[seqfan] Re: Pieces of cake sequence A265286 and a question

[Allan Wechsler]

Max, you can trick a MILP solver into giving all solutions by using as your
objective function a weighted sum of all the integer variables, with
irrational numbers as the weights. (Of course this can only be approximated
in floating point.) One solution optimizes this objective. Set that
solution aside and add a constraint that requires the objective to be less
than this optimum, and re-solve. This will give a second-best solution. By
continually adjusting the new constraint downward, you will list all the
solutions.

On Fri, Jan 22, 2016 at 7:21 PM, Max Alekseyev <maxale at gmail.com> wrote:

> Neil,
>
> Frankly speaking, I do not know how to get all the solutions for a MILP
> problem. I'm using Sage (powered with Gurobi solver) here and it reports
> only one solution.
>
> Instead, I've tried to minimize the smallest piece in a solution and found
> that 1/120 is the smallest possible piece (in a 9-piece solution) for n=5.
> This does not guarantee though that the denominator cannot go above 120
> (e.g., this does not eliminate a possibility of a part being, say, 7/240).
> Now it's running for n=6.
>
> Regards,
> Max
>
>
> On Fri, Jan 22, 2016 at 4:38 PM, Neil Sloane <njasloane at gmail.com> wrote:
>
> > OK, then I will add the question to A265286.
> >
> > But your integer programming approach make it possible to find ALL the
> > solutions for small n,
> > right, so in principal the question could be answered?
> >
> >
> >
> > Best regards
> > Neil
> >
> > Neil J. A. Sloane, President, OEIS Foundation.
> > 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> > Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
> > Phone: 732 828 6098; home page: http://NeilSloane.com
> > Email: njasloane at gmail.com
> >
> >
> > On Fri, Jan 22, 2016 at 4:34 PM, Max Alekseyev <maxale at gmail.com> wrote:
> >
> > > Neil,
> > > I'm sorry I misread your question. I do not know the answer and believe
> > > your question is a hard one, provided that we don't even know how to
> > > compute A265286 efficiently.
> > > Regards,
> > > Max
> > >
> > > On Fri, Jan 22, 2016 at 4:31 PM, Max Alekseyev <maxale at gmail.com>
> wrote:
> > >
> > > > Hi Neil,
> > > > The second example in A265286 negatively answers your question.
> > > > Regards,
> > > > Max
> > > >
> > > > On Fri, Jan 22, 2016 at 3:45 PM, Neil Sloane <njasloane at gmail.com>
> > > wrote:
> > > >
> > > >> Rainer R. mentioned Max Alekseyev's lovely sequence A265286.
> > > >>
> > > >> Given n, look for the smallest set of fractions {f_1, f_2, ..., f_M}
> > in
> > > >> the
> > > >> range 0 to 1 such that for each k with 1 <= k <= n, we can partition
> > the
> > > >> f_i into k groups whose sums are equal. For n=5 the minimal M is 9,
> > and
> > > a
> > > >> solution is
> > > >> {1/60, 1/30, 1/20, 1/12, 7/60, 2/15, 1/6, 1/5, 1/5}
> > > >>
> > > >> OK, now look at all the solutions for a given value of n,
> > > >> with M (the minimal value) parts. Now ask, what is the minimal
> > > >> denominator?
> > > >>
> > > >> Is it always A003418(n) = LCM{1,2,...,n}?
> > > >>
> > > >> If not, we get a new sequence: given n, first minimize the number of
> > > >> parts,
> > > >> then minimize the biggest denominator
> > > >>
> > > >> Max, do you know the answer?
> > > >>
> > > >> _______________________________________________
> > > >>
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> > > >>
> > > >
> > > >
> > >
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