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A198467 Consider triples a<=b where (a^2+b^2-c^2)/(c-a-b) = -1, ordered by a and then b; sequence gives b values.

%I A198467 #8 Jul 07 2016 23:48:50
%S A198467 3,6,10,7,15,10,21,28,11,36,14,22,45,27,55,15,21,66,18,25,78,45,91,15,
%T A198467 19,34,52,105,22,39,120,136,23,50,76,153,26,56,85,171,36,46,190,27,40,
%U A198467 51,69,210,30,36,76,115,231,126,253,31,91,276,34,58,99,300
%N A198467 Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) = -1, ordered by a and then b; sequence gives b values.
%C A198467 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 A198467 A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.
%H A198467 Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>
%e A198467 3*2 + 3*2 = 4*3
%e A198467 4*3 + 6*5 = 7*6
%e A198467 5*4 + 10*9 = 11*10
%e A198467 6*5 + 7*6 = 9*8
%e A198467 6*5 + 15*14 = 16*15
%o A198467 (True BASIC)
%o A198467 input k
%o A198467 for a = (abs(k)-k+4)/2 to  40
%o A198467 for b = a to (a^2+abs(k)*a+2)/2
%o A198467   let t = a*(a+k)+b*(b+k)
%o A198467    let c =int((-k+ (k^2+4*t)^.5)/2)
%o A198467     if c*(c+k)=t then print a; b; c,
%o A198467 next b
%o A198467 print
%o A198467 next a
%o A198467 end
%Y A198467 Cf. A103606, A198453-A198469.
%K A198467 nonn
%O A198467 1,1
%A A198467 _Charlie Marion_, Dec 19 2011