[seqfan] Re: No right isosceles triangles in a square grid

[Neil Sloane]

Looking at the isosceles-triangle avoiding problem, but for an m X n
rectangular grid, where m>1, n>1,
and without much data, I wildly conjecture that the answer is m+n-2.
And that this can be achieved by taking the left edge and the top edge,
removing the
intersection, and then moving Xs in the top row down a bit to avoid any
Pythagorean accidents.

Can anyone supply some more data for the rectangular case?

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, Apr 22, 2016 at 6:15 PM, <israel at math.ubc.ca> wrote:

> That comment would work for the case of right isosceles triangles
> (A271906), but there it looks like A271906(n) = 3n-7 for n >= 6.
>
> I can confirm that A271906(11) >= 26, with a solution
>     0     0     1     0     0     0     0     0     0     0     1
>     1     0     1     0     0     0     0     0     0     0     1
>     1     0     0     0     0     0     0     0     0     0     1
>     0     0     0     0     1     1     0     0     0     0     1
>     0     0     0     1     0     0     0     0     0     0     1
>     0     0     0     1     0     0     0     0     0     0     1
>     0     0     0     0     0     0     0     0     0     0     1
>     0     0     0     0     0     0     0     0     0     0     1
>     0     0     0     0     0     0     0     0     0     1     0
>     0     0     0     0     0     0     0     0     1     0     0
>     1     1     1     1     1     1     1     1     0     0     0
>
> and A271906(12) >= 29, with a solution
>
>     0     0     0     0     1     1     1     1     1     1     1     1
>     0     0     0     1     0     0     0     0     0     0     0     0
>     0     0     1     0     0     0     0     0     0     0     0     0
>     0     1     0     0     0     0     0     0     0     0     0     0
>     1     0     0     0     0     0     0     0     0     0     0     0
>     1     0     0     0     0     0     0     0     1     0     0     0
>     1     0     0     0     0     0     0     0     1     0     0     0
>     1     0     0     0     0     0     0     0     0     0     1     0
>     1     0     0     0     0     1     1     0     0     0     0     0
>     1     0     0     0     0     0     0     0     0     0     0     1
>     1     0     0     0     0     0     0     1     0     0     0     1
>     1     0     0     0     0     0     0     0     0     1     1     0
>
> I'm not sure that these are optimal, but my searches for better solutions
> have come up empty so far.
>
> Cheers,
> Robert
>
>
> On Apr 22 2016, Rob Pratt wrote:
>
> This comment seems wrong; " By using two edges of the grid minus their
>> intersection we get a(n) >= 2n-2. - N. J. A. Sloane, Apr 22 2016"
>>
>> For example, the cells (1,5), (4,1), and (9,1) lie in such a set but form
>> an isosceles triangle.
>>
>> -----Original Message----- From: SeqFan [mailto:
>> seqfan-bounces at list.seqfan.eu] On Behalf Of Neil Sloane Sent: Friday,
>> April 22, 2016 2:50 PM To: Sequence Fanatics Discussion list <
>> seqfan at list.seqfan.eu> Subject: [seqfan] Re: No right isosceles
>> triangles in a square grid
>>
>> If you did not find any clever solutions for n <= 11, I bet we have  a(n)
>> =
>> 2n-2 for all n>1.
>>
>> 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, Apr 22, 2016 at 1:35 PM, Giovanni Resta <g.resta at iit.cnr.it>
>> wrote:
>>
>> Il 22/04/2016 17:34, Neil Sloane ha scritto:
>>>
>>> 1, 2, 4, 6, 8, 10, 12, 14, 14, 16, 20
>>>>
>>>>>
>>>>> Could there be a typo? Did you mean to say
>>>>
>>>> Yes, sorry. I did make a mess merging results from Mathematica,
>>> lp_solve and cplex.
>>> It is 1,2,4,6,8,10,12,14,16,18 (I'm checking again a(11)=20 because I
>>> deleted the result),  i.e.,  it seems simply a(1)=1 and a(n)=2n-2 up to
>>> a(10).
>>>
>>> I'm going to check again the results when I prepare the pictures. Sorry
>>> for the error,
>>>
>>> Giovanni
>>>
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>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>
>>>
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