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A198457 Consider triples (a, b, c) where a <= b < c and (a^2+b^2-c^2)/(c-a-b) = 2, ordered by a and then b; sequence gives a, b and c values in that order.

%I A198457 #40 Feb 16 2025 21:24:52
%S A198457 3,6,7,4,4,6,5,16,17,6,10,12,7,8,11,7,30,31,8,18,20,9,14,17,9,48,49,
%T A198457 10,12,16,10,28,30,11,70,71,12,18,22,12,40,42,13,16,21,13,30,33,13,96,
%U A198457 97,14,25,29,14,54,56,15,22,27,15,40,43,15,126,127,16,20,26
%N A198457 Consider triples (a, b, c) where a <= b < c and (a^2+b^2-c^2)/(c-a-b) = 2, ordered by a and then b; sequence gives a, b and c values in that order.
%C A198457 See A198453.
%C A198457 Because either all sides or only one side of a Pythagorean (-+2)-triangle ABC is even their sum is always even. Thus csc(C) = -(a+b+c+k)/k is an integer. So ((a+2)^2 + (b+2)^2 - (c+2)^2)|(2*(a+2)*(b+2)) resp. (a^2 + b^2 - c^2)|(2*a*b). - _Ralf Steiner_, Sep 18 2019
%D A198457 A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.
%H A198457 David A. Corneth, <a href="/A198457/b198457.txt">Table of n, a(n) for n = 1..9999</a>
%H A198457 Ron Knott, <a href="https://r-knott.surrey.ac.uk/pythag/pythag.html">Pythagorean Triples and Online Calculators</a>.
%e A198457 3*5 +  6*8  =  7*9;
%e A198457 4*6 +  4*6  =  6*8;
%e A198457 5*7 + 16*17 = 17*18;
%e A198457 6*8 + 10*12 = 12*14;
%e A198457 7*9 +  8*10 = 11*13;
%e A198457 7*9 + 30*32 = 31*33.
%o A198457 (True BASIC)
%o A198457 input k
%o A198457 for a = (abs(k)-k+4)/2 to 40
%o A198457    for b = a to (a^2+abs(k)*a+2)/2
%o A198457       let t = a*(a+k)+b*(b+k)
%o A198457       let c = int((-k+(k^2+4*t)^.5)/2)
%o A198457       if c*(c+k)=t then print a;b;c,
%o A198457    next b
%o A198457    print
%o A198457 next a
%o A198457 end
%Y A198457 Cf. A103606, A198453-A198456, A198458-A198468.
%K A198457 nonn,tabf
%O A198457 1,1
%A A198457 _Charlie Marion_, Nov 09 2011
%E A198457 More terms from _David A. Corneth_, Sep 22 2019