[seqfan] Re: Squares packings

[Fred Lunnon]

<< Sounds good for the upper bound.  I have also confirmed the lower bound up
through n = 100. >>

  Unclear: does your program establish that  _NO_ solution exists without
some 1x1 tile for edge length  n mod 6 in {1, 5}  &  n < 100 ?

  If so, this suggests an interesting conjecture ...    WFL



On 10/12/20, Rob Pratt <robert.william.pratt at gmail.com> wrote:
> Sounds good for the upper bound.  I have also confirmed the lower bound up
> through n = 100.
>
> On Mon, Oct 12, 2020 at 3:06 AM Jack Grahl <jack.grahl at gmail.com> wrote:
>
>> If we have a solution with 1 small square for n=11 and n=13, doesn't that
>> imply a solution with 1 small square for all larger numbers 6k+1 and 6k-1
>> (ie all for which the solution is greater than zero)?
>>
>> Simply form an L-shape with width 6, to extend a solution for 6k+1 to a
>> solution for 6(k+1)+1. The L shape is made up of a 6x6 square, and two
>> 6x(6k+1) strips. Since any number >1 can be formed by a sum of 2's and
>> 3's,
>> we can always make the strips. Same for 6k-1 of course.
>>
>> Jack Grahl
>>
>> On Sun, 11 Oct 2020, 20:25 Rob Pratt, <robert.william.pratt at gmail.com>
>> wrote:
>>
>> > Via integer linear programming, I get instead
>> > 1 0 0 0 4 0 3 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 1
>> >
>> > Here's an optimal solution for n = 17, with the only 1x1 appearing in
>> cell
>> > (12,6):
>> >   1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
>> > 1 2 2 3 3 38 38 38 4 4 5 5 6 6 7 7 8 8
>> > 2 2 2 3 3 38 38 38 4 4 5 5 6 6 7 7 8 8
>> > 3 9 9 10 10 38 38 38 11 11 12 12 13 13 14 14 15 15
>> > 4 9 9 10 10 39 39 39 11 11 12 12 13 13 14 14 15 15
>> > 5 16 16 17 17 39 39 39 18 18 19 19 40 40 40 41 41 41
>> > 6 16 16 17 17 39 39 39 18 18 19 19 40 40 40 41 41 41
>> > 7 42 42 42 20 20 21 21 22 22 23 23 40 40 40 41 41 41
>> > 8 42 42 42 20 20 21 21 22 22 23 23 24 24 25 25 26 26
>> > 9 42 42 42 27 27 43 43 43 44 44 44 24 24 25 25 26 26
>> > 10 45 45 45 27 27 43 43 43 44 44 44 28 28 29 29 30 30
>> > 11 45 45 45 31 31 43 43 43 44 44 44 28 28 29 29 30 30
>> > 12 45 45 45 31 31 1 46 46 46 47 47 47 32 32 48 48 48
>> > 13 33 33 34 34 35 35 46 46 46 47 47 47 32 32 48 48 48
>> > 14 33 33 34 34 35 35 46 46 46 47 47 47 36 36 48 48 48
>> > 15 49 49 49 50 50 50 51 51 51 52 52 52 36 36 53 53 53
>> > 16 49 49 49 50 50 50 51 51 51 52 52 52 37 37 53 53 53
>> > 17 49 49 49 50 50 50 51 51 51 52 52 52 37 37 53 53 53
>> >
>> > Clearly, even 2n can be partitioned into 2x2 only, and 3n can be
>> > partitioned into 3x3 only.  Otherwise, it appears that the minimum is 1
>> for
>> > n >= 11.
>> >
>> > On Sun, Oct 11, 2020 at 3:11 AM <michel.marcus at free.fr> wrote:
>> >
>> > >
>> > > Hello Seqfans,
>> > >
>> > >
>> > > "Images du CNRS" French site 4th problem for September 2020 is:
>> > > .
>> > > What is the minimum number of 1X1 pieces to fill a 23X23 square with
>> 1X1,
>> > > 2X2, and 3X3 pieces.
>> > > Problem is at http://images.math.cnrs.fr/Septembre-2020-4e-defi.html.
>> > > Solution is at http://images.math.cnrs.fr/Octobre-2020-1er-defi.html
>> > > (click in Solution du 4e défi de septembre)
>> > >
>> > >
>> > > I wondered what we get for other squares and painstakingly obtained
>> > > for
>> > > n=1 up to n=23:
>> > > 1 0 0 0 4 0 3 0 0 0 1 0 1 0 0 0 4 0 4 0 0 0 1
>> > >
>> > >
>> > > Do you see some mistakes? Is it possible to extend it?
>> > >
>> > >
>> > > Thanks. Best.
>> > > MM
>> > >
>> > > --
>> > > Seqfan Mailing list - http://list.seqfan.eu/
>> > >
>> >
>> > --
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>>
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