The question is rather vague (I couldn't phrase it more clearly though) so let me explain. Say you have a circle, and you put four equidistant points on it. You then have two types of neighbours: nearest neighbours and second nearest neighbours. This would be $1,2; 2,3; 3,4; 4,1$ and $1,3; 2,4$. Third nearest neighbours don't exist, or at least I don't count them because they are already nearest neighbours. So in this case of 4 points we have two different types of neighbours, the first group contains 4 and the second group contains 2 elements.
Now, my question is, is there a system for $N$ equidistant points on a circle? It doesn't seem obvious to me, but it should be there. Lets take the example of 6: nearest neighbours are $1,2; 2,3; 3,4; 4,5; 5,6; 6,1$, 2nd NN are $1,3; 2,4; 3,5; 4,6; 5,1; 6,2$, 3rd nearest are $1,4; 2,5; 3,6$.
Okay, so now there's 3 groups. The first two have the same size, the last is half. If we do the same for 8 (I'll leave out the numbers) you get 4 groups and the last group has size 4. So to me the pattern seems to be you have N/2 types of neighbours, the first N-1 of which have size N, and the last has size N/2. Does this work for general $N$? I haven't tried for odd; maybe here one has to use the $floor$ function?
For even $N$, given one point, there are $N-1$ other points, which is an odd number. There are two at each distance up to $\frac N2-1$ and one at distance $\frac N2$
For odd $N$, given one point, there are $N-1$ other points, which is an even number. There are two at each distance up to $\frac {N-1}2$