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A198456 Consider triples a<=b where (a^2+b^2-c^2)/(c-a-b) = 1, ordered by a and then b; sequence gives c values.

%I A198456 #29 Feb 16 2025 21:25:12
%S A198456 3,6,10,8,15,11,21,28,13,36,16,23,45,28,55,18,23,66,21,27,78,46,91,20,
%T A198456 23,36,53,105,26,41,120,136,28,52,77,153,31,58,86,171,40,49,190,33,44,
%U A198456 54,71,210,36,41,78,116
%N A198456 Consider triples a<=b<c where (a^2+b^2-c^2)/(c-a-b) = 1, ordered by a and then b; sequence gives c values.
%C A198456 See A198453.
%C A198456 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 A198456 A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.
%H A198456 Ron Knott, <a href="https://r-knott.surrey.ac.uk/pythag/pythag.html">Pythagorean Triples and Online Calculators</a>.
%H A198456 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 A198456 2*3 + 2*3 = 3*4
%e A198456 3*4 + 5*6 = 6*7
%e A198456 4*5 + 9*10 = 10*11
%e A198456 5*6 + 6*7 = 8*9
%e A198456 5*6 + 14*15 = 15*16
%e A198456 6*7 + 9*10 = 11*12
%o A198456 (True BASIC)
%o A198456 input k
%o A198456 for a = (abs(k)-k+4)/2 to  40
%o A198456 for b = a to (a^2+abs(k)*a+2)/2
%o A198456   let t = a*(a+k)+b*(b+k)
%o A198456    let c =int((-k+ (k^2+4*t)^.5)/2)
%o A198456     if c*(c+k)=t then print a;b;c,
%o A198456 next b
%o A198456 print
%o A198456 next a
%o A198456 end
%Y A198456 Cf. A000217, A156682, A198453-A198455, A198457-A198469, A333530, A333531.
%K A198456 nonn
%O A198456 1,1
%A A198456 _Charlie Marion_, Oct 26 2011