On Wed, Oct 30, 2013 at 2:51 AM, Don Kelly <[email protected]> wrote:

> You are right but it is a bit of serendipity  What I wanted is (a*b+
> b*c+c*a ) % c a b .
> it doesn't matter how the pairs are formed in the numerator ac+ba+cb is
> the same.
>

Like this?

   F=: +/ .* (</~i.3) +/ .* ] % */

That's probably awful for speed, but it does express the idea.

Thanks,

-- 
Raul
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