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

Max Alekseyev maxale at gmail.com
Sun Jan 24 20:52:26 CET 2016


Neil,

Minimizing the smallest piece is not directly related to denominators. For
example, even if the smallest piece for n=6 is 1/120 (as I've established),
this does not eliminate a possibility of a denominator being larger than
120 (e.g., there may be a piece 7/240 or alike).

As of sequence L(n) = min max denominator, my result does not imply L(6) =
120 either. E.g., there is a solution with the denominator = 60:
[1/30, 1/30, 1/20, 1/15, 1/15, 1/12, 1/12, 7/60, 2/15, 1/6, 1/6]

So, L(6) <= 60.

Regards,
Max


On Sun, Jan 24, 2016 at 1:27 PM, Neil Sloane <njasloane at gmail.com> wrote:

> Max, Very interesting! I have taken the liberty of adding that comment to
> A265286.
>
> So if L(n) denotes the min max denominator, your result is that L(6) = 120
> (instead of
> LCM{1..6} = 60).
>
> I would guess that L(1)=1, L(2)=2,  L(3)=6, L(4)=12,  L(5)=60. Do you know
> if this is correct?
>
>
> 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 Sun, Jan 24, 2016 at 10:33 AM, Max Alekseyev <maxale at gmail.com> wrote:
>
> > For n=6, the minimum smallest piece in an 11-piece solution equals 1/120,
> > like in
> > [1/120, 1/40, 1/30, 7/120, 3/40, 11/120, 13/120, 1/8, 17/120, 1/6, 1/6]
> >
> > It would take about 37 days of single-core computations in Sage+Gurobi.
> > Luckily I run on a 40-core system and it took only about a day to
> establish
> > the above result.
> >
> > Regards,
> > Max
> >
> > 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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