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Feb 18, 2017 at 19:53 comment added Max Alekseyev @StefanKohl: I'm not sure what is interesting to you. Personally I find it quite interesting that $m_k$ is fully determined by $m_0$ (i.e., the smallest non-fixed point of $\sigma$) and otherwise does not depend on $\sigma$.
Feb 18, 2017 at 16:14 vote accept Stefan Kohl
Feb 18, 2017 at 15:06 history edited Max Alekseyev CC BY-SA 3.0
added 97 characters in body
Feb 18, 2017 at 15:01 history answered Max Alekseyev CC BY-SA 3.0