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Nov 25, 2016 at 19:38 comment added LSpice In the notation of @GeoffRobinson's answer, we have $d_n(G) = \lvert G\rvert^{-n}c_{n + 1}(G)$, and the recursive formula Lemma 4.1 seems to be the same as his $c_{n + 1}(G) = \sum_{x \in G} c_n(\mathrm C_G(x))$.
Sep 29, 2012 at 7:30 history answered François Brunault CC BY-SA 3.0