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%I #10 Jul 07 2016 23:48:49
%S 7,6,17,12,11,31,20,17,49,16,30,71,22,42,21,33,97,29,56,27,43,127,26,
%T 37,72,161,32,46,90,31,67,199,56,110,37,81,241,36,46,67,132,59,287,42,
%U 54,79,156,41,69,113,337,92,182,47,131,391,40,46,72,106,210,449,45,52
%N Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) =2, ordered by a and then b; sequence gives c 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 Knott’s 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 3*5 + 6*8 = 7*9
%e 4*6 + 4*6 = 6*8
%e 5*7 + 16*17 = 17*18
%e 6*8 + 10*12 12*14
%e 7*9 + 8*10 = 11*13
%e 7*9 + 30*32 = 31*33
%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 15 2011
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