# [seqfan] Re: [math-fun] Re: Squares packings

Fred Lunnon fred.lunnon at gmail.com
Tue Oct 13 03:39:22 CEST 2020

```  Doh --- what's worse, JG had just beaten me to making the the exact
same remark myself!

WFL

On 10/13/20, Rob Pratt <robert.william.pratt at gmail.com> wrote:
> (B) is already settled by Jack Grahl’s argument.
>
>> On Oct 12, 2020, at 9:11 PM, Fred Lunnon <fred.lunnon at gmail.com> wrote:
>>
>> ﻿  I don't know --- I was hoping that we have some seasoned packer of
>> squares
>> on hand who can tell us what might be known.
>>
>>  In the meantime, just to set up a formal aunt Sally:
>> Conjecture: For natural  n mod 6 in {1, 5} :
>> (A) no packing of a  n x n  box by  1 x 1,  2 x 2,  3 x 3  tiles
>> exists with no  1 x 1  tile;
>> (B) there exists some packing with only a single  1 x 1  tile.
>>
>> WFL
>>
>>
>>
>>> On 10/13/20, rcs at xmission.com <rcs at xmission.com> wrote:
>>> Is it established that large 6k+-1 squares require any 1x1s?  --Rich
>>>
>>> -----
>>> Quoting Fred Lunnon <fred.lunnon at gmail.com>:
>>>> << 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
>>>>>>>>
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>>>>>>>>
>>>>>>>
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```