An extremal graph problem 2

With respect to the problem described, Jan Haugland and I have been corresponding a little bit. We managed to show that 2/(f(d)+2) <= f(d+1)/f(d) < 1 for all d. We think lim f(d+1)/f(d) exists. This follows if lim f(d) = 0, which we presume would not be too difficult to resolve. Jan thinks f(d) grows roughly like Alog(d)/d.

Considering planar graphs embedded on a sphere, we see that g(d+2) <= 2e by going through all the faces. This inequality gives f(d) >= 2/(d+2). The following construction gives a better bound on f(d).

Given d, let n be the smallest prime power such that d <= n^3-1. Now take a projective plane of order n and consider its point-block incidence graph G. G has v = 2(n^2+n+1) vertices, e = (n+1)(n^2+n+1) edges, and girth 6. Now, e-v = n^3-1 >= d. Since f is a decreasing function, we now have f(d) > 6/((n+1)(n^2+n+1)) = (6/d)*(1-2/(n+1)) = (6/d)*(1-2/(d^(1/3)+1)).