TITLE: Patterns of the zeros of the Faulhaber polynomials (modified)
QUESTION [37 upvotes]: Faulhaber polynomial of order $p \in \Bbb{N}$ is defined as the unique polynomial of degree $p+1$ satisfying
$$ S_{p}(n) = \sum_{k=1}^{n} k^p $$
for $n = 1, 2, 3, \cdots$. For example,
\begin{align*}
S_0(x) &= x, \\
S_1(x) &= \frac{x(x+1)}{2}, \\
S_2(x) &= \frac{x(x+1)(2x+1)}{6}, \\
S_3(x) &= \frac{x^2 (x+1)^2}{4}.
\end{align*}
In order to grasp some intuition on the partial decomposition of $1/S_p (x)$, I tried plotting the complex zeros of $S_p (x)$. The following graphics shows the distribution of the zeros of $S_{800}(x)$.

(The precision of the calculated zeros $z_j$ of $S_{800}(z)$ above satisfy $|f(z_j)| \leq 10^{-300}$.)
It turns out that they exhibits a very neat, yet still a strange pattern as seen above.
So far I have never heard of the topic related to this pattern, and I want to know (out of curiosity) if there are some results concerning the pattern of zeros of $S_p(x)$.
Addendum. The previous fuzzy graphics turned out to be a result of numerical error. Now I removed those errors.

REPLY [4 votes]: As an addendum to Antonio Vargas's answer, let's prove that the roots of $S_p$ are indeed symmetrically distributed around $-1/2$, or in other words that if $r$ is a root, then so is $-1-r$. A somewhat more precise result is that $S_p(-1-x) = S_p(x)$ when $p$ is odd, and $S_p(-1-x) = -S_p(x)$ when $p$ is even (thus confirming that the polynomial $T(x) := S_p(x-1/2)$ is an odd function (has only odd powers of $x$) if $p$ is even, and is an even function (has only even powers of $x$) if $p$ is odd). 
The key principle is that a polynomial is the zero polynomial iff it has infinitely many roots. For example, $f(x) = g(x)$ if $f(n) = g(n)$ for all $n \in \mathbb{N}$, and a polynomial $h$ is constant if $h(n) = h(0)$ for all $n \in \mathbb{N}$. If $h$ is a polynomial such that its first difference $(\Delta h)(x) := h(x) - h(x-1)$ satisfies $(\Delta h)(n) = 0$ for all $n \in \mathbb{N}$, then $h$ is constant with constant value $h(0)$. 
Thus we have $(\Delta S_p)(x) = S_p(x) - S_p(x-1) = x^p$ and so $S_p(-x) - S_p(-1-x) = (-1)^p x^p$. For fixed $p$, put $g(x) := S_p(-1-x)$, so that $g(x-1) = S_p(-1-(x-1)) = S_p(-x)$. Thus 
$$\Delta g(x) = g(x) - g(x-1) = S_p(-1-x) - S_p(-x) = (-1)^{p+1} x^p$$ 
which equals $\Delta S_p(x)$ if $p$ is odd, and $-\Delta S_p(x)$ if $p$ is even. Thus in the case where $p$ is odd, $\Delta(g-S_p) = 0$; applying the principle above, $g$ and $S_p$ differ by a constant: $S_p(-1-x) = S_p(x) + c$ for some constant $c$. For $x=0$, we note that $S_p(0) = 0$, and $S_p(0) - S_p(-1) = 0^p = 0$ so $S_p(-1) = 0$. It follows that $c = 0$, and we conclude $S_p(-1-x) = S_p(x)$ if $p$ is odd. 
The case where $p$ is even is wholly similar. There we derive $\Delta(g + S_p) = 0$, so $g(x) + S_p(x) = c$ for some constant $c$, and we argue as before that $c = 0$, and so in this case $S_p(-1-x) = -S_p(x)$.