Hi Christian, This makes sense. baj(sigma) = \sum_{i in des(sigma)} i*(n-i) so pi.baj_index() + pi.reverse().baj_index() = \sum_{i =1}^n i*(n-i) which I hope works out to what you said it does.
About your other question: It seems unusual that, pi.insertion_tableau().cocharge() == n * pi.inverse().number_of_descents() - pi.insertion_tableau().major_index() But, cocharge is very similar to the maj statistic in the permutation case so this is not totally unreasonable that you are counting a weight for each descent of the inverse of the permutation. -Mike On Monday, 4 June 2012 15:00:51 UTC-4, Christian Stump wrote: > > Hi Mike, > > > your statistic made its way already. > > some dependency the finder tells you: did you know that pi.baj_index() > + pi.reverse().baj_index() == \binom{n+2}{3} ? > > Best, Christian > On Monday, 4 June 2012 15:00:51 UTC-4, Christian Stump wrote: > > Hi Mike, > > > your statistic made its way already. > > some dependency the finder tells you: did you know that pi.baj_index() > + pi.reverse().baj_index() == \binom{n+2}{3} ? > > Best, Christian > -- You received this message because you are subscribed to the Google Groups "sage-combinat-devel" group. To view this discussion on the web visit https://groups.google.com/d/msg/sage-combinat-devel/-/xU6FH3hKKNkJ. To post to this group, send email to sage-combinat-devel@googlegroups.com. To unsubscribe from this group, send email to sage-combinat-devel+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/sage-combinat-devel?hl=en.