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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 $B_n(t,-1)$?

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 $B_n(t,-1)$?

Let $(q)_n=(1-q)(1-q^2)\cdots(1-q^n)$ with $(q;q)_0:=1$. 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 $B_n(t,-1)$?

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 $B_n(t,-1)$?

Let $(q)_n=(1-q)(1-q^2)\cdots(1-q^n)$ with $(q;q)_0:=1$. Define a $q$-exponential by $$e(z;q)=\sum_{n\geq0}\frac{z^n}{(q)_n}.$$ 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 $$\sum_{n\geq1}B_n(t,q)\frac{z^n}{(q)_n}=\frac{(e(z;q)-e(tz;q))\cdot(e(tz;q)+te(z;q))}{e(2tz;q)-te(2z;q)}.$$ 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 $B_n(t,-1)$?

Let $(q)_n=(1-q)(1-q^2)\cdots(1-q^n)$ with $(q;q)_0:=1$. 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 $B_n(t,-1)$?