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%I #49 Mar 14 2017 11:42:19
%S 6,24,30,54,60,84,96,120,150,180,210,210,216,240,270,294,330,336,384,
%T 480,486,504,540,546,600,630,720,726,750,756,840,840,840,864,924,960,
%U 990,1014,1080,1176,1224,1320,1320,1344,1350,1386,1470,1500,1536,1560,1620
%N List of ordered areas of Pythagorean triangles.
%C All terms are divisible by 6.
%C Let k be even, k > 2, q = (k/2)^2 - 1, and b = (kq)/2. Then, for any k, b is a term of a(n). In other words, for any even k > 2, there is at least one such integer q > 2 that b = (kq)/2 and b is a term of a(n), while hypotenuse c = q + 2 (proved by _Anton Mosunov_). - _Sergey Pavlov_, Mar 02 2017
%C Let x be odd, x > 1, k == 0 (mod x), k > 0, y = (x-1)/2, q = ky + (ky/x), b = (kq)/2. Then b is a term of a(n), while hypotenuse c = q + k/x. As a special case of the above equation (k = x), for each odd k > 1 there exist such q and b that q = (k^2 - 1)/2, b = (kq)/2, and b is a term of a(n), while hypotenuse c = q + 1. - _Sergey Pavlov_, Mar 06 2017
%D Albert H. Beiler, Recreations in the Theory of Numbers, The Queen of Mathematics Entertains, 2nd Ed., Chpt. XIV, "The Eternal Triangle", pp. 104-134, Dover Publ., NY, 1964.
%D Andrew Granville, Solution to Problem 90:07, Western Number Theory Problems, 1991-12-19 & 22, ed. R. K. Guy.
%H T. D. Noe, <a href="/A009111/b009111.txt">Table of n, a(n) for n = 1..1000</a>
%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html#periarea">Pythagorean Triangles</a>
%H Supriya Mohanty and S. P. Mohanty, <a href="http://www.fq.math.ca/Scanned/28-1/mohanty.pdf">Pythagorean Numbers</a>, Fibonacci Quarterly 28 (1990), 31-42.
%F Theorem: The number of pairs of integers a > b > 0 with ab(a^2-b^2) < n^2 is Cn + O(n^(2/3)) where C = (1/2)*Integral_{1..infinity} du/sqrt(u^3-u). [Granville] - _N. J. A. Sloane_, Feb 07 2008
%e 6 is in the sequence because it is the area of the 3-4-5 triangle.
%t t = {}; nn = 200; mx = Sqrt[2*nn - 1] (nn - 1)/2; Do[x = Sqrt[n^2 - d^2]; If[x > 0 && IntegerQ[x] && x > d && d*x/2 <= mx, AppendTo[t, d*x/2]], {n, nn}, {d, n - 1}]; t = Sort[t]; t (* _T. D. Noe_, Sep 23 2013 *)
%Y Cf. A009112, A024365.
%K nonn
%O 1,1
%A _David W. Wilson_
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