More from Warren: ---------- Forwarded message ---------- From: Warren D Smith <warren.wds@gmail.com> Date: Wed, Dec 20, 2017 at 12:44 AM Subject: Re: [math-fun] Cutting a pie ... To: James Propp <jamespropp@gmail.com> I can't bring myself to believe F(n) obeys any linear upper bound; I think it must have a superlinear lower bound. Indeed F(2*n) >= 2*F(n)+n-1 holds for n=1,2,3,4 and perhaps 5, and it seems plausible F(2*n)>=2*F(n)+K*n holds for K>=0, with indeed K lower bounded by some positive constant on average... which would imply a n*log(n) style lower bound. So my conjecture is F(n) asymptotically is both superlinear and subquadratic, but I have no idea where it lies within that range. Very annoying! -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)