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

[Rob Pratt]

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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