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

Problem 6.1.
[T. Amdeberhan] Consider the formal product $$F(t,x,z):=\prod_{j=0}^{\infty}(1tx^j)^{z1}.$$ (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) $$F(t,x,2)=\sum_{n\geq0}Z(S_n,(1x)^{1},\dots,(1x^n)^{1})t^n.$$ (b) If $t=x$ then we get NekrasovOkounkov’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 righthand side in $$\prod_{j\geq0}(1tx^j)^{z1}=?$$
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 hooklength 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 $(q;q)_n=(1q)(1q^2)\cdots(1q^n)$ 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: $$\binom{\alpha}a_q+\sum_k\binom{\alpha}k_q2^{\alphak}B_{k,b}(q)= \binom{\alpha}b_q +\sum_k\binom{\alpha}k_q2^{\alphak}B_{k,a}(q).$$
QUESTIONS:
(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 nonnegative integers (plus $x+y+z>0$ for the latter).
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.