I'm searching for an algorithm to determine a list of all integer partitions of a number $n$ into a fixed number $m$ of summands (say $n=6$ and $m=4$), for instance to be stored into a list of (vectors-of-length $m$), but with the additional restriction that simple rotations are omitted.
I've already managed the algorithm for the list of vectors, but it still contains rotations. For instance for $n=4$ and $m=6$ :
list index and marker for occurence of rotations
-------------------------------------------
0 0 0 6 | 1
0 0 1 5 | 2
0 1 0 5 | 3
1 0 0 5 | 4
0 0 2 4 | 5
0 1 1 4 | 6
1 0 1 4 | 7
0 2 0 4 | 8
1 1 0 4 | 9
2 0 0 4 | 10
0 0 3 3 | 11 = 20
0 1 2 3 | 12
1 0 2 3 | 13
0 2 1 3 | 14
1 1 1 3 | 15
2 0 1 3 | 16
0 3 0 3 | 17
1 2 0 3 | 18
2 1 0 3 | 19
3 0 0 3 | 20 = 11
0 2 2 2 | 21 = 23 = 26
1 1 2 2 | 22 = 25
2 0 2 2 | 23 = 21 = 26
1 2 1 2 | 24
2 1 1 2 | 25 = 22
2 2 0 2 | 26 = 21 = 23
Q1: How could I identify (and remove) the rotations?
Q2: Is there a mathematical model which allows to identify such rotations?
One idea, to use instead of an entry $e$ the primenumber $p_e = \text{primes}(e)$ and then to compare the numbers which are builded from the product over a row, is intriguing at first sight, but removes the information of the order along the rows.
I think the most practically idea might be to
1) identify first all vectors which have exactly one maximum. They cannot be rotations by construction of the list.
2) Then of the remaining do a list_1 of the first differences along the rows (circular over the borders).
Find all with one maximum and rotate them such that the maximum is in the first position, then compare.
3) Then find all with two maxima and do the second difference.
4),5),... Proceed analoguously.
But this is merely an idea and I do not really know whether this will be working at all. A mathematical model would really help here, but I could not imagine one...
Disclaimer: I've searched MSE and MO for "integer partitions" but none found which deals with the problem of rotations of summands
[Update] I've found the numbers for the numbers $n$ and number of summands $m$ in the OEIS as A241926. We find there:
The number of necklaces with n black beads and k white beads.
{Turning over the necklace is not allowed (the group is cyclic not dihedral)}
But this does not provide/I don't find a solution for the population of the actual lists. [\update]