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 E n are the integers defined by E n x n sec x + tan x (11) n! In 1879, D André [1, 2] gave a combinatorial interpretation for the Euler numbers E n These numbers count the number of permutations π π 1 π 2 π n of elements in the set [n] : {1, 2,, n} such that the sign of π i π i+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 [7, 8, 9, 10, 11, 13, 21, 29] 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 [16, 17, 18, 28] 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 [24], [4, Chapter 13]: where y 1 q n2 z n (q; q) n ( 1 + n1 1 (1 q)ψ 2 (q), y 2 0, y 3 zq 1 1 j1 qjn y j ), (12) q + q 3 ( 1)q 1 (q; q) 3 ψ 2 (q) n1 1 q 1 (1 q) 3 ψ 6 (q), y 4 y 1 y 3 1 (13)
2 TIM HUBER AND AE JA YEE The functions ψ(q) and (α; q) n appearing in (12) and (13) are defined by ψ(q) : n 1 q n(n+1)/2, (α; q) n : 1 αq j, q < 1 In [19], Ismail and Zhang observed that the polynomials appearing in the expansion (12), as the coefficients of (q; q) 1 j ψ 2 (q) in y j, are symmetric about the middle coefficient(s) We explain this symmetry in Sections 4, 5, 7, and 8 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 [22, 23] 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 [22, 23] Quotients of the series considered in the present paper also have beautiful continued fraction representations [15, 22] For nonnegative integers A, B, C, D, consider the following q-analogue of tan x: f (q) x (q; q) j0 ( 1) n q An2 +Bn x (q; q) ( 1) n q Cn2 +Dn x (14) (q; q) When (A, B, C, D) (0, 0, 0, 0), f (q) is the q-tangent number T (q) of F H Jackson [20] In [17], Huber proves that the coefficients T(q) o of (q; q) 1 ψ 2 (q) in y in (12) are f (q) for (A, B, C, D) (1, 1, 1, 0) In this paper, we discuss q-tangent numbers corresponding to (A, B, C, D) given in the following table Let τ αβ represent the probability that a word from {1, 2, 3, } of length defined in the preceeding paragraph satisfies the inequality conditions αβ f (A, B) (C, D) Probability T (0, 0) (0, 0) τ, > τ < T o (1, 1) (1, 0) τ <> T e (1, 0) (1, 0) τ >< The column on the right contains the numbers considered by Prodinger for which the series (14) is an associated generating function [22, Theorem 22] For each value of A, B, C and D, the quotient (14) induces a corresponding recursion relation for the function f (q) From these formulas, we obtain the following related polynomials For a polynomial p(q), let ˆp(q) denote the dual of p(q) (see [22, Remark 33]) Several of the dual polynomials occur in connection with probabilities from [22]
COMBINATORICS OF GENERALIZED q-euler NUMBERS 3 (A, B) (C, D) f Relevant Relations Probability (0, 1) (0, 1) T des T(q) des q n T (q) (2, 1) (2, 1) ˆT ˆT (q) T (q) des (2, 0) (2, 2) ˆT ˆT (q) des q T(q) des o o (1, 0) (1, 1) ˆT ˆT (q) q n T(q) o τ e e (1, 1) (1, 1) ˆT ˆT (q) q n 1 T(q) e τ In Section 2, a well-known arithmetic interpretation of the classical q-tangent numbers T (q) is discussed We provide an elementary proof of this interpretation that demonstrates fundamental ideas used throughout the paper A new q-analogue T(q) des is also discussed in the same section In Section 3, we deduce combinatorial interpretations for new q-analogues of the secant numbers appearing in [22, 23] defined by ( g (q) x (q; q) ) 1 ( 1) n q Cn2 +Dn x (q; q) (C, D) g Relevant Relations Probability (0, 0) S σ > (0, 1) qs des S des (q) q n 1 S (q) (1, 0) qs o, S e σ <> (2, 1) Ŝ Ŝ (q) q n( 1) S (q 1 ) σ < (2, 2) Ŝ des Ŝ (q) q n( 1) 1 S des (q 1 ) (1, 1) Ŝ o Ŝ o (q) q 1 n S o (q) The values σ αβ denote the probability that a given word of length, under the aforementioned hypotheses, satisfies the inequality conditions αβ It should be noted that the generating function for the polynomials S(q) o appearing in Section 3 differs by a factor of q 1 from the corresponding generating function for σ <> in [22] In Sections 4 and 5, we describe combinatorics of the new q-tangent numbers T(q) o and T(q) e Our arithmetic interpretations explain the symmetry arising among the coefficients of these polynomials In Section 6, we deduce arithmetic interpretations for second-order tangent numbers T (q), (2) T o (2) (q), T e (2) (q) obtained by squaring the generating functions discussed in the previous sections We include similar interpretations for second order q-secant numbers In [16], it is shown that the coefficients of (q; q) 1 ψ 4 (q) in y of (12) are scalar multiples of T o (2) (q) We indicate in section 7 how closed formulas for the q-euler numbers may be derived in terms of the Bell polynomials We also comment on an application of our results to a conjecture made by Ismail and Zhang [19, Conjecture 43] concerning a more general class of polynomials appearing in (12) within corresponding expansions of y j, j 1 Before proceeding, we introduce some necessary definitions and notation A pair (π i, π j ) is called an inversion of the permutation π π 1 π 2 π n if i < j and π i > π j We denote by inv(π) the number of inversions of the permutation π The descent set D(π) is defined by {i π i > π i+1 }, and des(π) denotes the size of D(π) For a σ
4 TIM HUBER AND AE JA YEE permutation π, we define π o π 1 π 3 π 5 and π e π 0 π 2 π 4, where π 0 Two kinds of half descents of π are defined by des(π o ) and des(π e ) Define the q-binomial coefficient by [ [ (q; q) n n n, if 0 n k, : (q; q) k] k] k (q; q) n k q 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 [27, p 132] Lemma 11 q inv(π) π [ n k], where the sum is over all permutations π with D(π) {k} A more instructive view of Lemma 11 follows by defining P n (k), for a given n and k n, to be the set of all permutations π on [n] such that π 1 < π 2 < π 3 < < π k 1 < π k, π k+1 < π k+2 < < π n 1 < π n Then it follows from Lemma 11 that π P (k) n q inv(π) [ n k] We denote by A n the set of up-down alternating permutations π π 1 π 2 π n on the set [n] with π 1 < π 2 > π 3 <, and we denote by Ān the set of down-up alternating permutations π on the set [n] with π 1 > π 2 < π 3 > 2 The classical q-euler numbers Jackson s q-analogues of the sine and cosine functions [20] are ( 1) n x ( 1) n x and (q; q) /(1 q) (q; q) /(1 q) By considering quotients of these functions, we arrive at a q-analogue of the tangent numbers, T (q), defined by T (q)x ( 1) n x ( 1) n x / (21) (q; q) (q; q) (q; q) If we replace x by x(1 q) in (21) and let q 1, the corresponding identity reduces to the relation obtained by equating odd parts on each side of equation (11) Multiplying
COMBINATORICS OF GENERALIZED q-euler NUMBERS 5 both sides of (21) by the denominator on the right side and equating coefficients of x, we obtain n + 1 T (q) ( 1) k 1 T 2(n k)+1 (q) + ( 1) n (22) k1 The following combinatorial interpretation of the polynomials T (q) is well known [6, 14, 26, 27] We include a proof based upon the recursion (22) as an aid to the reader, since later results in the paper have proofs that are similar in nature Theorem 21 For a nonnegative integer n, we have T (q) q inv(π) π A Proof Let For n 0, it is clear that f (q) π A q inv(π) f 1 (q) 1 T 1 (q) For any positive integer n, we will prove that f (q) satisfies the recurrence (22) For a positive integer k n, let A (k) be the set of permutations π on [ + 1] such that π 1 < π 2 < < π 1 < π, π +1 < π +2 > π +3 < < π > π Figure 1 shows the conditions for π A (k) π π 1 π π 1 2 π +2 π +4 π π +1 π +3 π 1 π π A (k) Figure 1 A (k) From Lemma 11 and the definition of f (q), we see that [ + 1 q inv(π) ] f 2(n k)+1 (q) (23) For a positive integer k n, we denote by B (k) the set of permutations π on [ + 1] such that π 1 < π 2 < π 3 < < π > π +1 < π +2 > π +3 < < π > π Figure 2 shows the conditions for π B (k)
6 TIM HUBER AND AE JA YEE π π π π π +2 π +4 1 1 2 π π +1 π +3 π 1 π Figure 2 B (k) We now compute the generating function for permutations π B (k) q inv(π) From the definitions of A (k) and B (k), we see that for any k, 1 k n, A (k) B (k) B (k+1), where B (n+1) {π π 1 < π 2 < < π < π } Thus q inv(π) q inv(π) By iterating (24), we deduce n q inv(π) ( 1) k 1 π B (1) k1 π A (k) π A (k) π B (k+1) q inv(π) + ( 1) n q inv(π) (24) π B (n+1) q inv(π) (25) Note that B (1) A Therefore, it follows from (23) that (25) is equivalent to n + 1 f (q) ( 1) k 1 f 2(n k)+1 (q) + ( 1) n, which completes the proof k1 The following theorem gives the generating function for alternating permutations π in A by weight inv(π) + des(π) Theorem 22 Define Then T des (q)x (q; q) T des (q) ( 1) n q n x (q; q) ( 1) n q n x (q; q) π A q inv(π)+des(π)
COMBINATORICS OF GENERALIZED q-euler NUMBERS 7 Proof Recalling the definition of T (q) given by (21) and noting that des(π) n for any π A, we have π A q inv(π)+des(π) q n T (q) By comparing the recurrence relations for T(q) des and T (q), we see that T des q n T (q) (q) The dual functions corresponding to T (q) and T des (q) are discussed in the following theorem Theorem 23 Define and Then we have ˆT (q)x (q; q) ˆT des (q)x (q; q) ( 1) n q 2 +n x (q; q), ( 1) n q 2 n x (q; q) ( 1) n q 2 x (q; q) ( 1) n q 2 x (q; q) ˆT (q) T (q) and ˆT des (q) q T des (q) Proof We first note that the polynomials ˆT (q) satisfy ˆT 1 (q) 1 and n 1 + 1 ˆT (q) ( 1) + 1 n k 1 q 2(n k)2 (n k) ˆT+1 (q) + ( 1) n q 2 +n (26) k0 For any alternating permutation π A, define π by π π π π 1 π 2 π 1 Then π is clearly an up-down alternating permutation Furthermore Thus inv(π) + inv( π) n( + 1) π A q inv(π) Therefore, it suffices to show that π A q n() inv( π) q n() T (q 1 ) ˆT (q) q n() T (q 1 )
8 TIM HUBER AND AE JA YEE We now show that q n() T (q 1 ) satisfies the same recursion as ˆT (q) Substitute q k(+1) T +1 (q 1 ) for ˆT +1 in (26) Then n 1 + 1 ( 1) + 1 n k 1 q 2(n k)2 (n k) q k(+1) T +1 (q 1 ) + ( 1) n q 2 +n k0 k0 q n 1 + 1 ( 1) + 1 n k 1 q 2+n T +1 (q 1 ) + ( 1) n q 2+n q 1 q n() T (q 1 ), where the last equality follows from the recursion formula (22) for T (q) It follows from the recurrences for ˆT des and ˆT that ˆT (q) q n des ˆT (q) des This identity is equivalent to ˆT (q) q T(q) des since ˆT (q) T (q) q n T des (q) By equating coefficients of x in the generating functions for T (q) and ˆT (q) we obtain a special case of a formula due to Gauss [3, p 37] Corollary 24 For any nonnegative integer n, n q 2j2 +j + 1 n 2j + 1 j0 j0 q 2j2 j + 1 2j The classical q-secant numbers enumerate alternating permutations on the set [] for n 1 by the number of inversions Theorem 25 Define Then ( S (q)x (q; q) S (q) ) 1 ( 1) n x (27) (q; q) π A q inv(π) For a proof of Theorem 25, see [5, 25, 26, 27] The following theorem gives the generating function for alternating permutations π in A by weight inv(π) + des(π) Theorem 26 Define Then ( S des (q)x (q; q) S des (q) ) 1 ( 1) n q n+1 x (q; q) π A q inv(π)+des(π)
COMBINATORICS OF GENERALIZED q-euler NUMBERS 9 Proof Recalling the definition of S (q) given by (27) and noting that des(π) n 1 for any π A, we have π A q inv(π)+des(π) q n 1 S (q) By comparing the recursions for S des (q) and S (q), we see that S des (q) q n 1 S (q) The dual functions of S (q) and S des (q) are discussed in the following theorem Theorem 27 Let and Then Ŝ (q) π Ā ( Ŝ (q)x (q; q) ( Ŝ des (q)x (q; q) ( 1) n q n( 1) x (q; q) ) 1 ) 1 ( 1) n q (n 1) x (q; q) q inv(π) and Ŝ des (q) π Ā q inv(π) des(π) Proof We first note that the polynomials Ŝ(q) satisfy Ŝ0(q) 1 and n 1 Ŝ (q) ( 1) n k 1 q (n k)(2(n k) 1) Ŝ (q), for n 1 (28) k0 For any up-down alternating permutation π A, define π by π π π 1 π 2 π 2 π 1 Then π is clearly a down-up alternating permutation Furthermore inv(π) + inv( π) n( 1) Thus q inv(π) q n( 1) inv(π) π Ā π A q n( 1) S (q 1 ) The first statement in the theorem is equivalent to Ŝ (q) q n( 1) S (q 1 ) (29)
10 TIM HUBER AND AE JA YEE We now show that q n( 1) S (q 1 ) satisfies the same recursion as Ŝ(q) Substitute q k( 1) S (q 1 ) for Ŝ in (28) Then n 1 ( 1) n k 1 q (n k)(2(n k) 1) q k( 1) S (q 1 ) k0 k0 q n 1 ( 1) n k 1 q 2 n S (q 1 ) q 1 q n( 1) S (q 1 ), where the last equality follows from the recursion formula for S (q) We now show that Ŝ des (q) q inv(π) des(π), which is equivalent to k0 π Ā Ŝ des (q) q (n 1) S (q 1 ) since des(π) n for a permutation π Ā and Ŝ(q) q n( 1) S (q 1 ) by (29) We now show that q (n 1) S (q 1 ) satisfies the same recursion as Ŝdes (q), namely n 1 Ŝ des (q) ( 1) n k 1 q 2(n k)(n k 1) Ŝ des (q), for n 1 (210) Substitute q (k 1) S (q 1 ) for Ŝdes ] n 1 k0 [ k0 q in (210) Then ( 1) n k 1 q 2(n k)(n k 1) q (k 1) S (q 1 ) n 1 ( 1) n k 1 q (n 1) S (q 1 ) q 1 q (n 1) S (q 1 ), where the last equality follows from the recursion formula for S (q) 3 New q-secant Numbers The following theorem provides a combinatorial interpretation for a new class of secant numbers Recall the definition of half descent des(π o ) for a permutation π Theorem 31 Define ( S(q)x o (q; q) ) 1 ( 1) n q n2 +1 x (q; q)
COMBINATORICS OF GENERALIZED q-euler NUMBERS 11 Then, for n 1, Proof Let S(q) o q inv(π)+des(πo) π A g (q) q inv(π)+des(πo) π A From the definitions of inv(π) and des(π o ), it is clear that g 2 (q) 1 S o 2(q) The polynomials S o (q) satisfy S o 0(q) q 1 and n 1 S(q) o ( 1) n k 1 q (n k)2 S(q), o for n 1 (31) k0 We define g 0 (q) q 1 We will show that the polynomials g (q) satisfy (31) for n > 1 For a positive integer k n, let A (k) be the set of permutations π on [] such that π 1 < π 2 > π 3 < > π 1 < π, π +2 > π +4 > > π > π 1 > π 3 > > π +1 Figure 3 shows the conditions for π A (k) π 2 π 1 π 3 π 2 π 3 π 1 π π +2 π +4 π π +3 +1 π 2 π π 3 π 1 Figure 3 A (k) Note that there are (n k) 2 (n k) inversions in π +1 π +2 π for the permutation π A (k) Thus, from Lemma 11 and the definition of g, we see that π A (k) q inv(π)+des(π 1π 3 π 1 )+(n k) q (n k)2 g (q) (32)
12 TIM HUBER AND AE JA YEE For a positive integer k < n, we decompose A (k) into disjoint subsets as follows: A (k) {π π > π +2 } {π π < π +2 } {π π > π +2 > π 1 } {π π > π 1 > π +2 } {π π +2 > π > π 1 } {π π > π +2 > π +1 > π 1 } {π π > π +2 > π 1 > π +1 } {π π > π 1 > π +2 > π +1 } {π π +2 > π > π 1 > π +1 } {π π +2 > π > π +1 > π 1 } {π π +2 > π +1 > π > π 1 } {π π > π +2 > π +1 > π 1 } {π π > π 1 > π +1 } {π π +2 > π > π +1 > π 1 } {π π +2 > π +1 > π > π 1 } :B (k) C (k) D (k) E (k) (33) Note that B (k) is the set of alternating permutations π on [] such that π 1 < π 2 > π 3 < > π 1 < π, π > π +2 > > π > π 1 > π 3 > > π +1 > π 1, from which it is clear that B (k) π B (k) is a subset of A (k 1) Figure 4 shows the conditions for π 2 π 1 π 3 π 2 π 2 π π π 3 π π 3 π 1 +3 +1 π 1 π π +2 π +4 We define B (n) then Figure 4 B (k) A For a permutation π B (k) with 1 < k n, if π 3 > π 1, π 2 > π 3 > π 1, which shows that such π satisfy the conditions of C (k 1) Thus B (k) {π π 3 > π 1 } {π π 3 < π 1 } C (k 1) {π π 3 < π 1 } (34) We now compute the generating function for permutations π B (k) q inv(π)+des(π 1π 3 π 1 )+(n k) Let π B (k) If π 3 > π 1, then inv(π) + des(π 1 π 3 π 1 ) + (n k) inv(π) + des(π 1 π 3 π 3 ) + (n k + 1)
However, if π 3 < π 1, then COMBINATORICS OF GENERALIZED q-euler NUMBERS 13 inv(π) + des(π 1 π 3 π 1 ) + (n k) inv(π) + des(π 1 π 3 π 3 ) + (n k + 1) 1 In this case, we look for a permutation σ such that Let m be defined by inv(π) + des(π 1 π 3 π 1 ) inv(σ) + des(σ 1 σ 3 σ 3 ) + 1 π +m max{π +j π +j < π 2, j 1} There exists such an m since π is an alternating permutation, so that π 1 < π 2 It follows that π 1 π +m < π 2 We switch π 2 and π +m, and denote the resulting partition by π Switching π 2 with π +m results in a decrease of the inversion number, namely inv(π) inv( π) + 1 Moreover, since π 2i+1 π 2i+1 for i < k 1 and π 3 < π 1, Thus des(π 1 π 3 π 1 ) des( π 1 π 3 π 3 ) inv(π) + des(π 1 π 3 π 1 ) inv( π) + des( π 1 π 3 π 3 ) + 1 If π 2 were switched with π +m for m 0, then π 1 < π +m and π π +m Hence, from the definition of π, we see that π 1 < π 2 > π 3 < < π 2, π > π +2 > > π > π > π 1 > > π +1 > π 1, π > π 2 > π 1 > π 3, which shows π D (k 1) If π 2 and π 1 were switched, namely m 1, then π 2 < π i for i Hence, from the definition of π, we see that π 1 < π 2 > π 3 < < π 2, π > π +2 > > π > π > π 1 > > π +1 > π 1, π > π 1 > π 2 > π 3, which shows π E (k 1) Thus, by (33) and (34), for any k, 1 < k n, {π π B (k), π 3 > π 1 } { π π B (k), π 3 < π 1 } C (k 1) A (k 1) D (k 1) \ B (k 1) E (k 1)
14 TIM HUBER AND AE JA YEE Therefore, q inv(π)+des(π 1π 3 π 1 )+(n k) π 3 >π 1 π 3 >π 1 π A (k 1) q inv(π)+des(π 1π 3 π 3 )+(n k+1) + q inv(π)+des(π 1π 3 π 3 )+(n k+1) + q inv(π)+des(π 1π 3 π 3 )+(n k+1) π 3 <π 1 π 3 <π 1 π B (k 1) q inv(π)+des(π 1π 3 π 3 )+(n k) q inv( π)+des( π 1 π 3 π 3 )+(n k+1) q inv(π)+des(π 1π 3 π 3 )+(n k+1) (35) By iterating (35), for any n > 1, we deduce n 1 q inv(π)+des(πo) ( 1) n k 1 π B (n) k1 k1 π A (k) + ( 1) n 1 π B (1) q inv(π)+des(π 1π 3 π 1 )+(n k) q inv(π)+n 1, which is equivalent to n 1 g (q) ( 1) n k 1 q (n k)2 g (q) + ( 1) n 1 q n2 1 n 1 ( 1) n k 1 q (n k)2 g (q), k0 where the second equality holds since g 0 (q) q 1 Theorem 32 Define Then, for n 1, ( Ŝ(q)x o (q; q) Ŝ o (q) q 1 n S o (q) ) 1 ( 1) n q n(n 1) x (q; q) Proof Using the recurrences satisfied by S(q) o and Ŝo (q), we can easily prove that Ŝ(q) o q 1 n S(q) o for n 1 We omit the details The other half descent des(π e ) is discussed in the following theorem
COMBINATORICS OF GENERALIZED q-euler NUMBERS 15 Theorem 33 Define Then we have S e (q) π A q inv(π)+des(πe) (36) S e (q) qs o (q) Proof For a permutation π A, define the map σ by σ(π) ( + 1 π )( + 1 π 1 ) ( + 1 π 2 )( + 1 π 1 ) Then σ(π) is an alternating permutation in A with inv(π) inv(σ(π)) and des(π o ) + 1 des(σ(π) e ), since σ(π) e ( )( + 1 π 1 )( + 1 π 3 ) ( + 1 π 3 )( + 1 π 1 ) Therefore, it follows that S(q) e qs(q) o It follows from Theorems 31 and 33 that ( S(q)x e (q; q) ) 1 ( 1) n q n2 x (q; q) 4 New q-tangent numbers associated with odd indices Define T o (q) by T o (q)x (q; q) ( 1) n q n2 +n x (q; q) ( 1) n q n2 x (q; q) Then, by the definition of S(q) o in Theorem 31, T(q)x o ( 1) n q n2 +n x (q; q) (q; q) Thus T o (q) n k0 qs o (q)x (q; q) + 1 ( 1) n k q (n k)2 +(n k)+1 S(q) o (41) In [17], Huber proves that the coefficients (q; q) 1 ψ 2 (q) in y in (12) are T o (q), whose combinatorial interpretation is given the following theorem Theorem 41 For each nonnegative integer n, we have T(q) o q inv(π)+des(πo) π A
16 TIM HUBER AND AE JA YEE Proof Let For n 0, it is clear that f (q) π A q inv(π)+des(πo) f 1 (q) 1 T o 1 (q) For any positive integer n, we will show that f (q) satisfies the equation (41) For a positive integer k n, let A (k) be the set of permutations π on [ + 1] such that π 1 < π 2 > π 3 < > π 1 < π, π +2 > π +4 > > π > π > π 1 > > π +1 (42) Figure 5 shows the conditions for π A (k) From Lemma 11 and Theorem 31, we π 2 π 1 π 3 π π 2 π 3 π 1 π +2 π +4 π π +3 +1 π 2 π π 3 π π 1 see that for k n, π A (k) Figure 5 A (k) q inv(π)+des(π 1π 3 π 1 ) We decompose A (n) into disjoint subsets as follows: + 1 q (n k)2 S(q) o (43) A (n) {π π > π } {π π < π } {π π > π > π 1 } {π π > π 1 > π } {π π < π } :B (n) D (n) E (n) (44) Note that A B (n) D (n) For k < n, we decompose A (k) into disjoint subsets as follows: A (k) {π π > π +1 } {π π < π +1 } {π π > π +1 > π 1 } {π π > π 1 > π +1 } {π π < π +1 } {π π > π +2 > π +1 > π 1 } {π π +2 > π > π +1 > π 1 } {π π > π 1 > π +1 } {π π < π +1 } :B (k) C (k) D (k) E (k) (45)
COMBINATORICS OF GENERALIZED q-euler NUMBERS 17 Note that, since the permutations in B (k) satisfy the conditions of (42), B (k) is the set of alternating permutations π on [ + 1] such that π 1 < π 2 > π 3 < < π, π > π +2 > > π > π > π 1 > > π 1 Figure 6 shows the conditions for π B (k) π 2 π 1 π 3 π π π +2 2 π 3 π π +3 +1 π 1 π +4 π 2 π π 3 π π 1 Figure 6 B (k) Furthermore, for any positive k, 1 < k n + 1, B (k) {π π 3 > π 1 } {π π 3 < π 1 } D (k 1) {π π 3 < π 1 }, where B (n+1) A For k > 1, we now compute the generating function for permutations π B (k) q inv(π)+des(π 1π 3 π 1 ) Let π B (k) If π 3 > π 1, namely π D (k 1), then inv(π) + des(π 1 π 3 π 1 ) inv(π) + des(π 1 π 3 π 3 ) + 1 However, if π 3 < π 1, then inv(π) + des(π 1 π 3 π 1 ) inv(π) + des(π 1 π 3 π 3 ) In this case, we look for a permutation σ such that inv(π) + des(π 1 π 3 π 1 ) inv(σ) + des(σ 1 σ 3 σ 3 ) + 1 Let the positive integer m be defined by π +m max{π +j π +j < π 2, j 1} Since π 1 < π 2, there exists such an m We switch π +m with π 2 and denote the resulting partition by π Switching π 2 with π +m results in a decrease of the inversion number, namely inv(π) inv( π) + 1 Moreover, if π 2 was switched with π +m for m > 1, it is trivial that des(π 1 π 3 π 1 ) des( π 1 π 3 π 3 ),
18 TIM HUBER AND AE JA YEE since π 2i+1 π 2i+1 for i < k and π 3 < π 1 If π 2 was switched with π 1, since π 3 < π 1 < π 2, Thus, in either case, des(π 1 π 3 π 1 ) des( π 1 π 3 π 3 ) inv(π) + des(π 1 π 3 π 1 ) inv( π) + des( π 1 π 3 π 3 ) + 1 From the definition of π, if π 2 was switched with π 1, then π 1 < π 2 > π 3 < < π 4 > π 3 < π 2, π > π +2 > > π > π > π 1 > > π 1, π 2 < π 1, which shows that π E (k 1) If π 2 was switched with π +m for m > 1, then it follows from the maximality of π +m that π 1 < π 2 > π 3 < < π 4 > π 3 < π 2, π > π +2 > > π > π > π 1 > > π 1, π > π 2 > π 1 > π 3, which shows π C (k 1) Thus, for any k, 1 < k n + 1, {π π B (k), π 3 > π 1 } { π π B (k), π 3 < π 1 } C (k 1) D (k 1) E (k 1) A (k 1) \ B (k 1) (46) where the last equality of (46) follows from (45) Therefore, q inv(π)+des(π 1π 3 π 1 ) π 3 >π 1 π 3 >π 1 q π A (k 1) q inv(π)+des(π 1π 3 π 3 )+1 + q inv(π)+des(π 1π 3 π 3 )+1 + q inv(π)+des(π 1π 3 π 3 ) π 3 <π 1 π 3 <π 1 π B (k 1) q inv(π)+des(π 1π 3 π 3 ) q inv( π)+des( π 1 π 3 π 3 )+1 q inv(π)+des(π 1π 3 π 3 ) (47)
By iterating (47), we deduce π B (n+1) q inv(π)+des(πo) COMBINATORICS OF GENERALIZED q-euler NUMBERS 19 n ( 1) n k k1 which is equivalent to f (q) π A (k) n k0 q inv(π)+des(π 1π 3 π 1 )+(n k)+1 + ( 1) n + 1 ( 1) n k q (n k)2 +(n k)+1 S(q) o by (43) and the definitions of B (n+1), B (1), and S o 0(q) q 1 Define ˆT o (q) by ˆT o (q)x (q; q) ( 1) n q n2 x (q; q) ( 1) n q n2 n x Theorem 42 For a nonnegative integer n, we have ˆT o (q) q n T o (q) (q; q) π B (1) q inv(π)+n, Proof The theorem follows from the recursions satisfied by T o (q) and ˆT o (q) 5 New q-tangent numbers associated with even indices Let T e (q) be the polynomial satisfying T e (q)x (q; q) ( 1) n q n2 x (q; q) ( 1) n q n2 x (q; q) From the generating function for T(q), e it follows that n 1 + 1 T(q) e ( 1) n q n2 + ( 1) + 1 n k 1 q (n k)2 T+1(q) e (51) k0 Theorem 51 For a nonnegative integer n, T(q) e q inv(π)+des(πe) π A
20 TIM HUBER AND AE JA YEE Proof Let For n 0, it is clear that f (q) π A q inv(π)+des(πe) f 1 (q) 1 T e 1 (q) For any positive integer n, we will show that f (q) satisfies the recurrence (51) For a nonnegative integer k n, let A (k) be the set of permutations π on [ + 1] such that π 1 < π 2 > π 3 < < π > π +1, π +2 > π +4 > > π > π > π 1 > > π +3 Figure 7 shows the conditions for π A (k) π 2 π 1 π 3 π π 2 π 3 π 1 π +1 π +2 π +4 π 2 π π π 3 π π 1 +3 π A (k) Figure 7 A (k) From Lemma 11 and the definition of f, we see that q inv(π)+des(π 0π 2 π 4 π + 1 ) q + 1 (n k)2 f +1 (q) (52) For k < n, we decompose A (k) into disjoint subsets as follows: A (k) {π π +1 < π +2 } {π π +1 > π +2 } {π π +1 < π +2 < π } {π π +1 < π < π +2 } {π π +1 > π +2 } {π π +1 < π +3 < π +2 < π } {π π +3 < π +1 < π +2 < π } {π π +1 < π < π +2 } {π π +1 > π +2 } :B (k) C (k) D (k) E (k) (53) Note that B (k) is the set of alternating permutations π on [ + 1] such that π 1 < π 2 > π 3 < < π > π +1, π > π +2 > > π > π > π 1 > > π +3 > π +1 Figure 8 shows the conditions for π B (k)
COMBINATORICS OF GENERALIZED q-euler NUMBERS 21 π 2 π 1 π 3 π π +2 π +4 π 2 π 2 π π 3 π π 1 π π 3 π π 1 +3 +1 Figure 8 B (k) We define B (n) A For a permutation π B (k) with 1 k n, if π 2 < π, then π 1 < π 2 < π, which shows that such π satisfies the conditions of D (k 1) Thus B (k) {π π 2 < π } {π π 2 > π } D (k 1) {π π 2 > π } We now compute the generating function for permutations π B (k) q inv(π)+des(π 0π 2 π 4 π ) Let π B (k) If π 2 < π, then inv(π) + des(π 0 π 2 π 4 π ) inv(π) + des(π 0 π 2 π 4 π 2 ) However, if π 2 > π, then inv(π) + des(π 0 π 2 π 4 π ) inv(π) + des(π 0 π 2 π 4 π 2 ) + 1 In this case, we look for a permutation σ such that Let m be defined by inv(π) + des(π 0 π 2 π 4 π ) inv(σ) + des(σ 0 σ 2 σ 4 σ 2 ) π +m min{π +j π +j > π 1, 0 j + 1} There exists such an m since π is an alternating permutation, namely π 1 < π So, π +m π < π 2 We switch π +m with π 1 and denote the resulting partition by π Switching π 1 with π +m results in increasing of an inversion, namely Moreover, since π 2i π 2i for i < k, Thus inv(π) + 1 inv( π) des(π 0 π 2 π 4 π 2 ) des( π 0 π 2 π 4 π 2 ) inv(π) + des(π 2 π 4 π ) inv(π) + des(π 0 π 2 π 4 π 2 ) + 1 inv( π) + des( π 0 π 2 π 4 π 2 )
22 TIM HUBER AND AE JA YEE If π 1 was switched with π +m for m > 0, then π +j < π Hence, from the definition of π, we see that π 1 < π 2 > π 3 < < π 2 > π 1, π > π +2 > > π > π > π 1 > > π +1, π 2 > π > π 1 > π +1, which shows, from (53), that π C (k 1) If π 1 and π were switched, namely m 0, then π 1 > π i for i > Hence, from the definition of π, we see that π 1 < π 2 > π 3 < < π 2 > π 1, π > π +2 > > π > π > π 1 > > π +1, π 2 > π 1 > π > π +1 (54) Note that, since E (k 1) A (k 1), we see that E (k 1) {π π 2 > π 1 > π > π +1 } Therefore, (54) implies π E (k 1) Hence, by (53), for any k, 1 k n Therefore, {π π B (k), π 2 < π } { π π B (k), π 2 > π } C (k 1) D (k 1) E (k 1) A (k 1) \ B (k 1) q inv(π)+des(π 0π 2 π 4 π ) π 2 <π π 2 <π π A (k 1) By iterating (55), we deduce π B (n) q inv(π)+des(πe) n 1 k0 q inv(π)+des(π 0π 2 π 4 π 2 ) + q inv(π)+des(π 0π 2 π 4 π 2 ) + q inv(π)+des(π 0π 2 π 4 π 2 ) ( 1) n k 1 π A (k) π 2 >π π 2 >π π B (k 1) q inv(π)+des(π 0π 2 π 4 π ) q inv(π)+des(π 0π 2 π 4 π 2 )+1 q inv( π)+des( π 0 π 2 π 4 π 2 ) q inv(π)+des(π 0π 2 π 4 π 2 ) (55) + ( 1) n π B (0) q inv(π),
COMBINATORICS OF GENERALIZED q-euler NUMBERS 23 which is equivalent to n 1 + 1 f (q) ( 1) + 1 n k 1 q (n k)2 f +1 (q) + ( 1) n q n2 k0 Theorem 52 Define Then, for n 1, ˆT e (q)x (q; q) ( 1) n q n2 +n x (q; q) ( 1) n q n2 n x ˆT e (q) q n 1 T e (q) (q; q) Proof Note that n 1 + 1 ˆT (q) e ( 1) n q n2 +n + ( 1) + 1 n k 1 q (n k)2 (n k) e ˆT +1 (q) (56) k0 To prove Theorem 52, multiply both sides of (56) by q n+1 to obtain, for n 1, q n+1 e + 1 ˆT (q) ( 1) n q (n+1)2 + ( 1) 1 n 1 q n2 +1 (57) n 1 + 1 + ( 1) + 1 n k 1 q (n k)2 +k+1 e ˆT +1 (q) Note that ( 1) n q (n+1)2 + + 1 1 k1 ( 1) n 1 q n2 +1 ( 1) n 1 q n2 +1 (1 + q + + q 1 ) [ + 1 ( 1) n q n2 + 1 ] ( 1) n 1 q n2 (58) Inserting (58) into (57), we see that the recursion (51) for T(q) e is identical to the recursion for q n+1 e ˆT (q) in (57) 6 Symmetries of the q-euler numbers In [19], Ismail and Zhang conjectured that the polynomials T o (q) are symmetric about the middle coefficient(s) We prove their conjecture in the following theorem Theorem 61 The polynomials T (q), T o (q), and T e (q) are symmetric about the middle coefficient(s)
24 TIM HUBER AND AE JA YEE Proof For each alternating permutation π π 1 π 2 π, the permutation π π π π 1 is also an alternating permutation Recall that des(π e ) π 0 π 2 π and des(π e ) π 0 π π 2 From the definition of π, it follows that n( + 1) inv(π) + inv(π), (n + 1) inv(π) + des(π o ) + inv(π) + des(π o ), (n + 1) + 1 inv(π) + des(π e ) + inv(π) + des(π e ) Therefore, the inversion map π π is a bijection between {π inv(π) k} and {π inv(π) n( + 1) k}, {π inv(π) + des(π o ) k} and {π inv(π) + des(π o ) (n + 1) k}, {π inv(π) + des(π e ) k} and {π inv(π) + des(π e ) (n + 1) + 1 k} If k n( + 1)/2, then {π inv(π) k} {π inv(π) n( + 1) k} Otherwise, the two sets are disjoint It follows that, if n is even, the coefficients of T are symmetric about the term {π inv(π) n( + 1)/2} q n()/2 If n is odd, the coefficients are symmetric about the terms corresponding to q n()/2 and q n()/2 +1 The coefficients of T(q) o and T(q) e can similarly be seen to be symmetric about the middle term(s) The q-secant numbers discussed in Section 3 are symmetric about the middle coefficient Theorem 62 The polynomials S o (q) and Ŝo (q) are symmetric about the middle coefficient Proof It suffices to show that S o (q) is symmetric First note that the alternating permutation with the least weight is π 1 3 2 5 4 7 ( 1) ( 2) with inv(π) n 1 and des(π o ) 0; while the alternating permutation with the largest weight is π ( 1) ( 3) ( 2) 1 2 with inv(π) 2 and des(π o ) n 1 Thus S o (q) q n 1 + + q 2 n 1 It is clear that S2(q) o 1 is symmetric [ Suppose S o (q) is symmetric about 1 qk2 for x k < n Since the q-binomial coefficient is symmetric, y] q (n k)2 S(q) o is symmetric The exponent of the middle term is (n k) 2 + (k 2 ) + k 2 1 n 2 1
COMBINATORICS OF GENERALIZED q-euler NUMBERS 25 Therefore, S o (q) is symmetric about q n2 1 Note: It would be interesting find a combinatorial proof of Theorem 62 analogous to that of Theorem 61 We note in passing that the polynomials S (q), S des (q) and their dual polynomials Ŝ(q), Ŝdes (q) appearing in Theorems 25, 26, and 27 are the only polynomials considered in this paper that are not symmetric about the middle coefficient(s) Definition A polynomial p : C C of degree n is said to be reciprocal if ( ) 1 p(z) ±z n p (61) z The following corollary follows from Theorems 23, 32 42, 52, and 61 ˆT des Corollary 63 The polynomials (q), ˆT (q), o ˆT (q), e and Ŝo (q) are reciprocal More precisely, if f(q) { ˆT (q), ˆT (q), o ˆT (q), e Ŝo (q) n 0}, and f has des degree n, then f(1/q) f(q)/q n 7 Higher Order q-euler Numbers The tangent numbers of order k are defined by the Taylor series coefficients in the expansion of tan k z about z 0 Since d dz tan2 z d2 tan z, dz2 we see that for n 1, the numbers d dz tan2 z d tan z z0 dz (71) z0 each enumerate the alternating permutations on [ + 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 A by α(π) inv(π) + max(π) 1, β(π) inv(π) + des(π o ) + max(π) 3 + sign(π max(π) 1 π max(π)+1 ), 2 γ(π) inv(π) + des(π e ) + max(π) 2, where max(π) denotes the index of + 1 in π Throughout the section, we denote π j i π iπ i+1 π j for a given permutation π
26 TIM HUBER AND AE JA YEE Theorem 71 Let T (q), T(q), o T(q) e denote the q-analogues of the tangent numbers defined by Theorems 21, 41, and 51, respectively Define T (2) (q), T o (2) (q), T e (2) (q) by ( ) T (2) 2 (q)z T (q)z, (q; q) (q; q) ( ) T o (2) 2 (q)z T (q)z o, (q; q) (q; q) Then T (2) (q) π A q α(π), T e (2) (q)z (q; q) T o(2) (q) ( T e (q)z (q; q) π A q β(π), ) 2 T e(2) (q) π A q γ(π) Proof For a permutation π in A, let π + 1 for some k, 1 k n Then inv(π) {(i, j) i < < j and π i > π j } + {(i, j) i < j < and π i > π j } Thus, + {(i, j) < i < j and π i > π j } + {(, j) < j} {(i, j) i < < j and π i > π j } + inv(π1 1 ) + inv(π ) + 2(n k) + 1 +1 α(π) {(i, j) i < < j and π i > π j } + inv(π 1 1 ) + inv(π +1 ) We now show that T (2) (q) is the generating function for permutations in A with weight α The polynomials T (2) (q) satisfy n T (2) (q) T 1 1 (q)t 2(n k)+1 (q) (72) k1 By Lemma 11, the q-binomial coefficient in the summand on the right hand 1 side of (72) counts the inversions between the two sub-permutations π1 1 and π +1, namely {(i, j) i < < j and π i > π j } ; T 1 (q) and T 2(n k)+1 (q) count the inversions of π1 1 and π +1, respectively Therefore, T (2) (q) is the generating function for permutations π in A with weight α(π) The arithmetic interpretations for the polynomials T o (2) (q) and T e (2) (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 Sections 2 and 3 The proof for each interpretation is similar to that of Theorem 71
COMBINATORICS OF GENERALIZED q-euler NUMBERS 27 Theorem 72 Let S (q) and S(q) o denote the q-analogues of the secant numbers defined by Theorems 25 and 31, respectively Define S (2) (q) and S o (2) (q) by Then ( S (2) (q)z (q; q) S o (2) (q)z (q; q) S (2) (q) S o (2) (q) ( q δ(π), π A q υ(π) π A ) 2 S (q)z, (q; q) ) 2 S o (q)z (q; q) π A q ω(π), where ( ) + 1 min(π) δ(π) inv(π) 2 inv(π min(π)+1 ) + 2 min(π) + 1, υ(π) δ(π) + des(π o ) 2 des((π min(π)+1 ) o) + n min(π) + 3 2 ω(π) δ(π) + des(π e ) 2 des((π min(π)+1 ) e) + n min(π) + 2 + sign(π max(π) 1 π max(π)+1 ) 2 + sign(π 1) + sign(π 1 1) 2 Proof For a permutation π in A, let π +1 1 for some k, 0 k n Then inv(π) {(i, j) i < + 1 < j and π i > π j } + {(i, j) i < j < + 1 and π i > π j } + {(i, j) + 1 < i < j and π i > π j } + {(i, + 1) i < + 1} {(i, j) i < + 1 < j and π i > π j } + inv(π 1 ) + inv(π +2 ) + Thus, ( ) 2(n k) δ(π) {(i, j) i < + 1 < j and π i > π j } + inv(π1 ) inv(π +2 ) + 2 {(i, j) i < + 1 < j and π i > π j } + inv(π1 ) + inv(π ), (73) where π +2 π π π +3 π +2 We now show that S (2) (q) is the generating function of permutations in A with weight δ The polynomials S (2) (q) satisfy S (2) (q) n k0 +2 S (q)s 2(n k) (q) (74)
28 TIM HUBER AND AE JA YEE By Lemma 11, the q-binomial coefficient in the summand on the right hand side of (74) counts the inversions between the two sub-permutations π 1 and π +2, namely {(i, j) i < + 1 < j and π i > π j } ; S (q) and S 2(n k) (q) count the inversions of π1 and π +1, respectively, since π 1 A and π +1 A 2(n k) Therefore, S (2) (q) is the generating function for permutations π in A with weight δ(π) To obtain the claimed arithmetic interpretation for the polynomials S o (2) (q), note that, with the convention that des( ) 0, where denotes the empty permutation, des(π π 1 π min(π)+2 ) + sign(π 1) 1 Hence, by (73), we find that min(π) 1 2 des(π min(π)+2 π min(π)+4 π ) υ(π) δ(π) + des(π o ) 2 des((π min(π)+2 ) o) + n min(π) + 3 2 min(π) 1 δ(π) + des(π 1 π 3 π min(π) 2 ) des(π min(π)+2 π min(π)+3 π ) + 2 δ(π) + des(π 1 π 3 π min(π) 2 ) + des(π π 1 π min(π)+2 ) + sign(π 1) 1 {(i, j) i < min(π) < j and π i > π j } + inv(π min(π) 1 1 ) + des((π min(π) 1 1 ) o ) + inv(π min(π)+1 ) + des((π min(π)+1 ) o) + sign(π 1) 1, where π min(π)+1 π π π +3 π min(π)+1 Therefore, since S0(q) o q 1, by Lemma 11, Theorem 31, and the definition of S o (2) (q), we see that S o (2) (q) π A q υ(π) By squaring the generating function for S(q) e defined by (36) and using the fact that S(q) e S(q), o we readily observe that the weight ω(π) corresponds to the same enumeration for A as υ(π) We omit the details 8 Concluding Remarks Define the Bell polynomials B n,k (x 1, x 2,, x n k+1 ) via the generating function [12] ( ) x m z m n exp u 1 + B n,k (x 1, x 2, x n k+1 ) uk z n m! n! m1 n1 k1 Then, by Faà Di Bruno s formula [12, p 137] and Theorem 25, ( S (q) (q; q) ()! 1 + v1 ( 1) v v!b,v (α 1,, α v+1 ) ), (81)
COMBINATORICS OF GENERALIZED q-euler NUMBERS 29 where α k { ( 1) k/2 k!/(q; q) k, if k is even, 0, if k is odd From (21) and (27), we obtain T (q) n j0 + 1 ( 1) 2j n j S 2j (q) (82) Closed formulas and relations for the other q-euler numbers can be similarly derived As mentioned in the introduction, the generalized tangent numbers T(q) o are the polynomials arising in the coefficient of ψ 2 (q) in (q; q) y of (12) Constant multiples of the second order extensions T o (2) (q) appear in (12) as the coefficient of ψ 4 (q) in the corresponding expansion of (q; q) y for n 2 [17, Theorem 34] Ismail and Zhang [19, Theorem 41] prove that each y j can be expressed as a polynomial in certain elliptic parameters over the field of rational functions in q The authors of [19] suggest that polynomials appearing in the numerators of these expansions, denoted by D r,s,t (q), have interesting combinatorial properties Our study of T(q) o explicitly addresses the combinatorics of the polynomials D r,0,1 (q) and D r,0,2 (q) Recursion formulas for y j appearing in [16, 17] show that, in general, the polynomials D r,s,t (q) arise as linear combinations of finite products j (T o j +1(q)) m j The results of the present paper explain the symmetry of the polynomials appearing in expansions for y j observed by the authors of [19] 9 Acknowledgements The authors wish to thank Bruce Berndt for his helpful suggestions, corrections and encouragement References [1] D André Développement de sec x et de tan x C R Acad Sci Paris, 88:965 967, 1879 [2] D André Mémoire sur les permutations alternées J Math Pur Appl, 6:167 184, 1881 [3] G E Andrews The theory of partitions Cambridge Mathematical Library Cambridge University Press, Cambridge, 1998 Reprint of the 1976 original [4] G E Andrews and B C Berndt Ramanujan s lost notebook Part I Springer, New York, 2005 [5] G E Andrews and D Foata Congruences for the q-secant numbers European J Combin, 1(4):283 287, 1980 [6] G E Andrews and I Gessel Divisibility properties of the q-tangent numbers Proc Amer Math Soc, 68(3):380 384, 1978 [7] L Carlitz Enumeration of up-down permutations by number of rises Pacific J Math, 45:49 58, 1973 [8] L Carlitz Enumeration of up-down sequences Discrete Math, 4:273 286, 1973 [9] L Carlitz Permutations, sequences and special functions SIAM Rev, 17:298 322, 1975 [10] L Carlitz and R Scoville Enumeration of up-down permutations by upper records Monatsh Math, 79:3 12, 1975 [11] R Chapman and L K Williams A conjecture of Stanley on alternating permutations Electron J Combin, 14(1):Note 16, 7 pp (electronic), 2007
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