Labeling the vertices of a polygon with 0's and 1's Suppose $P_n$ is the regular polygon with n vertices ($n\geq 5$). Let $V=\{v_1,\ldots,v_n\}$ be the vertex set. I would like to define a labeling function $\ell:V\to \{0,1\}$ so that $\sum_{i=1}^{n}\ell(v_i)$ is even and $\geq 4$. This means that a labeling assigns a 0 or a 1 to each vertex and that this must be done so the number of 1's is even and is at least 4.
Two such labelings are equivalent if there is a automorphism of $P_n$ that sends one labeling to the other. How many equivalence classes of such labelings are there are on $P_n$?

What you would want to do is find the Polya inventory polynomial in two colors for the Dihedral group $D_{n}$. Begin with the cycle index of $D_{n}$, $Z(D_{n})$:
$$Z(D_{n}) = \frac{1}{2n} \sum_{d|n} \phi(d) a_d^{n/d} + \left\{
     \begin{array}{lr}
       \frac{1}{2}a_1 a_2^{(n-1)/2} &n \textrm{ odd}\\
       \frac{1}{4}\left(a_2^{n/2}  + a_1^2 a_2^{(n-2)/2}\right) &n \textrm{ even}
     \end{array}\right.$$
Where $\phi$ is Euler's phi function. Next replace $a_i$ by $(b^i + w^i)$ to get the Polya inventory polynomial:
$$P(b,w) =  \frac{1}{2n} \sum_{d|n} \phi(d) (b^d + w^d)^{n/d} + \left\{
     \begin{array}{lr}
       \frac{1}{2}(b+w)(b^2 + w^2)^{(n-1)/2} &n \textrm{ odd}\\
       \frac{1}{4}\left((b^2 + w^2)^{n/2}  + (b+w)^2 (b^2+w^2)^{(n-2)/2}\right) &n \textrm{ even}
     \end{array}\right.$$
The coefficient of $b^i w^j$ in this polynomial counts the number of distinct bracelets with $n$ beads, $i$ of which are black and $j$ of which are white. To relate this to your polygons, let the black beads be the ones labeled with 1, and the white beads labeled 0. Then summing the coefficients of each $b^i w^j$ in $P(b,w)$ with $i$ even and $i\geq 4$ will give you the number of such polygons that you are looking for.
As you can see this value depends on the factorization of $n$. You may be able to get some nice results for certain $n$. For example, if $n=p$, where $p$ is an odd prime, the polynomial will be:
$$\begin{array} \\
P(b,w) &= \frac{1}{2p}\left((b+w)^p + (p-1)(b^p + w^p)\right) + \frac{1}{2}(b+w)(b^2 + w^2)^{(p-1)/2} \\
 &= \frac{1}{2p}\left(\sum_{i=0}^p \binom{p}{i}b^i w^{p-1} + (p-1)(b^p + w^p)\right) +  \frac{1}{2}(b+w)\sum_{i=0}^{(p-1)/2} \binom{(p-1)/2}{i} b^{2i} w^{p-1+i} \\
\end{array}$$
Taking only coefficients of even powers of $b$ at least 4, we get that the number of labelings is:
$$\begin{array} \\
&= \frac{1}{2p}\sum_{i=2}^{\lfloor p/2 \rfloor} \binom{p}{2i} +  \frac{1}{2}\sum_{i=2}^{(p-1)/2} \binom{(p-1)/2}{i} \\
&= \frac{1}{2p}\left(2^{p-1} - 1 - \frac{p(p-1)}{2}\right) + \frac{1}{2}\left(2^{(p-1)/2} - 1 - \frac{p-1}{2}\right) \\
&= \frac{1}{2p}\left(2^{p-1} - 1\right) - \frac{p-1}{4} + 2^{(p-3)/2} - \frac{1}{2} - \frac{p-1}{4} \\
&= \frac{1}{2p}\left(2^{p-1} - 1\right) + 2^{(p-3)/2} - \frac{p}{2}
\end{array}$$
This is a nice quick intro to the topic of Polya enumeration if you're interested in more.
