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A198468 Consider triples a<=b where (a^2+b^2-c^2)/(c-a-b) = -1, ordered by a and then b; sequence gives c values.

%I A198468 #8 Jul 07 2016 23:48:50
%S A198468 4,7,11,9,16,12,22,29,14,37,17,24,46,29,56,19,24,67,22,28,79,47,92,21,
%T A198468 24,37,54,106,27,42,121,137,29,53,78,154,32,59,87,172,41,50,191,34,45,
%U A198468 55,72,211,37,42,79,117,232,128,254,39,94,277,42,63,102,301
%N A198468 Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) = -1, ordered by a and then b; sequence gives c values.
%C A198468 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 A198468 A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.
%H A198468 Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>
%e A198468 3*2 + 3*2 = 4*3
%e A198468 4*3 + 6*5 = 7*6
%e A198468 5*4 + 10*9 = 11*10
%e A198468 6*5 + 7*6 = 9*8
%e A198468 6*5 + 15*14 = 16*15
%o A198468 (True BASIC)
%o A198468 input k
%o A198468 for a = (abs(k)-k+4)/2 to  40
%o A198468 for b = a to (a^2+abs(k)*a+2)/2
%o A198468   let t = a*(a+k)+b*(b+k)
%o A198468    let c =int((-k+ (k^2+4*t)^.5)/2)
%o A198468     if c*(c+k)=t then print a; b; c,
%o A198468 next b
%o A198468 print
%o A198468 next a
%o A198468 end
%Y A198468 Cf. A103606, A198453-A198469.
%K A198468 nonn
%O A198468 1,1
%A A198468 _Charlie Marion_, Dec 19 2011