Alex Fink & I are now able to give the right answers. The number of incongruent integer-edged Heron triangles whose circumdiameter is the product of n distinct primes each of shape 4k + 1 is (2*7^n - 3*3^n + 1)/6 This is A016161 in OEIS, tho this fact is not mentioned there. The number of these which are right triangles is (3^n - 1)/2, as stated before, and this is A003462. The number which are not right-angled is thus (2*7^n - 6*3^n + 4)/6 or 0, 8, 88, 720, 5360, ... or eight times A016212. R + Alex Fink. On Sun, 14 Aug 2005, Richard Guy wrote:
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.