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](http://mathworld.wolfram.com/LogNormalDistribution.html) 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.



Polynomial related to lognormal moments
$s$-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 $2s<r$. 


