Polynomial related to lognormal moments Consider the polynomial:
$$p(x) = \sum_{k=0}^{r}(-1)^{r-k} {r \choose k} x^{k(k-1) / 2}$$
I want to show that
$$p(x) = (x - 1)^{\lceil r/2 \rceil} \, q(x)$$
That is, $(x - 1)^{\lceil r/2 \rceil}$ is a factor of $p(x)$. Even better, find a formula for the quotient polynomial $q(x)$.
This problem arises when trying to compute the central moments of the log-normal distribution. Raw moments of this distribution are given by $M_r = \left< x^r \right> = e^{r\mu + \frac{1}{2} r^2 \sigma^2}$. So central moments are
$$C_r = \left< (x - M_1)^r \right> = \sum_{k=0}^{r} {r \choose k} M_k \, (-M_1)^{r-k} \\ =
e^{r(\mu + \frac{1}{2}\sigma^2)} \sum_{k=0}^{r} (-1)^{r-k} {r \choose k} e^{\frac{1}{2}k(k-1)\sigma^2}$$
You will recognize the sum in this expression as the polynomial in the problem above. Using Mathematica or a similar program, it's easy to test that the alleged property holds for any particular $r$ you care to test. For numerical work, it's valuable to be able to express $C_r$ in factored form because there are catastrophic cancellations in the original form.
Presumably this comes down to some binomial identity with which I am unfamiliar.
 A: $s$-th derivative of $p$ at 1 equals $$\sum_{k=0}^{r}(-1)^{r-k} {r \choose k} g_s(k),$$
where $g_s$ is a polynomial of degree $2s$. This equals 0 if $2s<r$.  
A: Here is a combinatorial proof. Let 
$$t(x)=p(x+1)=\sum_{k=0}^r (-1)^{r-k}\binom rk (1+x)^{\binom k2}.$$
We want to show that $t(x)$ is divisible by $x^{\left\lceil r/2\right\rceil}$.
The coefficient of $x^j$ in $t(x)$ is the number of graphs with vertex set $\{1,2,\dots, r\}$, $j$ edges, and no isolated vertices. This can be proved easily using inclusion-exclusion or properties of exponential generating functions. The coefficients of these polynomials, with this combinatorial interpretation, can be found in the OEIS as sequence A054548 or A276639.
Since a graph with $r$ vertices and no isolated vertices must have at least $\left\lceil r/2\right\rceil$ edges, $t(x)$ is divisible by $x^{\left\lceil r/2\right\rceil}$.
It's interesting to note that the cumulants of the log-normal distribution are related to the inversion enumerator for labeled trees.
Additional comment: 
Here's a more detailed explanation of the combinatorial interpretation. Let
$$
u_n(x) = \sum_G x^{e(G)},
$$
where the sum is over all graphs $G$ with vertex set $[n]:=\{1,2,\dots,n\}$ and $e(G)$ is the number of edges of $G$,
and let 
$$
t_n(x) = \sum_H x^{e(H)},
$$
where the sum is over all graphs $H$ with vertex set $[n]$ and no isolated vertices.
Then 
$$u_n(x) = \sum_{k=0}^n \binom nk t_k(x)$$
since any graph with vertex set $[n]$, and with $n-k$ isolated vertices, can be specified by choosing a $k$-subset $S$ of $[n]$, constructing a graph without isolated vertices with vertex set $S$, and leaving the elements of $[n]\setminus S$ as isolated vertices.
 This may be inverted to give
$$t_n(x) = \sum_{k=0}^n (-1)^{n-k}\binom nk u_k(x).$$
But $u_n(x) = (1+x)^{\binom n2}$, since a graph with vertex set $[n]$ may be specified by including or not including each of the $\binom n2$ possible edges. Therefore
$$t_n(x) = \sum_{k=0}^n (-1)^{n-k}\binom nk (1+x)^{\binom n2}.$$
