Pólya-Burnside Enumeration, Part 2

So, previously I have set up all the mathematical machinery needed to do the actual counting. So here's an example. It's not the most interesting example, but it has the advantage of being relatively short.

Let's say we want to count how many linear molecules are there of length n, given t choices for atoms (these choices must be dictated by chemistry, because mathematicians don't care). Different kinds of bonds aren't distinguished; what matters are the atoms themselves. For simplicity, let's just do this in two dimensions*.

In two dimensions, there are only two operations on a linear molecule that can possibly leave the molecule invariant: "do nothing", and rotation by π radians. So the group of interest is has two elements: G = {e, g}.

Let the set of elements we are interested in be all strings of t elements, of length n, regardless of whether they are invariant under the elements in G. (Or, if you like, consider the set of t-ary strings of length n, where by t-ary I mean a generalization of things like binary strings, ternary strings, etc., with t elements instead of 2 and 3 for binary and ternary, respectively). We want to count the number of orbits in X because otherwise we'll overcount; for example the "molecule" a-b-c is the same as c-b-a but not c-a-b.

It is pretty clear to see that
F(1) = tⁿ

For F(g), we need to think about what kind of strings are invariant under rotation by π radians. Well, essentially the molecule needs to be "palindromic", and read the same forwards and backwards. So for example
a-b-c-c-b-a
or
a-b-c-b-a
are invariant under g. In other words, molecules invariant under g can anything in the first half of the molecule, but the second half is determined by the first half. So, if n is even, there are tⁿ/2 possibilities in X, and if n is odd, there are t⁽ⁿ⁺¹⁾/2 possibilities in X.

Applying our theorem gives the number of orbits to be

[X/S] = \frac{1}{2}(t^n + t^{\frac{n}{2}})}

[X/S] = \frac{1}{2}(t^n + t^{\frac{n+1}{2}})}

for even and odd n, respectively.

Let's do a reality check. For a triatomic molecule with two choices of atoms (e.g. {N, O}), 6 molecules are possible. They are:

N-N-N
N-N-O
N-O-N
O-N-O
N-O-O
O-O-O

which is, happily, six molecules. Rather obviously, from a chemical viewpoint, they molecules might not actually exist in the geometry of interest, but it's a start.

Admittedly, most people don't care about linear molecules**. But any geometry will do. Apparently even crystal structures will do. It's much faster than counting in the naive way, that's for sure! (Pólya apparently did a great many number of examples in his paper; I really ought to track that English translation down...)

*Question for the day, of which I don't know the answer: Why can't we use point groups for the Pólya-Burnside enumerations? Or, why are mirror planes and inversions not acceptable when considering symmetry operations on a molecule? (For linear molecules, we can argue that there are an infinite number of mirrors and rotations about the z-axis, and the enumeration doesn't work for infinite sets, but most symmetries have finite mirror planes.)
**Ok, most people don't care about counting molecules like this, either, but hey, it's kind of cool!