I am trying to find a closed formula for the coefficient of an arbitrary term in product expansion n polynomials of the form $(x^1+x^2+x^3+x^4+...+x^a)(x^1+x^2+x^3+x^4+...+x^{(a+1)})(x^1+x^2+x^3+x^4+...+x^{(a+2)})$
I have solved it for the $2$ polynomial case and got a piecewise function (on accident from a different, seemingly unrelated problem with dice) but trying to find a way to generalize it, hopefully to $n$ such polynomials. Any tips or advice would be great!
Let's call your product $f(x)=\sum_{r=3}^{\infty}c_rx^r$, where we want a formula for $c_r$. I'll do the case $n=3$, you'll see the larger $n$ is, the messier it gets.
The main observation is, $$ (1-x)^3f(x)=(x-x^{a+1})(x-x^{a+2})(x-x^{a+3}) $$ So, $$ f(x)=x^3(1-x)^{-3}(1-x^a)(1-x^{a+1})(1-x^{a+2}) $$ Now by the binomial theorem, $$ (1-x)^{-3}=\sum_{r=0}^{\infty}{r+2\choose2}x^r $$ so we have to pick out the coefficient of $x^r$ in $$ x^3(1-x^a-x^{a+1}-x^{a+2}+x^{2a+1}+x^{2a+2}+x^{2a+3}-x^{3a+3})\sum_{r=0}^{\infty}{r+2\choose2}x^r $$ and that will be $$ {r-1\choose2}-{r-1-a\choose2}-{r-2-a\choose2}-{r-3-a\choose 2}+ $$
$$ \qquad\qquad+{r-2-2a\choose2}+{r-3-2a\choose2}+{r-4-2a\choose2}-{r-4-3a\choose2} $$