How many weak compositions of $n$ with $k$ parts up to reflection? The number of weak compositions of $n$ with $k$ parts is given by:
$$ \binom{n+k -1}{k-1}. $$
Consider the following identification: two compositions $(s_1 + \cdots + s_k), (t_1 + \cdots + t_k)$ are the same if $t_i = s_{k-i}$ for  $i = 1,\ldots,k$. In other words, we consider them equivalent if they are reflections of each other.
Under this identification, how many weak compositions of $n$ with $k$ parts are there?
Equivalently, how many palindromic weak compositions are there (i.e. weak compositions that read the same forward and backward)? 

This is a simple application of Polya Counting. Let us recall for a moment that the generating function of weak compositions of $n$ into $k$ parts is
$$Z(E_k)\left(\frac{1}{1-z}\right)$$
where $$Z(E_k) = a_1^k$$
is the cycle index of the identity group so that we get the OGF
$$\left(\frac{1}{1-z}\right)^k.$$
Now when we do coefficient extraction using the Newton binomial we obtain the count
$$[z^n] \left(\frac{1}{1-z}\right)^k = {n+ k-1\choose n},$$
as observed in the introduction.
Now to answer the question we need the cycle index of the group $F_k$ containing the identity and a single flip of the row of slots into which we distribute the elements of the composition. We have that when $k$ is even,
$$Z(F_k) = \frac{1}{2} a_1^k + \frac{1}{2} a_2^{k/2}$$
and when $k$ is odd,
$$Z(F_k) =  \frac{1}{2} a_1^k + \frac{1}{2} a_1 a_2^{(k-1)/2}.$$
Therefore the count $q_n$ when $k$ is even is given by
$$[z^n]Z(F_k)\left(\frac{1}{1-z}\right)
= \frac{1}{2} [z^n] \left(\frac{1}{1-z}\right)^k
+ \frac{1}{2} [z^n] \left(\frac{1}{1-z^2}\right)^{k/2}.$$
This gives 
$$q_n = \frac{1}{2} {n+k-1\choose n}
+ \frac{1}{2}\begin{cases} 0 & \text{if} \quad n \quad \text{is odd,}\\
{n/2 + k/2-1\choose n/2} & \text{if} \quad n \quad \text{is even.}
\end{cases}.$$
The count $q_n$ for odd $k$ is given by
$$[z^n]Z(F_k) \left(\frac{1}{1-z}\right)
= \frac{1}{2} [z^n] \left(\frac{1}{1-z}\right)^k
+ \frac{1}{2} [z^n] \frac{1}{1-z}\left(\frac{1}{1-z^2}\right)^{(k-1)/2}.$$
The second term in the sum simplifies to
$$(1+z)\left(\frac{1}{1-z^2}\right)^{(k-1)/2+1}.$$
This gives
$$q_n = \frac{1}{2} {n+k-1\choose n}
+ \frac{1}{2}\begin{cases} {(n-1)/2 + (k-1)/2\choose (n-1)/2}  & 
\text{if} \quad n \quad \text{is odd,}\\
{n/2 + (k-1)/2\choose n/2} & 
\text{if} \quad n \quad \text{is even.}
\end{cases}.$$
Here is a MSE Link to a chain of Polya Counting applications.
