Date: Sat, 19 May 2007 23:47:03 +0200
Comments from Thomas Wieder (wieder.thomas(AT)t-online.de)

At your request, I will try to explain the sequences A103446 and A025168 in words.
The combstruct command for the construction these sequences uses a 
substitute command.
This substitution makes the sequences rather difficult to understand.

I. Introduction

The sequence A103446 results from a set partition of n unlabeled elements,
a subsequent permutation of the subsets of each partition, and then a
partition of these permuted lists. In fact, there is a final permutation 
step, but this step results in no new sets for unlabeled elements.

In short, A103446 results from a partition-permutation-partition-sequence.

The sequence A025168 results from a set partition of n labeled elements,
a subsequent permutation of the subsets of each partition, a
partition of these permuted lists and finally a permutation of elements 
within lists.

In short, A025168 results from a 
partition-permutation-partition-permutation-sequence.

Now I will outline the steps in detail:


II. A103446

The sequence A103446 results from five steps.

Let {} denote a set and [] denote a list = permutation.
We will work on an initial list of n unlabeled elements.
As an example, let us consider n=3, the list is [*,*,*].

Step 1: Form the list [*,*,*].

Step 2 = Partition: Form all possible (unordered) partitions of the list 
into lists.
We get [*,*,*], [*,*][*], and [*][*][*].

Step 3 = Permutation: Permute these list.
 From [*,*,*] we get [*,*,*].
 From [*,*][*] we get [*,*][*], and [*][*,*].

Step 4 = Partition: Form sets of lists from these permutations.
 From [*,*,*] we get {[[*,*,*]]}.
 From [*,*][*] we get {[[*,*][*]]}, {[[*,*]][[*]]}.
 From [*][*,*] we get {[[*][*,*]]}, {[[*]][[*,*]]}.
 From [*][*][*] we get {[[*][*][*]]}, {[[*][*]][[*]]}, {[[*]][[*]][[*]]}.

Step 5 = Permutation: Permute the elements within a list.
For unlabeled elements, nothing happens, but see below for labeled elements.

Thus for n=3 we have in the unlabeled case:

{[[*,*,*]]},

{[[*,*],[*]]},
{[[*],[*,*]]},
{[[*,*]],[[*]]}
{[[*]],[[*,*]]},

{[[*],[*],[*]]},
{[[*],[*]],[[*]]},
{[[*]],[[*]],[[*]]}.



III. A025168

The labeled case = A025168 follows the same steps,
but the number of possibilities are higher within each permutation step.

We consider a list of n labeled elements.
As an example, let us consider n=2, the list is [1,2].

Step 1: Form the list [1,2].

Step 2 = Partition: Form all possible (unordered) partitions of the list 
into lists.
We get [1,2] and [1][2].

Step 3 = Permutation: Permute these list.
 From [1,2] we get [1,2].
 From [1][2] we get [1][2] and [2][1].

Step 4 = Partition: Form sets of lists from these permutations.
 From [1,2] we get {[[1,2]]}.
 From [1][2] we get {[[1][2]]} and {[[1]][[2]]}.
 From [2][1] we get {[[2][1]]} and {[[2]][[1]]} = {[[1]][[2]]}
and the same set from a partition step does count only once.

Step 5 = Permutation: Permute the elements within a list.
 From {[[1,2]]} we get {[[1,2]]} and {[[2,1]]}.
 From {[[1][2]]} we get {[[1][2]]} and {[[2][1]]} .
 From {[[1]][[2]]} we get {[[1]][[2]]}.

Thus for n=2 we have in the labeled case:

{[[1,2]]}
{[[2,1]]}

{[[1],[2]]}
{[[2],[1]]}

{[[1]],[[2]]}.


For n=3 we have in the labeled case:

{[[3,1,2]]},
{[[1,3],[2]]},
{[[3,2]],[[1]]},
{[[2,3],[1]]},
{[[1,3,2]]},
{[[2,3]],[[1]]},
{[[2],[3]],[[1]]},
{[[2],[1],[3]]},
{[[1],[3]],[[2]]},
{[[2,1,3]]},
{[[3,1],[2]]},
{[[1,2,3]]},
{[[3],[2]],[[1]]},
{[[2],[1,3]]},
{[[2],[3],[1]]},
{[[3,2],[1]]},
{[[1],[3,2]]},
{[[3],[1,2]]},
{[[3],[1],[2]]},
{[[2],[1]],[[3]]},
{[[1],[2,3]]},
{[[3],[2],[1]]},
{[[3]],[[1],[2]]},
{[[2]],[[3,1]]},
{[[1],[2],[3]]},
{[[3]],[[2,1]]},
{[[2],[3,1]]},
{[[2,3,1]]},
{[[2,1],[3]]},
{[[3,2,1]]},
{[[2]],[[3],[1]]},
{[[3]],[[1,2]]},
{[[3],[2,1]]},
{[[2]],[[1,3]]},
{[[2]],[[1]],[[3]]},
{[[1],[3],[2]]},
{[[1,2],[3]]}