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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.
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%I #8 Jul 07 2016 23:48:50

%S 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 19,34,52,105,22,39,120,136,23,50,76,153,26,56,85,171,36,46,190,27,40,

%U 51,69,210,30,36,76,115,231,126,253,31,91,276,34,58,99,300

%N 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 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 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*2 + 3*2 = 4*3

%e 4*3 + 6*5 = 7*6

%e 5*4 + 10*9 = 11*10

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

%e 6*5 + 15*14 = 16*15

%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, A198453-A198469.

%K nonn

%O 1,1

%A _Charlie Marion_, Dec 19 2011