Let us consider the set $[n]=\{1,\ldots,n\}$ and all of its partitions into exactly $m$ blocks, but let us allow each block to be internally ordered. For example, taking $n=6$ and $m=2$, we will distinguish between $\{(1,2),(3,5,4,6)\}$ and $\{(1,2),(5,3,4,6)\}$, but we will consider $\{(1,2),(3,5,4,6)\}$ and $\{(3,5,4,6),(1,2)\}$ as the same object.
For each partition $\pi$ of $[n]$ into $m$ parts as above, we will define a notion of weight as follows: for each part of $P$ of $\pi$, consider the number of elements in $P$ that are smaller (as positive integers) than the first element of $P$, this will be denoted by $w(P)$. Then, the weight of $\pi$ is defined by $$w(\pi) := \sum_{P\in \pi} w(P)$$
For instance, the weight of $\pi_1=\{(1,2),(3,5,4,6)\}$ is $w(\pi_1) = 0 + 0 = 0$, the weight of $\pi_2=\{(1,2),(5,3,4,6)\}$ is $w(\pi_2) = 0 + 2 = 2$, etc.
Denote by $W(\ell,n,m)$ the number of partitions $P$ of $[n]$ into $m$ parts and having weight $\ell$.
Problem: Show that for each $n,m$, the polynomial: $$P_{n,m}(x) := \sum_{\ell=0}^{n-m} W(\ell,n,m)\, x^{\ell}$$ has all of its complex roots lying on the unit circle $|z|=1$.
For example, if $n=6$ and $m=2$, one may calculate: $$P_{6,2}(x) = 274 + 404x + 444x^2+404x^3+274x^4.$$
For $n=7$ and $m=3$,
$$P_{7,3}(x)=1624 + 2954x + 3444x^2 + 2954x^3 + 1624x^4$$
I was originally interested in proving that $P_{n,m}(x)$ was log-concave, and I discovered that this unit-circle-rootedness pattern holds for all small values of $n$ and $m$.
