# [seqfan] Re: perms and set partitions

Joerg Arndt arndt at jjj.de
Mon Jun 8 04:20:14 CEST 2009

```* Joerg Arndt <arndt at jjj.de> [Jun 08. 2009 12:00]:
> [...]
>
> Is there a combinatorial interpretation of this table?

Well, m==0 are the permutations whose cycle form is a
valid set partition (elements in cycles increasing),
m==1 has one decreasing part in some cycle, etc.:

With n==4, we have (omitting fixed points from cycle form,
lex order):

0:    [ 0 1 2 3 ]    nt=0  ne=0  m=0
1:    [ 0 1 3 2 ]    nt=1  ne=1  m=0    (2, 3)
2:    [ 0 2 1 3 ]    nt=1  ne=1  m=0    (1, 2)
3:    [ 0 2 3 1 ]    nt=2  ne=2  m=0    (1, 2, 3)
4:    [ 0 3 1 2 ]    nt=2  ne=1  m=1    (1, 3, 2)
5:    [ 0 3 2 1 ]    nt=1  ne=1  m=0    (1, 3)
6:    [ 1 0 2 3 ]    nt=1  ne=1  m=0    (0, 1)
7:    [ 1 0 3 2 ]    nt=2  ne=2  m=0    (0, 1) (2, 3)
8:    [ 1 2 0 3 ]    nt=2  ne=2  m=0    (0, 1, 2)
9:    [ 1 2 3 0 ]    nt=3  ne=3  m=0    (0, 1, 2, 3)
10:    [ 1 3 0 2 ]    nt=3  ne=2  m=1    (0, 1, 3, 2)
11:    [ 1 3 2 0 ]    nt=2  ne=2  m=0    (0, 1, 3)
12:    [ 2 0 1 3 ]    nt=2  ne=1  m=1    (0, 2, 1)
13:    [ 2 0 3 1 ]    nt=3  ne=2  m=1    (0, 2, 3, 1)
14:    [ 2 1 0 3 ]    nt=1  ne=1  m=0    (0, 2)
15:    [ 2 1 3 0 ]    nt=2  ne=2  m=0    (0, 2, 3)
16:    [ 2 3 0 1 ]    nt=2  ne=2  m=0    (0, 2) (1, 3)
17:    [ 2 3 1 0 ]    nt=3  ne=2  m=1    (0, 2, 1, 3)
18:    [ 3 0 1 2 ]    nt=3  ne=1  m=2    (0, 3, 2, 1)
19:    [ 3 0 2 1 ]    nt=2  ne=1  m=1    (0, 3, 1)
20:    [ 3 1 0 2 ]    nt=2  ne=1  m=1    (0, 3, 2)
21:    [ 3 1 2 0 ]    nt=1  ne=1  m=0    (0, 3)
22:    [ 3 2 0 1 ]    nt=3  ne=2  m=1    (0, 3, 1, 2)
23:    [ 3 2 1 0 ]    nt=2  ne=2  m=0    (0, 3) (1, 2)

stats: 15, 8, 1, 0,   tot=24

```