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Let us $p_k(z)=\sum_{i=0}^{k-1}z^{i}.$ For any $m$ find coefficient of the term $z^m$ in the polynomial $\prod_{j=1}^n p_{2j}(z).$

Here are some attempts $$p_2=1+z$$ $$p_4=1+z+z^2+z^3$$ $$p_6=1+z+z^2+z^3+z^4+z^5$$ $$\vdots$$ $$p_{2n}=1+z+z^2+z^3+\ldots+z^{2n-1}$$ Coefficient of the term $z^0:$ $1$

Coefficient of the term $z^1:$ $n-1$

Coefficient of the term $z^2:$ $(n-1)+(n-1)(n-2)+((n-1)+(n-2)+\cdots+1)=(n-1)(1+n-2+\frac{n}{2})=\frac12(n-1)(3n-2)=(n-1)^2+\frac{n(n-1)}{2}$

Coefficient of the term $z^3:$ $2(n-1)^2$

Coefficient of the term $z^4:$ $2(n-2)+(n-2)(n-3)+(n-1)^2+\frac{(n-2)(n-1)}{2}=(n-1)^2+(n-1)(n-2)+\frac{(n-2)(n-1)}{2}=\frac{(n-1)(5n-8)}{2}$ Coefficient of the term $z^5:4(n-2)+(n-1)(n-2)=(n-2)(n+3)$

Coefficient of the term $z^6:5(n-2)+(n-3)(n-2)+\frac{(n-2)(n-1)}{2}=(n-2)(n+2)+\frac{(n-2)(n-1)}{2}=\frac{(n-2)(3n+3)}{2}$

Coefficient of the term $z^7:2(n-1)(n-3)+2(n-3)^2=4(n-2)(n-3)$ $$\ldots$$ What is generalized formula?

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$$\prod_{j=1}^{n}p_{2j}(z)=\prod_{j=1}^{n}\frac{1-z^{2j}}{1-z}=\sum_{r\geq 0}\binom{r+n-1}{n-1}z^r \prod_{j=1}^{n}(1-z^{2j})$$ so the coefficient of $z^m$ in the RHS is given by a weigthed sum (weigths are given by binomial coefficients) of partition numbers. Have a look at this related question.

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  • $\begingroup$ Then you want to find the m-th derivative and set z=0 (and adjust by m!). $\endgroup$ – marty cohen Sep 10 '15 at 21:11

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