Simplification of "reciprocal word 3by2" (aka invert_pi1)

[Paul Zimmermann]

       Dear Pawel,

> are you able also to generate a case for t0 == d0?

after thought, this is not possible. Indeed, if we write d=d1*beta+d0, we have
(p,t0) = (beta+v)*d % beta^2 at line 14, thus it would mean that
(beta+v)*d = beta^3 + d, thus d would divide beta^3, which implies that d is
a power of two. The only possible case is d1=beta/2 and d0=0. But when d0=0
the second part of the algorithm at lines 10-15 does no correction.

Paul