Degree of Polynomial in Centered Moments of Gamma$(n,1)$ I'm interested in the degree of the polynomial in $n$ of the expression for the $k$-th central moment
$$
E((X_n - n)^k)
$$
where $X_n$ is a Gamma$(n,1)$ random variable, that is, the sum of $n$ independent exponential variates of parameter $1$. It is easy to show that
$$
E((X_n - n)^2) = n; \quad E((X_n - n)^3) = 2n
$$
Via Mathematica I verify for $k$ up to $150$ that the resulting polynomial in the variable $n$ is of order $\lfloor k/2 \rfloor$. Alas, I have not found a general demonstration. Developing the expression, and using the expression for the raw moments of this distribution, one easily show that
$$
E((X_n - n)^k) = \sum_{i=0}^k \binom{k}{i} (-1)^{k-i} n^{k-i} \frac{(n+i-1)!}{(n-1)!}
$$
But I'm stuck there. I tried to express the ratio of factorials via the  Stirling numbers expansion of the Pochhammer symbol without success.
Also, Mathematica computes that 
$$
E((X_n - n)^k) = U(-k, 1-k-n, -n)
$$
where $U$ is the Tricomi confluent hypergeometric function.
Any help is welcomed
Thanks in advance!
 A: To find the degree of this polynomial
was Problem 11403 in the Dec. 2008 issue 
of the American Mathematical Monthly.
The function was there called 
$f_n(x)$ and its degree is indeed $\lfloor n/2\rfloor.$
In your notation we have $E[(X_n-n)^k]=f_k(n).$
Here is the first paragraph of the published solution
(March 2011):

The degree of $f_n$ 
  is $\lfloor n/2\rfloor.$ This follows immediately 
  from the stronger statement that the coefficient of
  $x^r$ in $f_n(x)$ is the number of derangements of 
  $[n]$ with $r$ cycles, since each cycle must
  have at least two elements. 
  Here $[n]=\{1,\dots,n\}$, and a derangement is a permutation
  with no fixed points.

The published solution also mentions other interesting combinatorial 
interpretations of $f_n(x)$.

Here is my never-before-published, inelegant solution.
For $n\geq 0$, the binomial coefficients satisfy
 ${n+1\choose i}={n\choose i}+{n\choose i-1}$
and ${n\choose i}=0$ unless $0\leq i\leq n$.
Thus, for any function $h(i)$ we get,
$$\sum_{i=0}^{n+1} {n+1\choose i} h(i) = \sum_{i=0}^n {n\choose i} [h(i)+ h(i+1)].\tag 1$$
For real $x$ and integers  $0\leq i\leq n$,
define $g(n,i)=(-x)^{n-i}\prod_{j=0}^{i-1}(x+j)$ so that
$f_n(x)= \sum_{i=0}^n {n\choose i} g(n,i)$.
Note that $f_0(x)\equiv 1$ and $f_1(x)\equiv 0$.
We have
$g(n+1,i)+ g(n+1,i+1)=i\,g(n,i)$, so plugging this into (1) gives
$$ \sum_{i=0}^{n+1} {n+1\choose i} g(n+1,i)= \sum_{i=0}^n {n\choose i} i\,g(n,i),\quad n\geq0.\tag 2$$
The binomial coefficients satisfy
 ${n\choose i} i =  n {n-1\choose i-1}$,
 while for $i\geq 1$ we easily check
 that $g(n,i)=g(n-1,i-1) (x+(i-1))$.
 Using these and equation (2) gives us, for $n\geq 1$,
\begin{eqnarray*}
 \sum_{i=0}^n {n\choose i}i\, g(n,i)
&=& n \sum_{i=1}^{n} {n-1\choose i-1} g(n,i) \\[3pt]
&=& n \sum_{i=1}^n {n-1\choose i-1} g(n-1,i-1) (x+(i-1)) \\[3pt]
&=& n \sum_{i=0}^{n-1} {n-1\choose i} g(n-1,i) (x+i) \\[3pt]
&=& n \left(x f_{n-1}(x) + f_{n}(x)\right).
\end{eqnarray*}
This shows that the polynomials $f_n$ satisfy the recurrence
$$f_{n+1}(x)=n\left(xf_{n-1}(x)+f_{n}(x)\right),\ n\geq 1.\tag 3$$
Here are the first few polynomials.
$$f_0(x)=1,\ f_1(x)= 0,\ f_2(x)= x,\ f_3(x)= 2x,\ f_4(x)= 3x^2+6x. $$
Induction now shows that $f_n$ has non-negative coefficients
so that there is no cancellation in (3), and hence
$$\deg(f_{n+1})=\max(\deg(f_{n-1})+1, \deg(f_{n})).\tag4$$
Using (4) and induction again we see that $\deg(f_n)=\lfloor n/2\rfloor$
for $n\geq 0$.
