# Linking formulas by Euler, Pólya, Nekrasov-Okounkov

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.

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