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A009111 List of ordered areas of Pythagorean triangles.

%I A009111 #53 Mar 12 2025 21:27:08
%S A009111 6,24,30,54,60,84,96,120,150,180,210,210,216,240,270,294,330,336,384,
%T A009111 480,486,504,540,546,600,630,720,726,750,756,840,840,840,864,924,960,
%U A009111 990,1014,1080,1176,1224,1320,1320,1344,1350,1386,1470,1500,1536,1560,1620
%N A009111 List of ordered areas of Pythagorean triangles.
%C A009111 All terms are divisible by 6.
%C A009111 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 A009111 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 A009111 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.
%H A009111 T. D. Noe, <a href="/A009111/b009111.txt">Table of n, a(n) for n = 1..1000</a>
%H A009111 Andrew Granville, Solution to Problem 90:07, <a href="https://westcoastnumbertheory.org/wp-content/uploads/2018/02/wcnt-problems-1991.pdf">Western Number Theory Problems, 1991-12-19</a> & 22, ed. R. K. Guy.
%H A009111 Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html#periarea">Pythagorean Triangles</a>
%H A009111 Supriya Mohanty and S. P. Mohanty, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/28-1/mohanty.pdf">Pythagorean Numbers</a>, Fibonacci Quarterly 28 (1990), 31-42.
%F A009111 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 A009111 6 is in the sequence because it is the area of the 3-4-5 triangle.
%t A009111 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 A009111 Cf. A009112, A024365.
%K A009111 nonn
%O A009111 1,1
%A A009111 _David W. Wilson_