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]]}