[seqfan] Re: Modular Partitions

[Gottfried Helms]

Am 30.04.2014 21:19 schrieb Jens Voß:
> 
> Hi there, sequence fans,
> 
> I was playing around with what I call "modular partition numbers": 
> Essentially different ways to write the neutral element of the group 
> Z/nZ as a sum of length k (for given n, k > 0).
> 
> For example, for n = 5 and k = 4, we have thepartitions
> 
> 0+0+0+0 = 0
> 0+0+1+4 = 5 = 0
> 0+0+2+3 = 5 = 0
> 0+1+1+3 = 5 = 0
> 0+1+2+2 = 5 = 0
> 0+2+4+4 = 10 = 0
> 0+3+3+4 = 10 = 0
> 1+2+3+4 = 10 = 0
> 1+3+3+3 = 10 = 0
> 3+4+4+4 = 15 = 0
> 
> so the number of 5-modular partitions of length 4 is 10.
> 
> I computed the the values for n + k < 20 (as a square array read by
> antidiagonals), and was somewhat surprised that this sequence isn't yet
> in the database (even though several of the rows resp. columns are).

Hi Jens,

 I just arrived at the same table of values by a slightly different
 view at the same problem but coming from another question (existence
 of cycles in the Collatz-problem).

 I'm looking at the following question

  given some positive integer number N and another number B.

  Let's look at the number of possibilites to express N by
  B nonnegative summands - just in the same way
  as you've expressed it above, but without the modular
  expression. Instead of this I try to omit solutions
  which are only circular shifts of each other.
  The following list occurs if I compute
  the partitioning of n=6 into m=4 summands; many obvious
  rotations are already omitted by the recursive routine:

  6  0  0  0
  5  1  0  0
  5  0  1  0
  5  0  0  1
  4  2  0  0
  4  1  1  0
  4  1  0  1
  4  0  2  0
  4  0  1  1
  4  0  0  2
  3  3  0  0  <==== X
  3  2  1  0
  3  2  0  1
  3  1  2  0
  3  1  1  1
  3  1  0  2
  3  0  3  0
  3  0  2  1
  3  0  1  2
  3  0  0  3   <=== X
  2  2  2  0   <=== Y
  2  2  1  1
  2  2  0  2   <=== Y
  2  1  2  1
  2  1  1  2
  2  0  2  2   <=== Y

 Here the two X's and the three Y's are
 multiple occurences (just with the cyclical
 shift) which my routine did not yet remove
 itself.
 After manually deleting that multiples
 I get the number of f(N,M) with the
 same results as of your table - which
 I only found after a query at the OEIS
 with my heuristic results. Thumbs up, OEIS! :-)

 What I do not yet see is, how your
 criterion of modularity translates to my
 criterion of cyclicitiness.

 late regards -

 Gottfried

 P.s. by the way: do you have an idea how to compute
 a correct table (not only its length) so that I can
 improve my routine?