# [seqfan] Re: Squares packings

Rob Pratt robert.william.pratt at gmail.com
Tue Oct 13 02:19:42 CEST 2020

```Correct. The minimum value is 1 for those n.

> On Oct 12, 2020, at 7:43 PM, Fred Lunnon <fred.lunnon at gmail.com> wrote:
>
> ﻿<< 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/
>>>>>
>>>>
>>>> --
>>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>>
>>>
>>> --
>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/

```