Reason for definition of precomputed reciprocals

[Paul Zimmermann]

       Niels,

> Using v = floor((B^2 - 1) / d) - B implies that
> 
>   (B + v) d = B^2 - k, with 1 <= k <= d
> 
> With the alternative smaller reciprocal you suggest, it seems that
> instead one would get
> 
>   (B + v) d = B^2 - k with d <= k < 2d

I agree with the above, and it seems the first is more accurate.
In fact the first is the best one if you want to avoid k=0
(which occurs only for d=B/2 when B=2^b, where it would give v=B which does not fit
in one limb).

Paul