition of π ¯ , we see that π ¯1 < π ¯2 > π ¯3 < · · · < π ¯2k−2 , π ¯2k > π ¯2k+2 > · · · > π ¯2n > π ¯2n+1 > π ¯2n−1 > · · · > π ¯2k+1 > π ¯2k−1 , π ¯2k > π ¯2k−1 > π ¯2k−2 > π ¯2k−3 , (k−1)
which shows π ¯ ∈ E2n+1 . Thus, by (3.4) and (3.5), for any k, 1 < k ≤ n, (k)
(k)
{π | π ∈ B2n , π2k−3 > π2k−1 } ∪ {¯ π | π ∈ B2n , π2k−3 < π2k−1 } (k−1)
= C2n
(k−1)
= A2n
(k−1)
∪ D2n
(k−1)
\ B2n
(k−1)
∪ E2n
.
Therefore, X q inv(π)+des(π1 π3 ···π2k−1 )+(n−k) (k)
π∈B2n
=
X (k) π∈B2n π2k−3 >π2k−1
=
X
X (k−1)
π∈A2n
q inv(π)+des(π1 π3 ···π2k−3 )+(n−k)
(k) π∈B2n π2k−3 π2k−1
=
X
q inv(π)+des(π1 π3 ···π2k−3 )+(n−k+1) +
q inv(π)+des(π1 π3 ···π2k−3 )+(n−k+1) −
q inv(¯π)+des(¯π1 π¯3 ···¯π2k−3 )+(n−k+1)
(k) π∈B2n π2k−3 1, we deduce n−1 X X X inv(π)+des(π1 π3 ···π2k−1 )+(n−k) q inv(π)+des(πo ) = (−1)n−k−1 q k=1
(n)
π∈B2n
(k)
π∈A2n
+ (−1)n−1
X
q inv(π)+n−1 ,
(1)
π∈B2n
which is equivalent to g2n (q) =
n−1 X 2n k=1
=
2k
n−1 X 2n k=0
2k
2
2 −1
(−1)n−k−1 q (n−k) g2k (q) + (−1)n−1 q n 2
(−1)n−k−1 q (n−k) g2k (q),
where the second equality holds since g0 (q) = q −1 . o o o Using the recurrences satisfied by S2n (q) and Sˆ2n (q), we can easily prove that Sˆ2n (q) = 1−n o q S2n (q) for n ≥ 1. We omit the details. For a permutation π = π1 π2 · · · , we define πe = π0 π2 π4 · · · , where π0 = ∞. Theorem 3.3. Define e S2n (q) =
X
q inv(π)+des(πe ) .
(3.7)
π∈A2n
Then we have e o S2n (q) = qS2n (q).
Proof. For a permutation π ∈ A2n , define the map σ by σ(π) = (2n + 1 − π2n )(2n + 1 − π2n−1 ) · · · (2n + 1 − π2 )(2n + 1 − π1 ). Then σ(π) is an alternating permutation in A2n with inv(π) = inv(σ(π))
and
des(πo ) + 1 = des(σ(π)e ),
since σ(π)e = (∞)(2n + 1 − π2n−1 )(2n + 1 − π2n−3 ) · · · (2n + 1 − π3 )(2n + 1 − π1 ). e o Therefore, it follows that S2n (q) = qS2n (q). It follows from Theorems 3.2 and 3.3 that !−1 2 ∞ ∞ e X X (−1)n q n x2n S2n (q)x2n . = (q; q) (q; q) 2n 2n n=0 n=0 o o Theorem 3.4. The polynomials S2n (q) and Sˆ2n (q) are symmetric about the middle coefficient. o Proof. It suffices to show that S2n (q) is symmetric. First note that the alternating permutation with the least weight is
π = 1 3 2 5 4 7 · · · (2n − 1) (2n − 2) 2n
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
13
with inv(π) = n − 1 and des(πo ) = 0; while the alternating permutation with the largest weight is π = (2n − 1) 2n (2n − 3) (2n − 2) · · · 1 2 with inv(π) = 2n2 − 2n and des(πo ) = n − 1. Thus 2 −n−1
o S2n (q) = q n−1 + · · · + q 2n
.
o S2k (q) is symmetric about q k It is clear that S2o (q) = 1 is symmetric. Suppose x k < n. Since the q-binomial coefficient is symmetric, y 2n (n−k)2 o q S2k (q) 2k
2 −1
for
is symmetric. The exponent of the middle term is (n − k)2 + (2nk − 2k 2 ) + k 2 − 1 = n2 − 1. 2 −1
o Therefore, S2n (q) is symmetric about q n
.
Note. Is there a combinatorial proof of Theorem 3.4 analogous to that of Theorem 5.3? Theorem 3.5. Let ∞ ˆ X S2n (q)x2n n=0
(q; q)2n
=
∞ X (−1)n q n(2n−1) x2n
(q; q)2n
n=0
!−1 .
Then Sˆ2n (q) =
X
q inv(π) ,
π
where the sum is over all alternating permutations π on [2n] with π1 > π2 < π3 > π4 < · · · < π2n−1 > π2n . Proof. We first note that the polynomials Sˆ2n (q) satisfy Sˆ0 (q) = 1 and n−1 X 2n ˆ S2n (q) = (−1)n−k−1 q (n−k)(2(n−k)−1) Sˆ2k (q), for n ≥ 1. 2k k=0
For any alternating permutation π ∈ A2n , define π ¯ by π ¯ = π2n π2n−1 π2n−2 · · · π2 π1 . Then π ¯ is clearly a down-up permutation. Furthermore inv(π) + inv(¯ π ) = n(2n − 1). Thus X π∈A2n
q inv(¯π) =
X
q n(2n−1)−inv(π)
π∈A2n
= q n(2n−1) S2n (q −1 ).
(3.8)
14
TIM HUBER AND AE JA YEE
The theorem is equivalent to Sˆ2n (q) = q n(2n−1) S2n (q −1 ). We now show that q n(2n−1) S2n (q −1 ) satisfies the same recursion as Sˆ2n (q). Substitute q k(2k−1) S2k (q −1 ) for Sˆ2k in (3.8). Then n−1 X 2n (−1)n−k−1 q (n−k)(2(n−k)−1) q k(2k−1) S2k (q −1 ) 2k q k=0 n−1 X 2 2n = (−1)n−k−1 q 2n −n S2k (q −1 ) 2k q−1 k=0 n(2n−1)
=q
S2n (q −1 ),
where the last equality follows from the recursion formula for S2n (q).
Note: The polynomials S2n (q) and Sˆ2n (q) appearing in Theorems 3.1 and 3.5 are the only polynomials considered in this paper that are not symmetric about the middle coefficient(s). 4. New q-tangent numbers associated with odd indices o Define T2n+1 (q) by 2 ∞ X (−1)n q n +n x2n+1
∞ o X (q)x2n+1 T2n+1
(q; q)2n+1
n=0
=
n=0 ∞ X n=0
Then, by the definition of
o S2n (q)
∞ o X (q)x2n+1 T2n+1 n=0
(q; q)2n+1
(q; q)2n+1 2
(−1)n q n x2n (q; q)2n
.
in Theorem 3.2,
2 ∞ ∞ o X (−1)n q n +n x2n+1 X qS2n (q)x2n = . (q; q) (q; q) 2n+1 2n n=0 n=0
Thus o T2n+1 (q)
=
n X 2n + 1 k=0
2k
(−1)n−k q (n−k)
2 +(n−k)+1
o S2k (q).
(4.1)
−2 o In [16], Huber proves that the coefficients (q; q)−1 2n+1 ψ (q) in y2n+1 in (1.2) are T2n+1 (q), whose combinatorial interpretation is given the following theorem. To define the corresponding statistic we denote, for a given permutation π = π1 π2 · · · π2n+1 , the permutation πo = π1 π3 π5 · · · π2n−1 π2n+1 .
Theorem 4.1. For each nonnegative integer n, we have X o T2n+1 (q) = q inv(π)+des(πo ) . π∈A2n+1
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
15
Proof. Let X
f2n+1 (q) =
q inv(π)+des(πo ) .
π∈A2n+1
For n = 0, it is clear that f1 (q) = 1 = T1o (q). For any positive integer n, we will show that f2n+1 (q) satisfies the equation (4.1). (k) For a positive integer k ≤ n, let A2n+1 be the set of permutations π on [2n + 1] such that π1 < π2 > π3 < · · · > π2k−1 < π2k , π2k+2 > π2k+4 > · · · > π2n > π2n+1 > π2n−1 > · · · > π2k+1 .
(4.2)
From Lemma 1.1 and Theorem 3.2, we see that for k ≤ n, X 2n + 1 (n−k)2 o inv(π)+des(π1 π3 ···π2k−1 ) q = S2k (q). q 2k
(4.3)
(k)
π∈A2n+1 (n)
We decompose A2n+1 into disjoint subsets as follows: (n)
A2n+1 ={π | π2n > π2n+1 } ∪ {π | π2n < π2n+1 } ={π | π2n > π2n+1 > π2n−1 } ∪ {π | π2n > π2n−1 > π2n+1 } ∪ {π | π2n < π2n+1 } (n)
(n)
(n)
(n)
(n)
=:B2n+1 ∪ D2n+1 ∪ E2n+1 .
(4.4) (k)
Note that A2n+1 = B2n+1 ∪ D2n+1 . For k < n, we decompose A2n+1 into disjoint subsets as follows: (k)
A2n+1 ={π | π2k > π2k+1 } ∪ {π | π2k < π2k+1 } ={π | π2k > π2k+1 > π2k−1 } ∪ {π | π2k > π2k−1 > π2k+1 } ∪ {π | π2k < π2k+1 } ={π | π2k > π2k+2 > π2k+1 > π2k−1 } ∪ {π | π2k+2 > π2k > π2k+1 > π2k−1 } ∪ {π | π2k > π2k−1 > π2k+1 } ∪ {π | π2k < π2k+1 } (k)
(k)
(k)
(k)
=:B2n+1 ∪ C2n+1 ∪ D2n+1 ∪ E2n+1 .
(4.5)
(k)
(k)
Note that, since the permutations in B2n+1 satisfy the conditions of (4.2), B2n+1 is the set of alternating permutations π on [2n + 1] such that π1 < π2 > π3 < · · · < π2k , π2k > π2k+2 > · · · > π2n > π2n+1 > π2n−1 > · · · > π2k−1 . Furthermore, for any positive k ≤ n, (k+1)
(k)
(k)
(k)
B2n+1 = B2n+1 ∪ C2n+1 ∪ D2n+1 ,
16
TIM HUBER AND AE JA YEE (n)
where B (n+1) = A2n+1 and C2n+1 = ∅. For k ≥ 1, we now compute the generating (k+1) function for permutations π ∈ B2n+1 X q inv(π)+des(π1 π3 ···π2k+1 ) . (k+1)
π∈B2n+1 (k+1)
(k)
Let π ∈ B2n+1 . If π2k−1 > π2k+1 , namely π ∈ D2n+1 , then inv(π) + des(π1 π3 · · · π2k+1 ) = inv(π) + des(π1 π3 · · · π2k−1 ) + 1. (k)
(k)
However, if π2k−1 < π2k+1 , namely π ∈ B2n+1 ∪ C2n+1 , then inv(π) + des(π1 π3 · · · π2k+1 ) = inv(π) + des(π1 π3 · · · π2k−1 ). In this case, we look for a permutation σ such that inv(π) + des(π1 π3 · · · π2k+1 ) = inv(σ) + des(σ1 σ3 · · · σ2k−1 ) + 1. Let the positive integer m be defined by π2k+m = max{π2k+j | π2k+j < π2k , 1 ≤ j ≤ 2(n − k) + 1}. Since π2k+1 < π2k , there exists such an m. We switch π2k+m with π2k and denote the resulting partition by π ¯ . Switching π2k with π2k+m results in a decrease of the inversion number, namely inv(π) = inv(¯ π ) + 1. Moreover, if π2k were switched with π2k+m for m > 1, it is trivial that des(π1 π3 · · · π2k+1 ) = des(¯ π1 π ¯3 · · · π ¯2k−1 ). If π2k were switched with π2k+1 , since π2k−1 < π2k+1 , des(π1 π3 · · · π2k+1 ) = des(¯ π1 π ¯3 · · · π ¯2k−1 ). Thus, in either case, inv(π) + des(π1 π3 · · · π2k+1 ) = inv(¯ π ) + des(¯ π1 π ¯3 · · · π ¯2k−1 ) + 1. (k)
From the definition of π ¯ , if π2k were switched with π2k+1 , namely π ∈ C2n+1 , then π ¯1 < π ¯2 > π ¯3 < · · · < π ¯2k−2 > π ¯2k−1 < π ¯2k , π ¯2k+2 > π ¯2k+4 > · · · > π ¯2n > π ¯2n+1 > π ¯2n−1 > · · · > π ¯2k+1 , π ¯2k < π ¯2k+1 , (k)
which shows that π ¯ ∈ E2n+1 . If π2k were switched with π2k+2 , then π ¯1 < π ¯2 > π ¯3 < · · · < π ¯2k−2 > π ¯2k−1 < π ¯2k , π ¯2k+2 > π ¯2k+4 > · · · > π ¯2n > π ¯2n+1 > π ¯2n−1 > · · · > π ¯2k+1 , π ¯2k+2 > π ¯2k > π ¯2k+1 > π ¯2k−1 ,
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
17
from the maximality of π2k+m ; if π2k were switched with π2k+m for m > 2, then π ¯1 < π ¯2 > π ¯3 < · · · < π ¯2k−2 > π ¯2k−1 < π ¯2k , π ¯2k+2 > π ¯2k+4 > · · · > π ¯2n > π ¯2n+1 > π ¯2n−1 > · · · > π ¯2k+1 , π ¯2k+2 > π ¯2k > π ¯2k+1 > π ¯2k−1 , since π2k+2 > π2k+m . Thus, if π2k were switched with π2k+m for m > 1, namely (k) (k) π ∈ B2n+1 , then π ¯ ∈ C2n+1 . Thus, for any k, 1 ≤ k ≤ n, (k+1)
(k+1)
{π | π ∈ B2n+1 , π2k−1 > π2k+1 } ∪ {¯ π | π ∈ B2n+1 , π2k−1 < π2k+1 } (k)
(k)
(k)
{¯ π | π ∈ B2n+1 } ∪ {¯ π | π ∈ C2n+1 } ∪ {π | π ∈ D2n+1 } (k)
(k)
(k)
= C2n+1 ∪ D2n+1 ∪ E2n+1 (k)
(k)
= A2n+1 \ B2n+1 ,
(4.6)
where the last equality of (4.6) follows from (4.5). Therefore, X q inv(π)+des(π1 π3 ···π2k+1 ) (k+1)
π∈B2n+1
=
X
X
q inv(π)+des(π1 π3 ···π2k−1 )+1
(k) π∈B2n+1 π2k−1 π2k+1
=
X
q inv(π)+des(π1 π3 ···π2k−1 ) +
q inv(π)+des(π1 π3 ···π2k−1 )+1 +
X
q inv(¯π)+des(¯π1 π¯3 ···¯π2k−1 )+1
(k) π∈B2n+1 π2k−1 π2k+1
X X inv(π)+des(π1 π3 ···π2k−1 ) = q q − q inv(π)+des(π1 π3 ···π2k−1 ) . (k) π∈A2n+1
(4.7)
(k) π∈B2n+1
By iterating (4.7), we deduce X q inv(π)+des(πo ) (n+1)
π∈B2n+1
=
n X
X X inv(π)+des(π1 π3 ···π2k−1 )+(n−k)+1 n (−1)n−k q q inv(π)+n , + (−1)
k=1
(k)
(1)
π∈A2n+1
π∈B2n+1
which is equivalent to f2n+1 (q) =
n X 2n + 1 k=0
2k (n+1)
(−1)n−k q (n−k)
(1)
2 +(n−k)+1
by (4.3) and the definitions of B2n+1 , B2n+1 , and S0o (q) = q −1 .
o S2k (q).
18
TIM HUBER AND AE JA YEE o Define Tˆ2n+1 (q) by 2 ∞ X (−1)n q n x2n+1
∞ ˆo X T2n+1 (q)x2n+1 n=0
(q; q)2n+1
=
n=0 ∞ X n=0
(q; q)2n+1 2
(−1)n q n −n x2n (q; q)2n
.
Theorem 4.2. For a nonnegative integer n, we have o o Tˆ2n+1 (q) = q −n T2n+1 (q). o o Proof. The theorem follows from the recurrences satisfied by T2n+1 (q) and Tˆ2n+1 (q).
5. New q-tangent numbers associated with even indices e Let T2n+1 (q) be the polynomial satisfying 2 ∞ X (−1)n q n x2n+1
∞ e X (q)x2n+1 T2n+1 n=0
(q; q)2n+1
=
n=0 ∞ X n=0
(q; q)2n+1 2
(−1)n q n x2n (q; q)2n
.
e (q), it follows that From the generating function for T2n+1 n−1 X 2 2n + 1 e n n2 e T2n+1 (q) = (−1) q + (−1)n−k−1 q (n−k) T2k+1 (q). 2k + 1
(5.1)
k=0
Recall πe for a permutation π defined in Section 3, namely πe = π0 π2 π4 · · · with π0 = ∞. Theorem 5.1. For a nonnegative integer n, X e T2n+1 (q) = q inv(π)+des(πe ) . π∈A2n+1
Proof. Let f2n+1 (q) =
X
q inv(π)+des(πe ) .
π∈A2n+1
For n = 0, it is clear that f1 (q) = 1 = T1e (q). For any positive integer n, we will show that f2n+1 (q) satisfies the recurrence (5.1). (k) For a nonnegative integer k ≤ n, let A2n+1 be the set of permutations π on [2n + 1] such that π1 < π2 > π3 < · · · < π2k > π2k+1 , π2k+2 > π2k+4 > · · · > π2n > π2n+1 > π2n−1 > · · · > π2k+3 .
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
From Lemma 1.1 and the definition of f2n+1 , we see that X 2n + 1 (n−k)2 inv(π)+des(π0 π2 π4 ···π2k ) q = q f2k+1 (q). 2k + 1
19
(5.2)
(k) π∈A2n+1
(k)
For k < n, we decompose A2n+1 into disjoint subsets as follows: (k)
A2n+1 ={π | π2k+1 < π2k+2 } ∪ {π | π2k+1 > π2k+2 } ={π | π2k+1 < π2k+2 < π2k } ∪ {π | π2k+1 < π2k < π2k+2 } ∪ {π | π2k+1 > π2k+2 } ={π | π2k+1 < π2k+3 < π2k+2 < π2k } ∪ {π | π2k+3 < π2k+1 < π2k+2 < π2k } ∪ {π | π2k+1 < π2k < π2k+2 } ∪ {π | π2k+1 > π2k+2 } (k)
(k)
(k)
(k)
=:B2n+1 ∪ C2n+1 ∪ D2n+1 ∪ E2n+1 .
(5.3)
(k)
Note that B2n+1 is the set of alternating permutations π on [2n + 1] such that π1 < π2 > π3 < · · · < π2k > π2k+1 , π2k > π2k+2 > · · · > π2n > π2n+1 > π2n−1 > · · · > π2k+3 > π2k+1 . (n)
(k)
We define B2n+1 = A2n+1 . For a permutation π ∈ B2n+1 with 1 ≤ k ≤ n, if π2k−2 < π2k , then π2k−1 < π2k−2 < π2k , (k−1)
which shows that such π satisfies the conditions of D2n+1 . Thus (k)
B2n+1 ={π | π2k−2 < π2k } ∪ {π | π2k−2 > π2k } (k−1)
=D2n+1 ∪ {π | π2k−2 > π2k }. (k)
We now compute the generating function for permutations π ∈ B2n+1 X q inv(π)+des(π0 π2 π4 ···π2k ) . (k)
π∈B2n+1 (k)
Let π ∈ B2n+1 . If π2k−2 < π2k , then inv(π) + des(π0 π2 π4 · · · π2k ) = inv(π) + des(π0 π2 π4 · · · π2k−2 ). However, if π2k−2 > π2k , then inv(π) + des(π0 π2 π4 · · · π2k ) = inv(π) + des(π0 π2 π4 · · · π2k−2 ) + 1. In this case, we look for a permutation σ such that inv(π) + des(π0 π2 π4 · · · π2k ) = inv(σ) + des(σ0 σ2 σ4 · · · σ2k−2 ). Let m be defined by π2k+m = min{π2k+j | π2k+j > π2k−1 , 0 ≤ j ≤ 2n − 2k + 1}.
20
TIM HUBER AND AE JA YEE
There exists such an m since π is an alternating permutation, namely π2k−1 < π2k . So, π2k+m ≤ π2k < π2k−2 . We switch π2k+m with π2k−1 and denote the resulting partition by π ¯ . Switching π2k−1 with π2k+m results in increasing of an inversion, namely inv(π) + 1 = inv(¯ π ). Moreover, since π2i = π ¯2i for i < k, des(π0 π2 π4 · · · π2k−2 ) = des(¯ π0 π ¯2 π ¯4 · · · π ¯2k−2 ). Thus inv(π) + des(π2 π4 · · · π2k ) = inv(π) + des(π0 π2 π4 · · · π2k−2 ) + 1 = inv(¯ π ) + des(¯ π0 π ¯2 π ¯4 · · · π ¯2k−2 ). If π2k−1 were switched with π2k+m for m > 0, then π2k+j < π2k . Hence, from the definition of π ¯ , we see that π ¯1 < π ¯2 > π ¯3 < · · · < π ¯2k−2 > π ¯2k−1 , π ¯2k > π ¯2k+2 > · · · > π ¯2n > π ¯2n+1 > π ¯2n−1 > · · · > π ¯2k+1 , π ¯2k−2 > π ¯2k > π ¯2k−1 > π ¯2k+1 , (k−1)
which shows, from (5.3), that π ¯ ∈ C2n+1 . If π2k−1 and π2k were switched, namely m = 0, then π2k−1 > πi for i > 2k. Hence, from the definition of π ¯ , we see that π ¯1 < π ¯2 > π ¯3 < · · · < π ¯2k−2 > π ¯2k−1 , π ¯2k > π ¯2k+2 > · · · > π ¯2n > π ¯2n+1 > π ¯2n−1 > · · · > π ¯2k+1 , π ¯2k−2 > π ¯2k−1 > π ¯2k > π ¯2k+1 . (k−1)
(5.4)
(k−1)
Note that, since E2n+1 ⊆ A2n+1 , we see that (k−1)
E2n+1 = {π | π2k−2 > π2k−1 > π2k > π2k+1 }. (k−1)
Therefore, (5.4) implies π ¯ ∈ E2n+1 . Hence, by (5.3), for any k, 1 ≤ k ≤ n (k)
(k)
{π | π ∈ B2n+1 , π2k−2 < π2k } ∪ {¯ π | π ∈ B2n+1 , π2k−2 > π2k } (k−1)
(k−1)
(k−1)
= C2n+1 ∪ D2n+1 ∪ E2n+1 (k−1)
(k−1)
= A2n+1 \ B2n+1 .
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
Therefore, X
21
q inv(π)+des(π0 π2 π4 ···π2k )
(k)
π∈B2n+1
X
=
(k) π∈B2n+1 π2k−2 π2k
X
q inv(π)+des(π0 π2 π4 ···π2k−2 ) +
(k) π∈B2n+1 π2k−2 π2k
q inv(π)+des(π0 π2 π4 ···π2k−2 ) −
(k−1) π∈A2n+1
X
q inv(π)+des(π0 π2 π4 ···π2k−2 ) .
(5.5)
(k−1) π∈B2n+1
By iterating (5.5), we deduce X q inv(π)+des(πe ) (n)
π∈B2n+1
=
n−1 X
X X inv(π)+des(π0 π2 π4 ···π2k ) n (−1)n−k−1 q q inv(π) , + (−1)
k=0
(k)
(0)
π∈A2n+1
π∈B2n+1
which is equivalent to f2n+1 (q) =
n−1 X 2n + 1 k=0
2k + 1
2
2
(−1)n−k−1 q (n−k) f2k+1 (q) + (−1)n q n .
Theorem 5.2. Define 2 ∞ X (−1)n q n +n x2n+1
∞ ˆe X T2n+1 (q)x2n+1 n=0
(q; q)2n+1
=
n=0 ∞ X n=0
(q; q)2n+1 2
(−1)n q n −n x2n (q; q)2n
.
Then, for n ≥ 1, e e (q) = q −n−1 T2n+1 (q). Tˆ2n+1
Proof. Note that 2 e Tˆ2n+1 (q) = (−1)n q n +n +
n−1 X 2n + 1 k=0
2k + 1
(−1)n−k−1 q (n−k)
2 −(n−k)
e Tˆ2k+1 (q).
(5.6)
22
TIM HUBER AND AE JA YEE
To prove Theorem 5.2, multiply both sides of (5.6) by q n+1 to obtain, for n ≥ 1, 2 2n + 1 n+1 ˆ e n (n+1)2 q T2n+1 (q) = (−1) q + (−1)n−1 q n +1 (5.7) 1 n−1 X 2 2n + 1 e (q). + (−1)n−k−1 q (n−k) +k+1 Tˆ2k+1 2k + 1 k=1
Note that n (n+1)2
(−1) q
2 2 2n + 1 + (−1)n−1 q n +1 = (−1)n−1 q n +1 (1 + q + · · · + q 2n−1 ) 1 2 2n + 1 n n2 = (−1) q + (−1)n−1 q n . 1
(5.8)
e (q) is identical to the Inserting (5.8) into (5.7), we see that the recursion (5.1) for T2n+1 n+1 ˆ e recursion for q T2n+1 (q) in (5.7). o In [18], Ismail and Zhang conjectured that the polynomials T2n+1 (q) are symmetric about the middle coefficient(s). We prove their conjecture in the following theorem. o e Theorem 5.3. The polynomials T2n+1 (q), T2n+1 (q), and T2n+1 (q) are symmetric about the middle coefficient(s).
Proof. For each alternating permutation π = π1 π2 . . . π2n+1 , the permutation π = π2n+1 π2n . . . π1 is also an alternating permutation. Recall that des(πe ) = π0 π2 · · · π2n and des(π e ) = π0 π2n · · · π2 . From the definition of π, it follows that n(2n + 1) = inv(π) + inv(π), 2n(n + 1) = inv(π) + des(πo ) + inv(π) + des(π o ), 2n(n + 1) + 1 = inv(π) + des(πe ) + inv(π) + des(π e ). Therefore, the inversion map π → π is a bijection between {π | inv(π) = k}
and
{π | inv(π) = n(2n + 1) − k},
{π | inv(π) + des(πo ) = k}
and
{π | inv(π) + des(πo ) = 2n(n + 1) − k},
{π | inv(π) + des(πe ) = k}
and
{π | inv(π) + des(πe ) = 2n(n + 1) + 1 − k}.
If k = n(2n + 1)/2, then {π | inv(π) = k} = {π | inv(π) = n(2n + 1) − k}. Otherwise, the two sets are disjoint. It follows that, if n is even, the coefficients of T2n+1 are symmetric about the term |{π | inv(π) = n(2n + 1)/2}| · q n(2n+1)/2 . If n is odd, the coefficients are symmetric about the terms corresponding to q bn(2n+1)/2c o e and q bn(2n+1)/2c+1 . The coefficients of T2n+1 (q) and T2n+1 (q) can similarly be seen to be symmetric about the middle term(s). Definition. A polynomial p : C → C of degree n is said to be reciprocal if 1 n p(z) = ±z p . z
(5.9)
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
23
The following corollary follows from Theorems 2.3, 4.2, 5.2, and 5.3. e o des (q) are reciprocal. More (q), and Tˆ2n+1 (q), Tˆ2n+1 Corollary 5.4. The polynomials Tˆ2n+1 des o e precisely, if f (q) ∈ {Tˆ2n+1 (q), Tˆ2n+1 (q), Tˆ2n+1 (q) | n ≥ 0}, and f has degree n, then
f (1/q) = f (q)/q n . 6. Higher Order q-Euler Numbers The tangent numbers of order k are defined by the Taylor series coefficients in the expansion of tank z about z = 0. Since d tan2 z = dz we see that for n ≥ 1, the numbers d2n 2 tan z = dz 2n z=0
d2 tan z, dz 2 d2n+1 tan z dz 2n+1 z=0
(6.1)
each enumerate the alternating permutations on [2n + 1]. Equivalently, the first and second order tangent numbers are identical. The q-extensions of second order q-tangent numbers, in contrast, generate polynomials distinct from those of first order. In the following theorem, we offer a combinatorial interpretation for the second order q-tangent numbers arising in the previous sections. Before we state the theorem, we define the permutation statistics α, β, and γ on A2n+1 by α(π) = inv(π) + max(π) − 2n − 1, 3 + sign(πmax(π)−1 − πmax(π)+1 ) , 2 γ(π) = inv(π) + des(πe ) + max(π) − 2n − 2,
β(π) = inv(π) + des(πo ) + max(π) − 2n −
where max(π) denotes the index of 2n + 1 in π. Throughout the section, we denote πij = πi πi+1 · · · πj for a given permutation π. o e Theorem 6.1. Let T2n+1 (q), T2n+1 (q), T2n+1 (q) denote the q-analogues of the tangent (2) (2) (2) numbers defined by Theorems 2.1, 4.1, and 5.1, respectively. Define T2n (q), T o 2n (q), T e 2n (q) by !2 ∞ (2) X T2n+1 (q)z 2n+1 X T2n (q)z 2n = , (q; q)2n (q; q)2n+1 n=0 n=0 !2 ∞ o 2n+1 o (2) 2n X X T (q)z T 2n (q)z 2n+1 = , (q; q)2n (q; q)2n+1 n=0 n=0 !2 ∞ (2) X X T e (q)z 2n+1 T e 2n (q)z 2n 2n = . (q; q) (q; q) 2n 2n+1 n=0 n=0
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TIM HUBER AND AE JA YEE
Then (2)
T2n (q) =
X
q α(π) ,
X
(2)
T o 2n (q) =
π∈A2n+1
q β(π) ,
(2)
T e 2n (q) =
π∈A2n+1
X
q γ(π) .
π∈A2n+1
Proof. For a permutation π in A2n+1 , let π2k = 2n + 1 for some k, 1 ≤ k ≤ n. Then inv(π) =|{(i, j) | i < 2k < j and πi > πj }| + |{(i, j) | i < j < 2k and πi > πj }| + |{(i, j) | 2k < i < j and πi > πj }| + |{(2k, j) | 2k < j}| 2n+1 =|{(i, j) | i < 2k < j and πi > πj }| + inv(π12k−1 ) + inv(π2k+1 ) + 2(n − k) + 1.
Thus, 2n+1 α(π) = |{(i, j) | i < 2k < j and πi > πj }| + inv(π12k−1 ) + inv(π2k+1 ). (2)
We now show that T2n (q) is the generating function for permutations in A2n+1 with (2) weight α. The polynomials T2n (q) satisfy n X 2n (2) T (q)T2(n−k)+1 (q). (6.2) T2n (q) = 2k − 1 q 2k−1 k=1
2n in the summand on the right hand By Lemma 1.1, the q-binomial coefficient 2k − 1 q
2n+1 side of (6.2) counts the inversions between the two sub-permutations π12k−1 and π2k+1 , namely
|{(i, j) | i < 2k < j and πi > πj }|; 2n+1 T2k−1 (q) and T2(n−k)+1 (q) count the inversions of π12k−1 and π2k+1 , respectively. There(2) fore, T2n (q) is the generating function for permutations π in A2n+1 with weight α(π). (2) (2) The arithmetic interpretations for the polynomials T o 2n (q) and T e 2n (q) are similarly derived. We omit the details.
In the next theorem, we present a corresponding interpretation for the second order q-secant numbers studied in Section 3. The proof for each interpretation is similar to that of Theorem 6.1. o Theorem 6.2. Let S2n (q) and S2n (q) denote the q-analogues of the secant numbers (2) (2) defined by Theorems 3.1 and 3.2, respectively. Define S2n (q) and S o 2n (q) by !2 ∞ (2) 2n 2n X X S2n (q)z S2n (q)z = , (q; q)2n (q; q)2n n=0 n=0 !2 ∞ (2) X X S o (q)z 2n S o 2n (q)z 2n 2n = . (q; q) (q; q) 2n 2n n=0 n=0
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
25
Then (2)
S2n (q) =
X
q δ(π) ,
π∈A2n+1 (2)
S o 2n (q) =
X
q υ(π) =
π∈A2n+1
X
q ω(π) ,
π∈A2n+1
where 2n + 1 − min(π) δ(π) = inv(π) − 2 · + − min(π) + 1, 2 min(π) + 3 2n+1 υ(π) = δ(π) + des(πo ) − 2 · des((πmin(π)+1 )o ) + n − 2 min(π) + 2 + sign(πmax(π)−1 − πmax(π)+1 ) 2n+1 ω(π) = δ(π) + des(πe ) − 2 · des((πmin(π)+1 )e ) + n − 2 + sign(π2n+1 − 1) + sign(π1 − 1) − 2. 2n+1 inv(πmin(π)+1 )
Proof. For a permutation π in A2n+1 , let π2k+1 = 1 for some k, 0 ≤ k ≤ n. Then inv(π) =|{(i, j) | i < 2k + 1 < j and πi > πj }| + |{(i, j) | i < j < 2k + 1 and πi > πj }| + |{(i, j) | 2k + 1 < i < j and πi > πj }| + |{(i, 2k + 1) | i < 2k + 1}| 2n+1 =|{(i, j) | i < 2k + 1 < j and πi > πj }| + inv(π12k ) + inv(π2k+2 ) + 2k.
Thus, δ(π) = |{(i, j) | i < 2k + 1 < j and πi > πj }| +
inv(π12k )
−
2n+1 inv(π2k+2 )
2(n − k) + 2
2n+1 ), = |{(i, j) | i < 2k + 1 < j and πi > πj }| + inv(π12k ) + inv(π2k+2
(6.3)
2n+1 where π2k+2 = π2n+1 π2n · · · π2k+3 π2k+2 . (2) We now show that S2n (q) is the generating function of permutations in A2n+1 with (2) weight δ. The polynomials S2n (q) satisfy n X 2n (2) S2n (q) = S (q)S2(n−k) (q). (6.4) 2k q 2k k=0 2n By Lemma 1.1, the q-binomial coefficient in the summand on the right hand 2k q 2n+1 side of (6.4) counts the inversions between the two sub-permutations π12k and π2k+2 , namely
|{(i, j) | i < 2k + 1 < j and πi > πj }|; 2n+1 S2k (q) and S2(n−k) (q) count the inversions of π12k and π2k+1 , respectively, since π12k ∈ A2k (2) 2n+1 and π2k+1 ∈ A2(n−k) . Therefore, S2n (q) is the generating function for permutations π in A2n+1 with weight δ(π).
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TIM HUBER AND AE JA YEE (2)
To obtain the claimed arithmetic interpretation for the polynomials S o 2n (q), note that, with the convention that des(∅) = 0, where ∅ denotes the empty permutation, des(π2n+1 π2n−1 . . . πmin(π)+2 ) + sign(π2n+1 − 1) − 1 =
2n − min(π) − 1 − des(πmin(π)+2 πmin(π)+4 · · · π2n+1 ). 2
Hence, by (6.3), we find that 2n+1 υ(π) =δ(π) + des(πo ) − 2 · des((πmin(π)+2 )o ) + n −
min(π) + 3 2
2n − min(π) − 1 2 = δ(π) + des(π1 π3 · · · πmin(π)−2 ) + des(π2n+1 π2n−1 · · · πmin(π)+2 ) + sign(π2n+1 − 1) − 1
=δ(π) + des(π1 π3 · · · πmin(π)−2 ) − des(πmin(π)+2 πmin(π)+3 · · · π2n+1 ) + min(π)−1
=|{(i, j) | i < min(π) < j and πi > πj }| + inv(π1
min(π)−1
) + des((π1
)o )
2n+1 2n+1 + inv(πmin(π)+1 ) + des((πmin(π)+1 )o ) + sign(π2n+1 − 1) − 1, 2n+1 = π2n+1 π2n · · · π2k+3 πmin(π)+1 . Therefore, since S0o (q) = q −1 , by Lemma where πmin(π)+1 (2)
1.1, Theorem 3.2, and the definition of S o 2n (q), we see that X (2) S o 2n (q) = q υ(π) . π∈A2n+1 e (q) defined by (3.7) and using the fact By squaring the generating function for S2n o e that S2n (q) = S2n (q), we readily observe that the weight ω(π) corresponds to the same enumeration for A2n+1 as υ(π). We omit the details.
7. Concluding Remarks Define the Bell polynomials Bn,k (x1 , x2 , · · · , xn−k+1 ) via the generating function [11] ! ∞ ∞ X n X X xm z m uk z n exp u =1+ Bn,k (x1 , x2 , xn−k+1 ) . m! n! m=1 n=1 k=1 Then, by Fa´a Di Bruno’s formula [11, p. 137] and Theorem 3.1, (q; q)2n S2n (q) = (2n)!
1+
2n X
! (−1)v v!B2n,v (α1 , . . . , α2n−v+1 ) ,
(7.1)
v=1
where ( (−1)k/2 k!/(q; q)k , if k is even, αk = 0, if k is odd. From (2.1) and (3.1), we obtain T2n+1 (q) =
n X 2n + 1 j=0
2j
(−1)n−j S2j (q).
(7.2)
Closed formulas and relations for the other q-Euler numbers can be similarly derived.
COMBINATORICS OF GENERALIZED q-EULER NUMBERS
27
o (q) are the As mentioned in the introduction, the generalized tangent numbers T2n+1 −2 polynomials arising in the coefficient of ψ (q) in (q; q)2n+1 y2n+1 of (1.2). Constant (2) multiples of the second order extensions T o 2n+1 (q) appear in (1.2) as the coefficient of ψ −4 (q) in the corresponding expansion of (q; q)2n y2n for n ≥ 2. Ismail and Zhang [18, Theorem 4.1] prove that each yj can be expressed as a polynomial in theta functions over the field of rational functions in q. The authors of [18] suggest that polynomials appearing in the numerators of these expansions, denoted by Dr,s,t (q), have interesting o (q) addresses the combinatorics of the combinatorial properties. Our study of T2n+1 polynomials Dr,0,1 (q) and Dr,0,2 (q). Recursion formulas for yj appearing in [15, 16] show that the polynomials Dr,s,t (q) arise as linear combinations of finite products Q mj o j (T2nj +1 (q)) . Therefore, the current paper shows that the polynomials Dr,s,t (q) have integer coefficients1. Ismail and Zhang conjecture [18, Conjecture 4.3] that the polynomials Dr,s,t (q) are symmetric about the middle coefficient(s). The symmetry o of the polynomials T2n (q) established in the current paper shows that, to prove the +1 general Ismail-Zhang conjecture, it suffices the form of the representation Q oto determine mj (q)) . From such formulas, arithmetic for Dr,s,t (q) in terms of the products j (T2n j +1 interpretations for the polynomials Dr,s,t (q) for all r, s, t ∈ N may be deduced.
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