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

Neil Sloane njasloane at gmail.com
Fri Jan 22 22:38:17 CET 2016


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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