Diliaoctadecagram

Given a regular $n$ -gon, label the vertices consecutively $v_0, v_1, ... , v_{n-1}$ . Form a regular $n$ -gram by choosing an integer $1 < k < n-1$ , and then drawing line segments between every $k$ -th vertex $v_0, v_k, v_{2k}, ...$ , cycling around if necessary. For instance, in the example provided in the problem statement, $k = 1$ for the dodecahedron, $k = 2$ for the 2-hexagon compound, $k = 3$ for the 3-square compound, and so on. If $k$ divides $n$ , we will form a compound of $k$ closed $\frac{n}{k}$ -gons. Notice that the values $k$ and $n-k$ result in the same $n$ -gram, because in both cases each segment jumps the same number of consecutive vertices, one just does so clockwise and the other counter-clockwise. So there are only $\frac{n}{2}$ distinct n-grams for $n$ even and $\frac{n-1}{2}$ $n$ -grams for $n$ odd. In particular, for $n = 2018$ we have 1009 $n$ -grams.