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!
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 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 $n\geq 0$.