How to caluclate the coefficients of finite product $$\prod\limits_{i=1}^{n}(z^i-1)=\sum\limits_{k=0}^{\frac{n(n+1)}{2}}a_{n,k}z^k$$ effectively?
I have figured out a recursive rule for $n>1$:
$$a_{n,k} = a_{n-1, k} - (-1)^na_{n-1,\frac{n(n-1)}{2}-k}$$
But I feel that there should be something much efficient than this.
And also there is a sequence in OEIS: https://oeis.org/A231599
Let $$P= \prod_{i=1}^n (z_i-1)\ \ \ \ \ \text{ and } \ \ \ \ \ Q=\prod_{i=1}^n (t-z_i) \ . $$ The main goal here is to compute the coefficients of $p(z)=P(z, \dots, z^n) \in \mathbb{C}[z]$. Think $Q$ as a one variable polynomial in $t$ and observe that $$Q(1)=(-1)^n \prod_{i=1}^n (z_i-1)= (-1)^n P(z_1, \dots, z_n)\ .$$ On the other hand, according to Vieta's theorem, we have that $$Q(t)= \sum_{i=0}^n\sigma_i(z_1, \dots , z_n)\cdot t^n $$ where $\sigma_1, \dots , \sigma_n$ denote the elementary symmetric functions in $z_1, \dots, z_n$. Summing up $$P(z_1, \dots, z_n)=(-1)^n\sum_{i=0}^n\sigma_i(z_1, \dots , z_n) \ , $$ from where the identity $$p(z)=(-1)^n \sum_{i=0}^n \sigma_i(z, \dots, z^n) \ .$$
In order to find the coefficient $a_k$ of $p(z)=a_0+ a_1z+ \dots +a_nz^n$ you have now to count (with signs!) how many times the monomial $z^k$ appears in the sum on the right. More precisely we have that $$a_k = \sum_{i=0}^k (-1)^i \ \# \big(z^k \text{ appears in } \sigma_i(z, \dots , z^n) \big) \ .$$ By inspecting the definition of the elementary symmetric functions one can easily see that $z^k$ appears in the expression $\sigma_i(z, \dots , z^n)$ exactly $A_{k,i}$ times, where $A_{k,i}$ denotes the number of positive integer solutions of the equation $k=x_1+ \dots +x_i$ with $x_1< \dots <x_i<n$. Summarising, we have the formula $$a_k = \sum_{i=0}^k (-1)^i A_{k,i} \ .$$ The point now is to understand the coefficients $A_{k,i}$. These numbers have been studied extensively in enumerative combinatorics but I have no reference on the top of my head.