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

[Max Alekseyev]

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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>> > >
>> > >
>> >
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