Very interesting (as J.P. wrote). It may not help anything, but I like to think of an Hadamard matrix geometrically. Consider the n-cube as [-1,1]^n. Then the rows (or columns) of an order-n Hadamard matrix are n vertices of the n-cube whose directions from the origin are perpendicular. —Dan
On Sep 7, 2016, at 7:57 AM, Veit Elser <ve10@cornell.edu> wrote:
The Hadamard matrix conjecture holds that such matrices exist for all orders that are divisible by 4. After surveying what’s been done on the classification/enumeration of Hadamard matrices (e.g. http://neilsloane.com/hadamard/) I’ve felt that what’s humbling about the conjecture is that we lack the knowledge to prove the existence of even a single Hadamard (at each possible order) when the evidence points to a very rapid growth in their number. Now I’m not so sure.
Let x = log_2(N), N = order of Hadamard (a multiple of 4), and y = (1/N^2)log_2(num(N)), where num(N) is the number of Hadamard matrices of order N. Sequence A206711 gives num(N) for N = 1, … , 32. If you plot y vs. x you get a very straight line: y = 0.78785 - 0.09458 x. Taking this literally, there should be a maximum in the number of Hadamard matrices at order N = 196, and beyond that the number plummets, vanishing at around N = 322. The available constructions (beyond this number) would then represent isolated points.