Let $(q;q)_n=(1-q)(1-q^2)\cdots(1-q^n)$ with $(q;q)_0:=1$. Define a $q$-exponential by $$e(z;q)=\sum_{n\geq0}\frac{z^n}{(q;q)_n}.$$ There is a notion of $q$-Eulerian polynomials, see the reference. I like to introduce $q$-Eulerian polynomial of type B via the generating function $$\sum_{n\geq1}B_n(t,q)\frac{z^n}{(q;q)_n}=\frac{(e(z;q)-e(tz;q))(e(tz;q)+te(z;q))}{e(2tz;q)-te(2z;q)}.$$ 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. Here are the first few terms: \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}

Claim. if $a, b\in\Bbb{N}$ and $\alpha=a+b+1$, then the symmetric relation holds: $$\binom{\alpha}a_q+\sum_k\binom{\alpha}k_q2^{\alpha-k}B_{k,b}(q)= \binom{\alpha}b_q +\sum_k\binom{\alpha}k_q2^{\alpha-k}B_{k,a}(q).$$


(a) I don't have a proof for my claim which seems very true though. Do you?

(b) Is there a combinatorial interpretation for these polynomials $B_n(t,q)$ or the Eulerian numbers $B_{n,k}(q)$? You might be inspired by the reference.

  • $\begingroup$ I proved (a), but I don't know anything about (b). $\endgroup$ – user82588 Oct 24 '16 at 15:26
  • $\begingroup$ @Nemo: do you like to share your proof? $\endgroup$ – T. Amdeberhan Oct 25 '16 at 11:56
  • $\begingroup$ Why type B? Is it somehow related to the ABCDEF nomenclature? $\endgroup$ – მამუკა ჯიბლაძე Oct 25 '16 at 13:59
  • $\begingroup$ Type B because there are types A & B in the classical literature ($q=1$). $\endgroup$ – T. Amdeberhan Oct 25 '16 at 15:38
  • $\begingroup$ Could you please provide a link? $\endgroup$ – მამუკა ჯიბლაძე Oct 26 '16 at 7:57

$\bf{Step~1}.$ $B_{n,a}(q)=B_{n,n-a}(q)$.

$\it{Proof}$. Write $$ \sum_{n\geq1}\dfrac{B_n(t,q)}{t^{n/2}}\frac{z^n}{(q;q)_n}=\frac{e(z/\sqrt{t};q)-e(z\sqrt{t};q)}{\dfrac{e(2z\sqrt{t};q)}{\sqrt{t}}-\sqrt{t}~e(2z/\sqrt{t};q)}\left(\dfrac{e(z\sqrt{t};q)}{\sqrt{t}}+\sqrt{t}~e(z/\sqrt{t};q)\right). $$ The rhs is symmetric under $t\to 1/t$, so $\dfrac{B_n(t,q)}{t^{n/2}}$ is also symmetric: $$ \dfrac{B_n(t,q)}{t^{n/2}}=t^{n/2}{B_n(1/t,q)}.\tag{1} $$ We know that $B_n(t,q)=B_{n,0}t^0+...$ starts with the power $t^0$. Due to $(1)$ this means that the highest power of $t$ in the polynomial $B_n(t,q)$ is $n$, and $B_{n,a}(q)=B_{n,n-a}(q)$ as required.

$\bf{Step~2}.$ $\displaystyle{\binom{\alpha}a_q+\sum_k\binom{\alpha}k_q2^{\alpha-k}B_{k,b}(q)= \binom{\alpha}b_q +\sum_k\binom{\alpha}k_q2^{\alpha-k}B_{k,a}(q)},~ a,b,\in \mathbb{N},~a+b+1=\alpha.$

$\it{Proof}$. According to q-binomial theorem $e(z;q)=\sum_{n\geq0}\frac{z^n}{(q;q)_n}=\frac{1}{(z;q)_\infty}$. Now we equate the coefficients of $z^n$ in both sides of \begin{align} \left(\sum_{l\geq0}\frac{(2tz)^l}{(q;q)_l}-t\sum_{l\geq0}\frac{(2z)^l}{(q;q)_l}\right)\cdot\sum_{k\geq1}B_k(t,q)\frac{z^k}{(q;q)_k}={(e(z;q)-e(tz;q))(e(tz;q)+te(z;q))}\\ =\left(\sum_{l\geq0}\frac{z^l}{(q;q)_l}-\sum_{l\geq0}\frac{(tz)^l}{(q;q)_l}\right)\left(\sum_{l\geq0}\frac{(tz)^l}{(q;q)_l}+t\sum_{l\geq0}\frac{z^l}{(q;q)_l}\right) \end{align} and obtain $$ \sum _{k=1}^n \binom{n}{k}_q\frac{B_k(t,q) \left(t^{n-k}-t\right) }{2^{k-n}}-\sum _{k=1}^n \left(1-t^k\right) \left(t^{n-k}+t\right) \binom{n}{k}_q=0. $$ The coefficient of $t^{a+1}$ in the above formula is (note that for $a\in \mathbb{N}$ we have $a+1\neq1$ and for $b=n-a-1\in \mathbb{N}$ we have $a+1\neq n$, so the 'boundary' terms with $t$ and $t^n$ don't make any contribution in the formula below) $$ -\sum _{k=1}^n\binom{n}{k}_q \frac{B_{k,a}(q) }{2^{k-n}}+\sum _{k=1}^n \binom{n}{k}_q\frac{B_{k,a+1+k-n}(q)}{2^{k-n}}+\binom{n}{a}_q-\binom{n}{b}_q=0. $$ Finally, taking into account $B_{k,a+1+k-n}(q)=B_{k,k-b}(q)=B_{k,b}(q)$ we have $$ \binom{n}{b}_q+\sum _{k=1}^n\binom{n}{k}_q \frac{B_{k,a}(q) }{2^{k-n}}=\binom{n}{a}_q+\sum _{k=1}^n \binom{n}{k}_q\frac{B_{k,b}(q)}{2^{k-n}}. $$

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  • $\begingroup$ Step 1 was proved in arxiv.org/pdf/1207.4045.pdf as well. I'll look into the rest. $\endgroup$ – T. Amdeberhan Oct 25 '16 at 15:42
  • $\begingroup$ @T.Amdeberhan , what logic did lead you to this conjecture? $\endgroup$ – user82588 Oct 25 '16 at 15:52
  • $\begingroup$ I tried generalizing classical symmetrical relations. There's already one such q-analogue for type A by Guo-Niu Han. $\endgroup$ – T. Amdeberhan Oct 25 '16 at 16:33
  • $\begingroup$ I checked the proof. Correct. $\endgroup$ – T. Amdeberhan Nov 2 '16 at 0:44

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