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Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) = 3, ordered by a and then b; sequence gives a, b and c values in that order.
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%I #8 Jul 07 2016 23:48:49

%S 2,3,4,3,7,8,4,12,13,5,18,19,6,6,9,6,11,13,6,25,26,7,15,17,7,33,34,8,

%T 42,43,9,10,14,9,15,18,9,52,53,10,30,32,10,63,64,11,36,38,11,75,76,12,

%U 14,19,12,19,23,12,27,30,12,88,89,13,102,103,14,57,59,14,117,118

%N Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) = 3, ordered by a and then b; sequence gives a, b and c values in that order.

%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 Knottā€™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 2*5 + 3*6 = 4*7

%e 3*6 + 7*10 = 8*11

%e 4*7 +12*15 = 13*16

%e 5*8 + 18*21 = 19*22

%e 6*9 = 6*9 = 9*12

%e 6*9 = 11*14 = 13*16

%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 26 2011