%I #8 Jul 07 2016 23:48:50
%S 3,7,12,18,6,11,25,15,33,42,10,15,52,30,63,36,75,14,19,27,88,102,75,
%T 117,18,23,42,65,133,150,30,39,168,22,27,60,92,187,102,207,42,54,228,
%U 22,26,31,81,250,51,135,273,147,297,30,35,105,322,45,66,84,348
%N Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) =3, ordered by a and then b; sequence gives b values.
%C The definition can be generalized to define Pythagorean k-triples a<=b<c where (a^2+b^2-c^2)/(c-a-b)=k, or where for some integer k, a(a+k) + b(b+k) = c(c+k).
%C If a, b and c form a Pythagorean k-triple, then na, nb and nc form a Pythagorean nk-triple.
%C A triangle is defined to be a Pythagorean k-triangle if its sides form a Pythagorean k-triple.
%C If a, b and c are the sides of a Pythagorean k-triangle ABC with a<=b<c, then cos(C) = -k/(a+b+c+k) which proves that such triangles must be obtuse when k>0 and acute when k<0. When k=0, the triangles are Pythagorean, as in the Beiler reference and Ron Knottās link. For all k, the area of a Pythagorean k-triangle ABC with a<=b<c equals sqrt((2ab)^2-(k(a+b-c))^2))/4.
%D A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.
%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>
%e 2*5 + 3*6 = 4*7
%e 3*6 + 7*10 = 8*11
%e 4*7 +12*15 = 13*16
%e 5*8 + 18*21 = 19*22
%e 6*9 = 6*9 = 9*12
%e 6*9 = 11*14 = 13*16
%o (True BASIC)
%o input k
%o for a = (abs(k)-k+4)/2 to 40
%o for b = a to (a^2+abs(k)*a+2)/2
%o let t = a*(a+k)+b*(b+k)
%o let c =int((-k+ (k^2+4*t)^.5)/2)
%o if c*(c+k)=t then print a; b; c,
%o next b
%o print
%o next a
%o end
%Y Cf. A103606, A198454-A198469.
%K nonn
%O 1,1
%A _Charlie Marion_, Nov 26 2011