[math-fun] Re: A016142

[Richard Guy]

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.