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A198465 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 A198465 #8 Jul 07 2016 23:48:50
%S A198465 3,3,4,4,6,7,5,10,11,6,7,9,6,15,16,7,10,12,7,21,22,8,28,29,9,11,14,9,
%T A198465 36,37,10,14,17,10,22,24,10,45,46,11,27,29,11,55,56,12,15,19,12,21,24,
%U A198465 12,66,67,13,18,22,13,25,28,13,78,79,14,45,47,14,91,92
%N A198465 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 A198465 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).  See A198453 for more about Pythagorean k-triples.
%D A198465 A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.
%H A198465 Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>
%e A198465 3*2 + 3*2 = 4*3
%e A198465 4*3 + 6*5 = 7*6
%e A198465 5*4 + 10*9 = 11*10
%e A198465 6*5 + 7*6 = 9*8
%e A198465 6*5 + 15*14 = 16*15
%o A198465 (True BASIC)
%o A198465 input k
%o A198465 for a = (abs(k)-k+4)/2 to  40
%o A198465 for b = a to (a^2+abs(k)*a+2)/2
%o A198465   let t = a*(a+k)+b*(b+k)
%o A198465    let c =int((-k+ (k^2+4*t)^.5)/2)
%o A198465     if c*(c+k)=t then print a; b; c,
%o A198465 next b
%o A198465 print
%o A198465 next a
%o A198465 end
%Y A198465 Cf. A103606, A198453-A198469.
%K A198465 nonn
%O A198465 1,1
%A A198465 _Charlie Marion_, Dec 19 2011