
EXAMPLE

a(1)=oo because for each dimension, d, the trivial Ferrers diagram given by the single node (1,1,1,...,1) is a totally symmetric ddimensional partition of 1.
For n > 2, a(n) < oo. This means that for n > 2, there are at most a finite number of dimensions, d, for which the number of totally symmetric ddimensional partitions of n is nonzero (and that for any dimension, d, there are at most a finite number of totally symmetric ddimensional partitions of n).
a(2)=1. Indeed the only totally symmetric partition of 2 occurs in dimension 1. The corresponding 1dimensional totally symmetric Ferrers diagram (TS FD) is given by the following two nodes (specified by the 1dimensional coordinates): (2) and (1).
a(8)=5.
There is one 1dimensional TS FD of 8:
{(8),(7),(6),(5),(4),(3),(2),(1)}
There are two 2dimensional TS FD of 8:
{(3,2),(2,3),(3,1),(2,2),(1,3),(2,1),(1,2),(1,1) and
{(4,1),(1,4),(3,1),(2,2),(1,3),(2,1),(1,2),(1,1)}
There is one 3dimensional TS FD of 8:
{(2,2,2),(2,2,1),(2,1,2),(1,2,2),(2,1,1),(1,2,1),(1,1,2),(1,1,1)}
There is one 7dimensional TS FD of 8:
{(2,1,1,1,1,1,1),(1,2,1,1,1,1,1),(1,1,2,1,1,1,1),(1,1,1,2,1,1,1),(1,1,1,1,2,1,1),(1,1,1,1,1,2,1),(1,1,1,1,1,1,2),(1,1,1,1,1,1,1)}
There are no TS FD of 8 of any other dimension. Hence a(8)=1+2+1+1=5.
a(72)=573
The TS FD of 72 are:
Dim 1: 1
Dim 2: 471
Dim 3: 85
Dim 4: 11
Dim 5: 3
Dim 6: 1
Dim 71: 1
(For n > 1) there is always exactly 1 TS FD of dimension 1 and 1 TS FD of dimension n1. If n > 2, these two dimensions are not equal, so there must be at least two TS FD. Hence a(n) >= 2 for n > 2.
