It seems too easy to be P8.... Consider $\prod_{i=1}^{n}\frac{x^{m+i}-1}{x^i-1}=x^n\cdot\prod_{i=1}^{n}\frac{x^{m+i-1}-1}{x^i-1}+\prod_{i=1}^{n-1}\frac{x^{m+i}-1}{x^i-1}$ and use induction.Done

Mr.Trick wrote: It seems too easy to be P8.... Consider $\prod_{i=1}^{n}\frac{x^{m+i}-1}{x^i-1}=x^n\cdot\prod_{i=1}^{n}\frac{x^{m+i-1}-1}{x^i-1}+\prod_{i=1}^{n-1}\frac{x^{m+i}-1}{x^i-1}$ and use induction.Done Anyway, I think most people are not satisfied with this solution; they want to find the model that corresponds to this generating function.

Trivial with cyclotomic polynomials...

Nguyen wrote: Mr.Trick wrote: It seems too easy to be P8.... Consider $\prod_{i=1}^{n}\frac{x^{m+i}-1}{x^i-1}=x^n\cdot\prod_{i=1}^{n}\frac{x^{m+i-1}-1}{x^i-1}+\prod_{i=1}^{n-1}\frac{x^{m+i}-1}{x^i-1}$ and use induction.Done Anyway, I think most people are not satisfied with this solution; they want to find the model that corresponds to this generating function. Now that I have got an idea. First, reduce all factors of $x-1$, the $h(x)$ we are hoping to find becomes $\frac{(1+x+x^2+...+x^n)...(1+x+x^2+...+x^{n+m-1})}{1(1+x)...(1+x+x^2+...+x^{m-1})}$. Denote $1(1+x)...(1+x+x^2+...+x^{m-1})$ by $F_m(x)$, it remains to show that $\frac{F_{m+n}(x)}{F_m(x)F_n(x)}$ is a polynomial with positive coefficients. See https://oeis.org/A008302, $F_{m+n}(x)$ is the generating function of permutations of 1 through $m+n$ counted by amount of inversions. We count the amount of inversions in another way. Divide the permutation into "first $m$ entries" and "last $n$ entries". Lemma: The amount of inversions between the two groups is equal to the sum of the first $m$ entries subtracted by the sum of 1 through $m$. (It's so obvious) After you have chosen which $m$ numbers take the first $m$ places and which $n$ numbers take the last $n$ places, you still get to choose independently the internal ordering of each group. This is like counting the amount of inversions using merge sort. Therefore, if $h(x)$ is a polynomial where the $k$-th coefficient is the amount of $m$-subsets of $\{1,2,...,m+n\}$ with sum $k+(1+2+...+m)$, then $F_{m+n}(x)=F_m(x)F_n(x)h(x)$.

https://mathworld.wolfram.com/q-BinomialCoefficient.html