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seen Jun 24 '13 at 16:56

May
31
awarded  Critic
Apr
14
comment On the proof of Robert Lipshitz's formula on Maslov index.
Why not just e-mail, Robert?
Mar
31
answered Difference between parallel transport and derivative of the exponential map
Feb
19
awarded  Enthusiast
Jan
29
answered Analytic curve on Riemann surface
Jan
28
revised Tubular neighborhoods of chains
added 36 characters in body
Jan
28
revised Tubular neighborhoods of chains
added 156 characters in body; deleted 103 characters in body
Jan
28
answered Tubular neighborhoods of chains
Jan
24
awarded  Scholar
Jan
24
awarded  Supporter
Jan
24
awarded  Student
Jan
21
awarded  Editor
Jan
21
revised Orientations for pseudoholomorphic curves with totally real boundary condition
added 773 characters in body
Jan
21
answered Orientations for pseudoholomorphic curves with totally real boundary condition
Oct
10
awarded  Teacher
Oct
10
answered Why are Gromov-Witten invariants of K3 surfaces trivial?
Jun
7
comment number of weighted trivalent trees
Thank you very much, everyone. Paul's answer in particular completely resolves my question. Here is a direct reformulation of his argument. Let $a_n$ be the weighted number of graphs with $n+1$ marked points divided by $n!$ and $f(x)=x+\sum_{n\ge2}a_n x^n$. From the natural recursion for $a_n$, $$f(x)=x+f(x)+(f(x)-1)\sum_{k=1}^{\infty}\frac{f(x)^k}{k} \quad\longrightarrow\quad (1-f(x))\ln(1-f(x))=-x.$$ Thus, $f(x)=1-e^{W(-x)}$. Using the $r=-1$ case of the above formula then gives $a_n=(n-1)^{(n-1)}/n!$.
Jun
7
accepted number of weighted trivalent trees
Jun
5
comment number of weighted trivalent trees
This is indeed related to moduli spaces (of stable maps). The structure coefficients in formulas for genus 0 Gromov-Witten invariants are sums over $N$ marked trivalent trees. These can be bounded by the number of weighted trees in my question. I can show this number is bounded above by $C^N\cdot N!$, which is good enough for me. However, instead of adding another half a page proving this, I thought I might be able to quote something from the literature (and then apply Stirling's formula to get $C^N\cdot N!$). Since valence - 3 is so natural for trivalent, I thought this formula were known...
Jun
5
asked number of weighted trivalent trees