Similar to integral Pythagorean triangles, but with reciprocal of integers: 1/x and 1/y are the legs and 1/z is the hypotenuse. Properties:
1) x*y is congruent to 0 mod z;
2) x = u*v *(u^2+v^2) * m /2 or x = (u^2-v^2)*(u^2+v^2)*m/4; y = (u^2-v^2)*(u^2+v^2)*m/4 or y = u*v *(u^2+v^2)*m /2; z = (u^2-v^2)*u*v*m /2; u and v are odd and relatively prime, u > v, m is an integer.
To use the equations: set n, evaluate (2*n^2+2*n+1), derive m = (2*n^2+2*n+1)*integer, then evaluate x, y, z.
3) Primitive integral solution, i.e., those solutions in which there is no factor common to x, y and z, are (15, 20, 12), (65, 156, 60), (136,255,120), (175, 600, 168), (369,1640, 360), (671, 3660, 660), ...
Primitive integral solution are those where m = 1.
4) If the triad (a, b, c) is a solution of a Pythagorean triangle, i.e., a^2 + b^2 = c^2, then (x = a*c, y = b*c, z= a*b) is a solution of (1/x^2 + 1/y^2 = 1/z^2). For Pythagorean triads see for example http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html [broken link]
5) First multiple solutions (more values for y and z given x ) are x = 60,120,180,195. Multiple solution can be derived by equations at point 2.
6) No prime is in the sequence.
Related topics: the Diophantine equation (1/x^i + 1/y^i = 1/z^i) has no integral solutions for i>2, as it is easy to demonstrate by means of the Fermat-Wiles theorem.
