I am experiencing some anxiety about the configuration you named after me. I fear that it's simply half of a flexing cuboctohedron, and therefore that B and F are collinear in all positions (as are D and H). But I don't know if I should be upset about it ... doesn't that mean that we have achieved four triples on only six directions? Wait a minute ... we proved that impossible. Now I'm all confused. On Fri, Jan 31, 2014 at 1:06 PM, Warren D Smith <warren.wds@gmail.com>wrote:
The following theorems are easy (n>1 assumed in them all):... 2: f(2*n-1) >= 2*f(n)
--more generally by the same reasoning, f(a+b-1) >= f(a)+f(b).
--I further conjecture that f(n+6)>=4+f(n) if n>3. I almost have a proof. Observe that Wechlser's configuration ABC+CDE+EFG+GHA showing f(8)>=4 is flexible, the angle between A and E can be flexed to become any value between 0 and 180 degrees (except 90). So we adjoin Wechsler's configuration to f(n) flexing and rotating it until A and E coincide with two old non-orthogonal directions in the f(n) configuration. This would prove it, except for the annoying possibility that another of the added directions would coincide with an old one causing a screw-up similar in nature to the forbidden 90-degree AE flex. I presume that fate can usually or always be avoided if n>3, but do not have a proof.
Anyhow, assuming this is true, would establish the lower bound on f(n) half of my conjecture that |2f-3n| stays bounded.