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A198458 Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) =2, ordered by a and then b; sequence gives a values. 0

%I #10 Jul 07 2016 23:48:49

%S 3,4,5,6,7,7,8,9,9,10,10,11,12,12,13,13,13,14,14,15,15,15,16,16,16,17,

%T 18,18,18,19,19,19,20,20,21,21,21,22,22,22,22,23,23,24,24,24,24,25,25,

%U 25,25,26,26,27,27,27,28,28,28,28,28,29,30,30,30,30,30,30,31,31,31

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

%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 3*5 + 6*8 = 7*9

%e 4*6 + 4*6 = 6*8

%e 5*7 + 16*17 = 17*18

%e 6*8 + 10*12 12*14

%e 7*9 + 8*10 = 11*13

%e 7*9 + 30*32 = 31*33

%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 15 2011

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