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6. Formal Products, Power Series, and Super Catalan Numbers

    1. Problem 6.1.

      [T. Amdeberhan] Consider the formal product $$F(t,x,z):=\prod_{j=0}^{\infty}(1-tx^j)^{z-1}.$$ (a) If $z=2$ then on the one hand we get Euler’s $$F(t,x,2)=\sum_{n\geq0}\frac{(-1)^nx^{\binom{n}2}}{(x;x)_n}t^n,$$ on the other we get Pólya’s formula (here the "cycle index decomposition", see Stanley’s EC2, p.19) $$F(t,x,2)=\sum_{n\geq0}Z(S_n,(1-x)^{-1},\dots,(1-x^n)^{-1})t^n.$$ (b) If $t=x$ then we get Nekrasov-Okounkov’s $$F(x,x,z)=\sum_{n\geq0}x^n\sum_{\lambda\vdash n}\prod_{\square\in\lambda}\left(1-\frac{z}{h_{\square}^2}\right).$$

      Question: Is there a unifying combinatorial right-hand side in $$\prod_{j\geq0}(1-tx^j)^{z-1}=?$$
          One possibility is perhaps a refinement in line with (b) involving hook-lengths.

      Even special cases, such as evaluation for specific numerical values of $t$ and/or $z$, would be interesting.

      Yet another direction is to find a hook-length expansion for $F(t,x,2)$. This could

      be fascinating because the result will connect the cycle index polynomial with hooks.
        • Problem 6.2.

          [T. Amdeberhan] 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. I like to introduce \it $q$-Eulerian polynomial of type B m ̊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:

          $$B_1(t,q)=1+t $$ $$B_2(t,q)=1+(2q+4)t+t^2$$ $$B_3(t,q)=1+(7q^2+7q+9)t+(7q^2+7q+9)t^2+t^3$$

          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 $B_{n,k}(q)$?
            • Problem 6.3.

              [T. Amdeberhan] Let $S_n$ be the permutation group and fix $k\in\mathbb{N}$. Then, $$\#\{\sigma\in S_n: \exists i, 1\leq i\leq n, \sigma(i)-i=k\}=\#\{\sigma\in S_n: \exists i, 1\leq i\lneq n, \sigma(i+1)-\sigma(i)=k\}.$$
                • Problem 6.4.

                  [T. Amdeberhan] Ira Gessel "dubbed" the name \mathit{ super Catalan} for the sequence $$S(m,n)=\frac{(2m)!(2n)!}{m!n!(m+n)!}$$ and offers an algebraic proof. Not combinatorial!

                  Note. The numbers $\frac{1}{2}S(m,n)$ are also integers.

                  I would like to extend the discussion by asking for a proof that the following numbers, which I call \mathit{super super Catalan numbers type 1}, are integers

                  $$S(x,y,z)=\frac{1}{3}\frac{(3x)!(3y)!(3z)!}{x!^2y!^2z!^2(x+y+z)!}$$ and the same question of integrality about the numbers, which I call \mathit{ super super Catalan numbers of type 2}, $$T(x,y,z)=\frac{x}3\frac{(3x)!(3y)!(3z)!}{x!^3y!^3z!^3(x+y+z)}$$ provided that $x, y, z$ are non-negative integers (plus $x+y+z>0$ for the latter).
                      It is evident that $$\frac{(3x)!(3y)!(3z)!}{x!^3y!^3z!^3}$$ are integral; the extra factor $x$ in the numerator of $T(x,y,z)$ is necessary, although $y$ or $z$ would do.

                  I don’t have a proof to these claims, but I am convinced of their truth. Even a generating function method is acceptable, yet a combinatorial proof is much desirable.

                      Cite this as: AimPL: Polyhedral geometry and partition theory, available at http://aimpl.org/polypartition.