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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 follow this link).

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.

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  • $\begingroup$ does Nekrasov-Okounkov formula really fall out of cycle index for permutation groups?? $\endgroup$ – john mangual Oct 21 '16 at 2:31
  • $\begingroup$ never mind -- NO formula generalizes the cycle index $\endgroup$ – john mangual Oct 21 '16 at 2:33
  • $\begingroup$ Neither one follows from the other. $\endgroup$ – T. Amdeberhan Oct 21 '16 at 2:36

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