
Re: Pythagorean triples
Posted:
Jun 26, 2006 9:12 AM


> Sorry, I should have specified 'an infinite number of > *nonsimilar*' triples, where one triple is similar > to another if each member is multiple of the > corresponding member.
Where do you see that my formula generates similar triples ????? They are not, of course !
However, there can't be similar triangles with b=a+1 because a, a+1 are allways coprime !
The list results into :
(3,4,5) (20,21,29) (119,120,169) (696,697,985) ...
just applying the recurrence relation from (0,1,1).
To get direct values without walking through recurrence relations, we also can use the formula
x(n) = (floor( (1+sqrt(2))(3+2sqrt(2))^n )  2 )/4
and for instance the huge triangle for n = 15 : 183648021599, 183648021600, 259717522849 can be obtained directly without calculating the 14 others
Regards.  Philippe C. chephip@free.fr

