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A198453 Consider triples a<=b where (a^2+b^2-c^2)/(c-a-b) = 1, ordered by a and then b; sequence gives a, b and c values in that order.

%I A198453 #81 Feb 16 2025 13:38:48
%S A198453 2,2,3,3,5,6,4,9,10,5,6,8,5,14,15,6,9,11,6,20,21,7,27,28,8,10,13,8,35,
%T A198453 36,9,13,16,9,21,23,9,44,45,10,26,28,10,54,55,11,14,18,11,20,23,11,65,
%U A198453 66,12,17,21,12,24,27
%N A198453 Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) = 1, ordered by a and then b; sequence gives a, b and c values in that order.
%C A198453 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 A198453 If a, b and c form a Pythagorean k-triple, then n*a, n*b and n*c form a Pythagorean n*k-triple.
%C A198453 A triangle is defined to be a Pythagorean k-triangle if its sides form a Pythagorean k-triple.
%C A198453 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.
%C A198453 For all k, the area of a Pythagorean k-triangle ABC with a<=b<c equals sqrt((2*a*b)^2-(k*(a+b-c))^2)/4.
%C A198453 Define a Pythagorean k-triple <a,b,c> to be primitive if and only if there are no integers r>1, s>0 such that <a,b,c> = <r*d,r*e,r*f> for some Pythagorean s-triple <d,e,f>. Thus, every Pythagorean 1-triple is primitive. For every k>1, the set of Pythagorean k-triples contains some non-primitive triples.
%C A198453 In particular, for d a proper divisor of k, it includes (k/d)*(a,b,c), where (a,b,c) is a Pythagorean d-triple. - _Franklin T. Adams-Watters_, Dec 01 2011
%D A198453 A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.
%H A198453 Ron Knott, <a href="https://r-knott.surrey.ac.uk/pythag/pythag.html">Pythagorean Triples and Online Calculators</a>
%e A198453 2*3 + 2*3 = 3*4
%e A198453 3*4 + 5*6 = 6*7
%e A198453 4*5 + 9*10 = 10*11
%e A198453 5*6 + 6*7 = 8*9
%e A198453 5*6 + 14*15 = 15*16
%e A198453 6*7 + 9*10 = 11*12
%o A198453 (True BASIC)
%o A198453 input k
%o A198453 for a = (abs(k)-k+4)/2 to  40
%o A198453 for b = a to (a^2+abs(k)*a+2)/2
%o A198453   let t = a*(a+k)+b*(b+k)
%o A198453    let c =int((-k+ (k^2+4*t)^.5)/2)
%o A198453     if c*(c+k)=t then print a;b;c,
%o A198453 next b
%o A198453 print
%o A198453 next a
%o A198453 end
%Y A198453 Cf. A103606, A198454-A198469.
%K A198453 nonn,tabf
%O A198453 1,1
%A A198453 _Charlie Marion_, Oct 25 2011