[seqfan] Re: Modular Partitions

[Neil Sloane]

Gottfried, Did you look at A241926? Perhaps that answers your questions?
Neil


On Wed, Aug 13, 2014 at 12:10 PM, Gottfried Helms <helms at uni-kassel.de>
wrote:

> 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?
>
>
>
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>



-- 
Dear Friends, I have now retired from AT&T. New coordinates:

Neil J. A. Sloane, President, OEIS Foundation
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
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