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

israel at math.ubc.ca israel at math.ubc.ca
Sat Apr 23 00:15:43 CEST 2016


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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>>
>
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