Warning -- I believe I've overestimated the number of Heron triangles. I was assuming that if c^2 = x^2 + y^2, where c is the circumdiameter, then all pairs of (x,y)s, considered as arctans of halves of angles subtended by sides of triangles at the circumcentre, generate integer sides. But if both (x1,y1) and (x2,y2) are primitive, then you get rational sides, but not necessarily integer ones. More later, when I've thought about it. Or perhaps someone will step in with the correct formula? Best, R. On Sat, 13 Aug 2005, Richard Guy wrote:
This (also) appears to be 3^(n-1) * (3^n - 1)/2.
VERY WILD surmise: It's the number of integer-sided Heron triangles whose circumdiameter is the product of n distinct primes of shape 4k + 1.
(3^n - 1)/2 of these are (believed to be) right triangles, so that the number of non-right ones is
(3^(n-1) - 1) * (3^n - 1)/2
A003462 and a new(?) sequence 0, 8, 104, 1040, 9680, ...
R.