
Re: Stone Cech
Posted:
Mar 20, 2013 5:47 PM


On Mar 20, 8:53 am, David C. Ullrich <ullr...@math.okstate.edu> wrote: > On Tue, 19 Mar 2013 21:25:44 0700 (PDT), Butch Malahide > <fred.gal...@gmail.com> wrote: > >On Mar 19, 8:49 pm, William Elliot <ma...@panix.com> wrote: > >> On Tue, 19 Mar 2013, David Hartley wrote: > >> > > > > Perhaps you could illustrate with the five different one to four point > >> > > > > point compactifications of two open end line segements. > > >> > > > (There are seven.) > [. . .] > >> Ok, seven nonhomeomophic finite Hausdorff compactications. > > >How many will there be if you start with n segments instead of 2? > > Surely there's no simple formula for that? > > The "obvious" approach seems like starting with the number > of partitions of 2n, which is already hopeless, and then
The number of partitions of 2n? As noted by Euler, p(2n) is just the coefficient of z^{2n} in the Maclaurin expansion of the function F(z) = 1/eta(z), where eta(z) is the infinite product eta(z) = (1  z)(1  z^2)(1  z^3)...(1  z^n).... From there you can work out the asymptotics etc. The recurrence relation is p(n) = p(n1) + p(n2)  p(n5)  p(n7) + p(n12 + p(n15)  p(n22)  ...
> somehow factoring out the ones that give homeomorphic > compactifications... hopeless squared.
In graphical terms, it's about enumerating the nonhomeomorphic graphs (undirected but with loops and multiple edges permitted) with n edges. No doubt this is much harder than the (solved) enumeration problem for nonisomorphic simple graphs with n vertices. It's just not immediately obvious to me that the question is "hopeless squared". Of course it's way too hard for me; that's why I posted the question instead of trying to figure it out myself.
> Is there a better way, or are you just making trouble again or what?
I wasn't necessarily expecting a *complete* answer, such as an explicit generating function. Maybe someone could give a partial answer, such as an asymptotic formula, or nontrivial upper and lower bounds, or a reference to a table of small values, or the ID number in the Encyclopedia of Integer Sequences, or just the value for n = 3. (I got 21 from a hurried hand count.)

