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

[Max Alekseyev]

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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