Conjugate m-dimensional partitions

[Max]

On 5/8/06, franktaw at netscape.net <franktaw at netscape.net> wrote:

> An m-dimensional partition can be considered as a set of points in an (m+1)
> dimensional corner.  (This is a generalized Ferrers diagram.)  As such, each
> is one of (m+1)! conjugate partitions induced by permutation of the axes.
> (Some of these may be the same, of course.)
>
> So, how many m-dimensional partitions are there, up to conjugacy?  For m =
> 1, we have A046682:
>
> 1,1,1,2,3,4,6,8,12,16,22,29,40,...
>
> and for m = 2, A000786:
>
> [1,]1,1,2,4,6,11,19,33,55,...
>
> For m = 3, I believe the sequence starts (this is hand-calculated):
>
> 1,1,1,2,4,7,13,23

I've got the following table:

m=0:    1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
m=1:    1, 1, 1, 2, 3, 4, 6, 8, 12, 16, 22, 29, 40, 52, 69, 90, 118,
151, 195, 248, 317
m=2:    1, 1, 1, 2, 4, 6, 11, 19, 33, 55, 95, 158, 267, 442, 731,
1193, 1947, 3137, 5039, 8026, 12726
m=3:    1, 1, 1, 2, 4, 7, 13, 25, 49, 93, 181, 351, 687, 1332, 2591,
5003, 9644, 18462, 35208, 66721, 125840
m=4:    1, 1, 1, 2, 4, 7, 14, 27, 55, 111, 232, 486, 1039, 2226, 4820,
10449, 22727
m=5:    1, 1, 1, 2, 4, 7, 14, 28, 57, 117, 251, 543, 1209, 2724, 6251, 14505
m=6:    1, 1, 1, 2, 4, 7, 14, 28, 58, 119, 257, 562, 1268, 2910, 6844
m=7:    1, 1, 1, 2, 4, 7, 14, 28, 58, 120, 259, 568, 1287, 2970
m=8:    1, 1, 1, 2, 4, 7, 14, 28, 58, 120, 260

> We can also ask how many infinite-dimensional partitions there are up to
> conjugacy.  There are infinitely many infinite-dimensional partitions of any
> n >= 2, but any m-dimensional partition of n can be reduced to at most n-2
> dimensions by conjugacy.

In other words, this sequence is given by the diagonal T(m,m+2) in the
table above and it happens to be:
1, 1, 1, 2, 4, 7, 14, 28, 58, 120, 260, 571, 1296, 2998 (the last
three values are guessed)

It is also interesting to notice that
T(m,m+3)=T(m,m+2)-1
T(m,m+4)=T(m+1,m+4)-2=T(m+2,m+4)-3
T(m,m+5)=T(m+1,m+5)-6 for m>=2
T(m,m+6)=T(m+1,m+6)-19 for m>=4
Perhaps, there exists a simple proof for that.

It may also happen that for any fixed k, T(m+1,m+k)-T(m,m+k)=C(k) for
all large enough m where C(k) is a constant depending on k but not on
m. If so, for k=1..6 we have
C(k): 0,0,1,2,6,19

Max