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
>
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