PensionDan<p>A more elegant path forward dawned on me (using <a href="https://mathstodon.xyz/tags/PascalsTriangle" class="mention hashtag" rel="tag">#<span>PascalsTriangle</span></a>!), so I won't be using the equation in the preceding post. <br />First, some comments & definitions. Cycles will refer to cycles of length 2 or more (not including 1-cycles, which we will regard as positions along the main diagonal that don't get permuted). Define the delta of two consecutive elements of a cycle a, b as b - a. Clearly, the sum of all the deltas of a cycle is zero, as each element is added and subtracted once. Therefore, every cycle has at least one negative delta among its pairs of consecutive elements.<br />Define an increasing cycle as a cycle with just one negative delta. Observe that an increasing cycle can be written with its elements in increasing order (the delta from the last element back to the first element is the negative one).<br />Consider these sets of positions NEZ and SWZ (see illustration) in an n x n matrix, which lie in the antidiagonal and the sub-antidiagonal (thus in the eligible positions for (\Omega^{xx}_{n}\)):<br />• NEZ; the northeast zigzag of (n-1) positions that begins at the upper right corner, ordered as positions \(a_{1,n}, a_{2,n}, a_{2,n-1}, a_{3,n-1},…a_{[(n+1)/2],[(n+1)/2]+1}\). <br />• SWZ; the southwest zigzag of (n-1) positions that begins at the lower left corner, ordered as positions \(a_{n,1}, a_{n,2}, a_{n-1,2}, a_{n-1,3},…a_{[(n+1)/2]+1,[(n+1)/2]}\).<br />We’ll describe a set of cycles C, and then show that permutations which are a product of disjoint cycles from C are vertices of \(\Omega^{xx}_{n}\). As we know that \(\Omega^{xx}_{n}\) has \(2^{n-1}\) vertices, if we can verify that there are \(2^{n-1}\) such permutations, then we will have characterized all the vertices of \(\Omega^{xx}_{n}\).</p>