login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

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.
2

%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