6
$\begingroup$

Let $(q)_n=(1-q)(1-q^2)\cdots(1-q^n)$ with $(q)_0:=1$. Define a $q$-exponential by $$e_q(z)=\sum_{n\geq0}\frac{z^n}{(q)_n}.$$ There is a notion of $q$-Eulerian polynomials of type $A$, see the reference. I like to introduce $q$-Eulerian polynomial of type B via the generating function $$\sum_{n\geq1}B_n(t,q)\frac{z^n}{(q)_n} =\frac{(e_q(z)-e_q(tz))\cdot(e_q(tz)+te_q(z))}{e_q(2tz)-te_q(2z)}.$$ Now, expand $B_n(t,q)$ as a polynomial $$B_n(t,q)=\sum_{k=0}^nB_{n,k}(q)t^k$$ and call $B_{n,k}(q)$ $q$-Eulerian numbers type B. The first few terms are: \begin{align} B_1(t,q)&=1+t, \\ B_2(t,q)&=1+(2q+4)t+t^2, \\ B_3(t,q)&=1+(7q^2+7q+9)t+(7q^2+7q+9)t^2+t^3. \end{align} Here is an earlier MO problem. This time, I'm interested in a specialized aspect of it. For instance, $B_n(t,1)$ become the ordinary Eulerian polynomials of type $B$ whose coefficients are listed at OEIS. On the other hand, the polynomials $B_n(t,-1)$ do not appear anywhere. You may look at the first few of these: $B_1(t,-1)=1, B_2(t,-1)=1+t, B_3(t,-1)=1+2t+t^2$, $$B_4(t,-1)=1+9t+9t^2+t^3 \qquad\text{and} \qquad B_5(t,-1)=1+12t+22t^2+12t^3+t^4.$$

QUESTION 1. Is there some interpretation of the coefficients in $B_n(t,-1)$?

QUESTION 2. Can you provide a proof for the unimodality of $B_n(t,-1)$?

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.