TITLE: Kostant partition function: asymptotics and specifics
QUESTION [9 upvotes]: Let $\Phi$ denote a root system and let $\mathfrak P$ denote the associated Kostant partition function. Thus $\mathfrak P(\lambda)$ is the number of ways of writing $\lambda$ as a sum of elements of $\Phi^+$. For example, if $\Phi=A_3$ then the longest root $\alpha_0$ can be written as $\alpha_0=(\alpha_1+\alpha_2+\alpha_3)$, $(\alpha_1)+(\alpha_2)+(\alpha_3)$, $(\alpha_1+\alpha_2)+\alpha_3$ or $\alpha_1 + (\alpha_2+\alpha_3)$. Thus $\mathfrak P(\alpha_0)=4$. There is a recursion $$\mathfrak P(\mu)=-\sum_{1\neq w\in W} (-1)^{l(w)}\mathfrak P(\mu+w(\rho)-\rho)$$ due to Kostant.
I can't find very many general results on $\mathfrak P$. 
One knows that $\mathfrak P((p-1)\rho)$ is the dimension of the $0$ weight space of the Steinberg module (see Jantzen, RAGS, II.10.12). I conjecture that one also has $\mathfrak P(r\rho)$ is the dimension of the $0$ weight space in $L_{\mathbb C}(r\rho)$, where $L_{\mathbb C}(r\rho)$ is the irreducible representation of high weight $r\rho$ for a complex Lie algebra with root system $\Phi$ (which may be zero if $r$ is odd). A reference even for this basic statement would be good!
[RE-EDIT: This paragraph, including the conjecture, is completely wrong. See Chuck Hague's and Jim Humphreys' remarks below. Of course one must have $\dim H^0(\lambda)_0\leq \mathfrak P(\lambda)$ for all $\lambda$, so this gives a lower bound.]
I'd also like some data on asymptotics. I have used the Maple `Coxeter/Weyl' package to compute $\mathfrak P(2\rho)$ for $A_n$ up to $n=5$. The numbers are $1,3,15,219,7834\dots$ and correspond to sequence A007081 on the OEIS. 
[EDIT: These calculated values are in fact the dimensions of the $0$ weight space of $H^0(2\rho)$. Since the conjecture that $\mathfrak P(2\rho)=\dim H^0(2\rho)_0$ is incorrect these do not coincide with the values of $\mathfrak P(2\rho)$ as claimed.]
I guess $\mathfrak P(2\rho) < n^n$. [EDIT: This looks very optimistic---suspect it's wrong for $n=4$.] Any ideas? What about $\mathfrak P (2r\rho)<(nr)^{nr}$ or something better? [EDIT: Similarly.] Lower bounds?

REPLY [2 votes]: I have some answers.
I couldn't find a computer program giving values of $\mathfrak P$ so I wrote one. Assuming it's giving the correct numbers, $\mathfrak P_{A_n}(2\rho)$ is 
$$1,3,26,898,128826,82462230$$ with natural logs
$$0,1.10,3.26,6.80,11.77,18.22.$$ The latter looks roughly quadratic to me. Note that the sequence does not come up on the OEIS.
As for proving some bounds... 
First of all, [Cline, Parshall, Scott; Reduced standard modules and cohomology] has a statement of a bound on page 5258. They state that it's easy to prove $\mathfrak P(2(h-1)\rho)\leq N!P(N(h-1))$ where $N=|\Phi^+|$ is the number of positive roots and $P$ is the usual partition function. Wikipedia tells me that $$P(n)\sim \frac{1}{4n\sqrt 3}\exp\left(\pi\sqrt{\frac{2n}{3}}\right)$$ so this gives $\log \mathfrak P(2(h-1)\rho)$ is $O(n^{3/2})$.
I think I can improve on this (just for type $A_n$ but it should be representative):
Observation 1: Looking up in the tables in Bourbaki, one gets $2\rho=n\cdot1\alpha_1+(n-1)\cdot2\alpha_2+\dots$. So the lowest coefficient is $n$ and the highest coefficient is roughly $(\frac{n+1}{2})^2$.
Observation 2: Once the coefficients of the non-simple roots in a sum of positive roots has been chosen, one completes the expression by adding in a determined number of simple roots to make up the difference. So $\mathfrak P(2r\rho)$ can be obtained by counting sets of positive non-simple roots whose sum is less than or equal to $2r\rho$ (in the dominance ordering).
Lower bounds: just use roots of height 2. Choose $\lfloor n/2\rfloor$ coefficients $b_i$ so that $$b_1(\alpha_1+\alpha_2)+b_2(\alpha_3+\alpha_4)+\dots\leq 2r\rho.$$ By Observation 2, each $b_i$ can be anything up to $rn$. Get $$\mathfrak P(2r\rho)\geq (rn)^{n/2}.$$ Probably not very good, but at least it shows it's exponential.
Upper bounds: There are $n(n-1)/2$ non-simple roots and by Obs 2 each can occur with any coefficient which at most $r(n+1)^2$. So get $$\mathfrak P(2r\rho)\leq (r(n+1))^{n(n-1)}.$$
Combining these two gives that $\log \mathfrak P(2\rho)$ having growth rate strictly more than linear and strictly less than cubic. The upper bound says that $\log \mathfrak P(2\rho)$ is $O(n^2\log n)$. The lower bound says that $\log \mathfrak P(2\rho)$ grows at least as fast as $n\log n$.
I think one can probably do better, particularly on the lower bound. I may have another look at it at some later stage. But basically I'm content with the statement that $\log \mathfrak P(2\rho)$ is roughly quadratic with the rank.