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TWO NEW TRIANGLES OF q INTEGERS VIA q EULERIAN POLYNOMIALS OF TYPE A AND B

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Niveau: Supérieur, Doctorat, Bac+8
TWO NEW TRIANGLES OF q-INTEGERS VIA q-EULERIAN POLYNOMIALS OF TYPE A AND B GUONIU HAN, FREDERIC JOUHET AND JIANG ZENG Abstract. The classical Eulerian polynomials can be expanded in the basis tk?1(1 + t)n+1?2k (1 ≤ k ≤ b(n+ 1)/2c) with positive integers coefficients. This formula implies both the symmetry and the unimodality of the Eulerian polynomials. In this paper, we prove a q-analogue of this expansion for Carlitz's q-Eulerian polynomials as well as a similar formula for Chow-Gessel's q-Eulerian polynomials of type B. We shall give some applications of these two formulae, which involve two new sequences of polynomials in q with positive integral coefficients. An open problem is to give a combinatorial interpretation for these polynomials. 1. Introduction The Eulerian polynomials An(t) := ∑n k=1 An,ktk?1 (see [FS70,Fo09, St97]) may be de- fined by ∑ k≥1 kntk = An(t)(1? t)n+1 (n ? N). It is well known (see [FS70]) that there are nonnegative integers an,k such that An(t) = b(n+1)/2c ∑ k=1 an,ktk?1(1 + t)n+1?2k.

  • han defined

  • integers satisfying

  • follows then

  • nonnegative integer

  • secant numbers

  • let a?n

  • fn

  • using doubloon model

  • integral coefficients


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Language English
TWO NEW TRIANGLES OF q -INTEGERS VIA q -EULERIAN POLYNOMIALS OF TYPE A AND B ´ ´ GUONIU HAN, FREDERIC JOUHET AND JIANG ZENG Abstract. The classical Eulerian polynomials can be expanded in the basis t k 1 (1 + t ) n +1 2 k (1 k ≤ ⌊ ( n + 1) 2 ) with positive integers coefficients. This formula implies both the symmetry and the unimodality of the Eulerian polynomials. In this paper, we prove a q -analogue of this expansion for Carlitz’s q -Eulerian polynomials as well as a similar formula for Chow-Gessel’s q -Eulerian polynomials of type B . We shall give some applications of these two formulae, which involve two new sequences of polynomials in q with positive integral coefficients. An open problem is to give a combinatorial interpretation for these polynomials.
1. Introduction The Eulerian polynomials A n ( t ) := P kn =1 A nk t k 1 (see [FS70, Fo09, St97]) may be de-fined by k X 1 k n t k =(1 A n t () t n ) +1 ( n N ) It is well known (see [FS70]) that there are nonnegative integers a nk such that ( n +1) 2 A n ( t ) = X a nk t k 1 (1 + t ) n +1 2 k (1.1) k =1 For example, for n = 1      4, the identity reads A 1 ( t ) = 1  A 2 ( t ) = 1 + t A 3 ( t ) = (1 + t ) 2 + 2 t 2  A 4 ( t ) = (1 + t ) 3 + 8 t (1 + t ) In particular, this formula implies both the symmetry and the unimodality of the Eulerian numbers ( A nk ) 1 k n for any fixed n . The coefficients a nk defined by (1.1) satisfy the following recurrence relation: a nk = ka n 1 k + 2( n + 2 2 k ) a n 1 k 1 (1.2) for n 2 and 1 k ≤ ⌊ ( n +1) 2 , with a 1 1 = 1, and a nk = 0 for k 0 or k > ( n +1) 2 . Date : March 15, 2012.
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