A006339

[Graeme McRae]

Sorry.  I hit "send" too soon...

Start with the formula in A004018.
Subtract the four lattice points on the axes.
Divide by four to get the results in the first quadrant.
Then divide by two to count a,b,c and b,a,c just once.

--Graeme

----- Original Message ----- 
From: "Graeme McRae" <g_m at mcraefamily.com>
To: "seqfan" <seqfan at ext.jussieu.fr>
Sent: Monday, January 16, 2006 12:04 PM
Subject: Re: A006339


> Start with A004018.  See the formula listed there.  Pythagorean triangles 
> can't have zero-length sides, so subtract 4 to eliminate the 4 points on 
> the axes.  Triangles can't have negative length sides, so divide that 
> number by 4 to get the solutions in the first quadrant only.  In your 
> case, 1105^2 = 5^2 13^2 17^2, so the number of lattice points is 
> 4(3)(3)(3).  The number of triangles is (4(3)(3)(3)-4)/4 = 13
> --Graeme
>
> ----- Original Message ----- 
> From: "David Wilson" <davidwwilson at comcast.net>
> To: "Sequence Fans" <seqfan at ext.jussieu.fr>
> Sent: Monday, January 16, 2006 11:22 AM
> Subject: Re: A006339
>
>
>> Can someone tell me how to compute, say, the number of distinct 
>> Pythagorean triangles with hypotenuse 1105?
>>
>>
>
>