On Fri, Aug 8, 2014 at 8:39 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 3:35 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 7:05 AM, Bill Gosper <billgosper@gmail.com> wrote:
If the four curvatures are k1, k2, k3, k4, and the bounding circles are k0 and k5, then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of signs to k0 and k5. --rwg
Found nothing for n=5,
k1^2 + k2 k3 + k2 k4 + k3 k5 == k1 k2 + k3^2 + k1 k4 + k2 k5 plus nine more, closing over D_5. Huh?
And apparently ten more like 0 == k0 k1 + k0 k2 - k1 k2 - k2^2 - k0 k3 + 2 k1 k3 + k3^2 - k0 k4 - 2 k2 k4 + k3 k4 + k1 k5 - k2 k5 + k3 k5 - k4 k5 - k1 k6 - k2 k6 + k3 k6 + k4 k6, where k0,k6 are the annulus.
And there seems to be a very weird one involving k0 and k6, which may well fail under scrutiny. --rwg
Failed. I've been finding these numerically (69D) using assorted transcendental homographic coefficients, figuring if I get a false positive, I'll be famous for finding a relation among π,e^-2, and EulerGamma, e.g. But I'm getting dozens of false positives. Boy am I confused.
but for n=6, {k1,...,k6} bounded by k0 and k7,
k1+k4=k2+k5=k3+k6 (opposite pairs)=3(k0-k7), *and* k1+k3+k5=k2+k4+k6. (Four relations.) --rwg
Apparently for composite cycle length n = j h, break the list
of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1). WFL> I can prove this via plane tetracyclic coordinates. The later conjectures can undoubtedly be decided in the same fashion, for small fixed n . Fred Lunnon Yeah, but in the States you need a prescription for that. Wait, you can do those weirdly unsymmetrical n=5's? How about n=7? I'm getting dozens of unlikely results like k3^2 + 2 k1 k4 + 2 k4 k5 == 2 k3 k4 + k4^2 + 2 k1 k5 (no k0,k8,k2,k6,k7!)