Hadamard product of specific type of power series I am consider the power series of the form $$F_n(t):=\frac{1}{\prod_{i=1}^n(1-t^i)}.$$ Given two power series $A(t)=\sum_{i\ge 0}{a_it^i}$ and $B(t)=\sum_{i\ge0}{b_it^i}$, their Hadamard product is defined to be $(A\star B)(t):=\sum_{i\ge0}{a_ib_it^i}$.
I am trying to compute $F_{n_1}(t)\star F_{n_2}(t)$ for two integers $n_1,n_2$. There is a technique in Stanley's Enumerative Combinatorics as well as here to do this by using residues of a contour integral.
This basic idea is to look at the function $G(t,s):=F_{n_1}(t)*F_{n_2}(s)$ (normal Cauchy product) and then do the substitution $G(re^{it},re^{-it})$ and rewriting this as $$\frac{P(r,z)}{Q(r,z)}$$, $z=e^{it}$, where $P,Q$ are polynomials.
Then one computes the sum of the residues of the poles (in terms of z) that have power series expansions around 0 of $$\frac{P(r,z)}{zQ(r,z)}$$ and the resulting function is the answer.
For the functions of type $F_n(t)$, when I attempt this calculation, it seems that the only poles are of the form $z=(1/r)^k$ for some natural number $k$.
This is you get $$\frac{1}{\prod_{i=1}^{n_1}{(1-(rz)^i)}}*\frac{1}{\prod_{i=1}^{n_2}{(1-(rz^{-1})^i)}}$$ and then rewriting this as a fraction of polynomials by multiplying top and bottom by $z^{n_2}$, where $n_2\ge1$.
The functions $z=(1/r)^k$ do not have a power series expansion around 0, so I seem to be getting an empty sum. I have obviously made a mistake somewhere, so I'd appreciate it if someone could tell me where.
 A: I'm not well familiar with the contour integral method, but here's what I know on the Hadamard product of rational functions.
Let $F_1(x) = \frac{P_1(x)}{Q_1(x)}$ and $F_2 = \frac{P_2(x)}{Q_2(x)}$, where $\deg P_1 < \deg Q_1$ and $\deg P_2 < \deg Q_2$. Let
$$
\begin{matrix}
Q_1(x) &=& (1-\lambda_1 x)\dots (1-\lambda_n x),\\
Q_2(x) &=& (1-\mu_1 x) \dots (1-\mu_m x).
\end{matrix}
$$
Then $F = F_1 \star F_2$ can be represented as $F(x) = \frac{P(x)}{Q(x)}$, where
$$
Q(x) = \prod\limits_{i,j}(1-\lambda_i \mu_j x)
$$
The polynomial $Q(x) = Q_1(x) \circ Q_2(x)$ itself is called the composed product of $Q_1(x)$ and $Q_2(x)$.
Note that $A \circ (B \cdot C) = (A \circ B) \cdot (A \circ C)$ for $B \cdot C$ being the Cauchy product of $B$ and $C$.
Due to this fact we can say that the composed product of the denominators of $F_{n_1}$ and $F_{n_2}$ is the total product of pairwise composed products of $(1-x^i)$ and $(1-x^j)$ for $1 \leq i \leq n_1$ and $1 \leq j \leq n_2$.
Then, one can prove that
$$
(1-x^i)\circ (1-x^j) = (1-x^{\operatorname{lcm}(i, j)})^{\gcd(i, j)},
$$
from which it follows that the denominator of $F$ is
$$
Q(x) = \prod\limits_{i=1}^{n_1} \prod\limits_{j=1}^{n_2} (1-x^{\operatorname{lcm}(i, j)})^{\gcd(i, j)}.
$$
Then, computationally, $P(x)$ can be found as
$$
P(x) \equiv F(x) Q(x) \pmod{x^{n_1 n_2}}.
$$
Unfortunately, I can't provide a more explicit formula for $P(x)$, but computational experiments suggest that

*

*$P(x)$ is usually not co-prime with $Q(x)$,

*$Q(x)$ is usually not divisible by $P(x)$.

Hence the result can't be expressed as $F(x) = \frac{1}{G(x)}$ for any polynomial $G(x)$ and the structure of $P(x)$ might be complicated to find it in a somewhat closed form.
