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

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13 pages
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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 integral 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 the variable 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=1An,kt k?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,kt k?1(1 + t)n+1?2k.

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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 ≤ b ( n + 1) / 2 c ) with positive integral 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 the variable 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 nk =1 A n,k t k 1 (see [FS70, Fo09, St97]) may be de-fined by X k n t k 1 A n t () t ) n +1 ( n N ) . = k 1 ( It is well known (see [FS70]) that there are nonnegative integers a n,k such that b ( n +1) / 2 c 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 (see for instance [Br08] for the definitions) 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 ≤ b ( n +1) / 2 c , with a 1 , 1 = 1, and a n,k = 0 for k 0 or k > b ( n +1) / 2 c .
Date : March 15, 2011.
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