I have seen in several places a useful formula that lets us calculate the clustering number of regular ring lattice graphs with even degree, but I have not found a convincing proof of it. Concretely, the formula is $C=\dfrac{3(d-2)}{4(d-1)}$, where $d$ is the degree of the graph and it is an even number.
The best proof of this fact which I have found it is the following one (extracted from http://www.hcs.harvard.edu/~cs134-spring2017/wp-content/uploads/2017/01/section2.pdf). 
However, I have several questions about it:
- I am not be able to understand why the numerator represents the number of neighbors of $v$ which are connected by an edge.
- Although the formula works in most cases, there exist regular ring lattice graphs where the number given by the formula doesn't coincide with the real clustering number: for example in a triangle or in a hexagon whose vertices have degree $4$. Why doesn't the proof discard these cases? What is the extra condition that we need impose on the graph so this formula works?
Any help will be welcome.
We're trying to count the number of edges which lie entirely within $S$. We can do this using the handshaking lemma: for each vertex in $S$, count how many edges meeting that vertex have their other end in $S$, and divide by $2$. (Work on the induced subgraph with vertex set $S$: the number of edges in this graph is half the sum of degrees.)
Now $v$ has $d$ neighbours in $S$ - all of its neighbours are in $S$ by definition. Next, look at the vertex immediately to the left of $v$. All its neighbours to the right are still in $S$, but its leftmost neighbour will be $1+d/2$ steps to the left of $v$, so not in $S$. This means it has $d-1$ neighbours in $S$, and similarly for the vertex to the right of $v$. For the vertex two steps left of $v$, its two leftmost neighbours are too far away from $v$ to be in $S$, so it has $d-2$ neighbours in $S$, as does the vertex two right of $v$. Continuing in this way, the sum of degrees is $d+2(d-1)+2(d-2)+\cdots$. The sum stops when we get to the vertices as far as possible away from $v$. These are $d/2$ steps away, and only $d/2$ of their neighbours are in $S$. So the degree sum is $d+2(d-1)+\cdots+2(d-d/2)$, and you halve this to get the number of edges within $S$. (The final $-d$ in the numerator is because we don't want to include the edges meeting $v$, even though these are within $S$.)
We have made an assumption in the proof above - if you look at a vertex in the left side of $S$, and go to a neighbour of that vertex which is further left, you can't get all the way around to the right side of $S$. This means the formula only works when this assumption is valid, i.e. $3d/2<n$, where $n$ is the total number of vertices. (So $n=6$ and $d=4$ just fails this test.)