Another math-fun member has kindly pointed out to me that an Hadamard matrix in dimension 4K implies the maximal number (n+1) of equidistant points among the vertices of the (4K-1)-dimensional cube... ...and that since the Hadamard conjecture is that an Hadamard matrix exists in every dimension of form 4K, it would if true mean that n+1 equidistant points exist in each dimension of form 4K-1. —Dan ----- The relationship is that if there is an n x n Hadamard matrix, then there exist n orthogonal vectors among the 2^n vertices {-1, 1}^n of the n-cube. These n vectors are equal length, so their endpoints are equidistant. I ask about a related but not identical situation: when there exist n+1 equidistant points among the 2^n points of {-1, 1}^n. And in general, *how large* is the largest subset of {-1, 1}^n consisting of equidistant points? —Dan Fred Lunnon wrote: ----- See https://en.wikipedia.org/wiki/Hadamard_matrix in particular, Sylvester construction. ----- On 8/9/18, Dan Asimov <dasimov@earthlink.net> wrote: Suppose the metric space Q_n is the vertices of the n-dimensional cube Q_n = {-1, 1}^n with the induced metric from Euclidean space, and the questions is: Question: --------- For which n do the points of Q_n include n+1 equidistant points, the maximum possible? (And in general, what is the size S(n) of the largest equidistant subset of Q_n?) If this is in OEIS, I didn't succeed in finding it. ----- ----- -----