OFFSET
1,1
COMMENTS
All terms are divisible by 6.
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
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
REFERENCES
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.
Andrew Granville, Solution to Problem 90:07, Western Number Theory Problems, 1991-12-19 & 22, ed. R. K. Guy.
LINKS
T. D. Noe, Table of n, a(n) for n = 1..1000
Ron Knott, Pythagorean Triangles
Supriya Mohanty and S. P. Mohanty, Pythagorean Numbers, Fibonacci Quarterly 28 (1990), 31-42.
FORMULA
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
EXAMPLE
6 is in the sequence because it is the area of the 3-4-5 triangle.
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved