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A SYMMETRICAL q EULERIAN IDENTITY

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A SYMMETRICAL q-EULERIAN IDENTITY GUO-NIU HAN, ZHICONG LIN, AND JIANG ZENG Abstract. We find a q-analog of the following symmetrical identity involving binomial coefficients (n m ) and Eulerian numbers An,m: ∑ k≥0 ( a+ b k ) Ak,a?1 = ∑ k≥0 ( a+ b k ) Ak,b?1, which was published by Chung, Graham and Knuth (J. of comb., Vol. 1, Number1, 29-38, 2010). We shall give two proofs using generating function and bijections, respectively. 1. Introduction The Eulerian polynomials An(t) are defined by the exponential generating function ∑ n≥0 An(t) zn n! = (1 ? t)ez ezt ? tez . (1.1) The classical Eulerian numbers An,k are the coefficients of the polynomial An(t), i.e., An(t) = ∑n k=0 An,ktk. Recently, Chung, Graham and Knuth [2] noticed that if we modify the value of A0(t), which is 1 by (1.1), by taking the convention that A0(t) = A0,0 = 0, then the following symmetrical identity holds: ∑ k≥0 (a + b k ) Ak,a?1 = ∑ k≥0 (a + b k ) Ak,b?1 (a, b > 0

  • exponential generating

  • combinatorial proof

  • over all ordered

  • eulerian polynomials

  • involving binomial

  • gessel's hook

  • let a0

  • symmetrical identity


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Language English
A SYMMETRICALqEULERIAN IDENTITY
GUONIU HAN, ZHICONG LIN, AND JIANG ZENG
Abstract.We find aqanalog of the following symmetrical identity involving binomial   n coefficients and Eulerian numbersAn,m: m     X X a+b a+b Ak,a1=Ak,b1, k k k0k0 which was published by Chung, Graham and Knuth (J. of comb., Vol. 1, Number1, 2938, 2010). We shall give two proofs using generating function and bijections, respectively.
1.Introduction TheEulerian polynomialsAn(t) are defined by the exponential generating function X n z z(1t)e An(t) =.(1.1) zt z n!ete n0 The classicalEulerian numbersAn,kare the coefficients of the polynomialAn(t), i.e., P n k An(t) =An,ktChung, Graham and Knuth [2] noticed that if we modify. Recently, k=0 the value ofA0(t), which is 1 by (1.1), by taking the convention thatA0(t) =A0,0= 0, then the following symmetrical identity holds:     X X a+b a+b Ak,a1=Ak,b1(>a, b 0).(1.2) k k k0k0 Equivalently, instead of (1.1), we define the Eulerian polynomials by the generating func tion X n z z tz z(1t)e ee An(t) =1 =.(1.3) zt z zt z n!ete ete n0 At the end of [2], the authors asked for, among other unsolved problems, aqanalog of (1.2). The aim of this paper is to give such an extension and provide two proofs, of which one is analytical and another one is combinatorial. We first introduce someqnotations. Theqshifted factorial (z;q)nis defined by (z;q)n:= Q n1 i (1zq) for any positive integernand (z;q)0The= 1. qexponential functione(z;q) i=0 is defined by X n z e(z;q) :=. (q;q)n n0 1