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 (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
$$F(x,x,z)=\sum_{n\geq0}x^n\sum_{\lambda\vdash n}\prod_{\square\in\lambda}\left(1-\frac{z}{h_{\square}^2}\right);$$
where $h_{\square}$ is the hook-length (for notations and references you may see *Theorems, problems and conjectures*).

My Question still waiting for an answer.Is there a unifying combinatorial right-hand side in $$\prod_{j\geq0}(1-tx^j)^{z-1}=?$$

**Remark 1.** One possibility is perhaps a refinement in line with (b) involving hook-lengths.

**Remark 2.** Even special cases, such as evaluation for specific numerical values of $t$ and/or $z$, would be interesting.

**Remark 3.** 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.

cycle index$\endgroup$ – john mangual Oct 21 '16 at 2:33