A016142

[Richard Guy]

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.
>