|
%I #22 Apr 26 2021 06:32:00
%S 2,5,9,6,14,9,20,27,10,35,13,21,44,26,54,14,20,65,17,24,77,44,90,14,
%T 18,33,51,104,21,38,119,135,22,49,75,152,25,55,84,170,35,45,189,26,39,
%U 50,68,209,29,35,75,114,230,125
%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).
%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 Knott’s link.
%C 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.
%C The definition amounts to saying that T_a+T_b=T_c where T_i denotes a triangular number (A000217). - _N. J. A. Sloane_, Apr 01 2020
%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>
%H J. S. Myers, R. Schroeppel, S. R. Shannon, N. J. A. Sloane, and P. Zimmermann, <a href="http://arxiv.org/abs/2004.14000">Three Cousins of Recaman's Sequence</a>, arXiv:2004:14000 [math.NT], April 2020.
%e 2*3 + 2*3 = 3*4
%e 3*4 + 5*6 = 6*7
%e 4*5 + 9*10 = 10*11
%e 5*6 + 6*7 = 8*9
%e 5*6 + 14*15 = 15*16
%e 6*7 + 9*10 = 11*12
%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. A000217, A156681, A198454-A198469, A333530, A333531.
%K nonn
%O 1,1
%A _Charlie Marion_, Oct 26 2011
|
