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

[Max Alekseyev]

Neil,

Example for n=5 should be
{1/60, 1/30, 1/20, 1/12, 7/60, 2/15, 1/6, 1/5, 1/5}
and it gives g(5) <= 12.

In fact, I confirm that
g(5) = 12
g(6) = 10 (somehow Gurobi solves this very quickly)

I'll see if I can get g(7)...

Regards,
Max



On Sun, Jan 24, 2016 at 8:05 PM, Neil Sloane <njasloane at gmail.com> wrote:

> Let me try that message again, correcting some errors: This replaces my
> previous message.
>
> Max, let me reveal what I really wanted!
>
> The original problem is:
>
> Given n, look for the smallest multiset 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}
>
> This involves fractions. If we multiply through by the LCM of the
> denominators we get an equivalent
> problem in integers: find the smallest number of integers F_1, ..., F_M
> (meaning minimize M)
> such that they can be partitioned into k clumps with equal sums for all k =
> 1..n.
>
> Now look at all solutions for a given n (having the minimal number of
> fractions), and define g(n) to be the min value of max{F_1, ...,f_M}.
>
> Then g(1)=1, g(2)=1 [use {1,1}], g(3)=2 [use {1,1,2,2}].
>
> From your example in A265286 for n=5 with M=9, that is,
> 1/30, 1/30, 1/20, 1/15, 1/15, 1/12, 1/12, 7/60, 2/15, 1/6, 1/6],
> after multiplying by the LCM, 60, we get
> 2,2,3,4,4,5,5,7,8,10,10,
> so g(5) <= 10.
>
> Your example for n=6, M=11,
> 1/30, 1/30, 1/20, 1/15, 1/15, 1/12, 1/12, 7/60, 2/15, 1/6, 1/6]
> shows that g(6) <= 10.
>
> Can you supply any other values of g(n)?
>
> >
> >
>
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