If anyone could confirm that I calculated the trajectory correctly (i.e., this is a path such that the travellers would perceive constant acceleration), I'd appreciate it. . .

There's a nice summary of the equations needed when you have an acceleration A that is constant from the point of view of the rocket, at
http://math.ucr.edu/home/baez/physic...SR/rocket.html.
The upshot is, the time it takes to go a distance d, where both quantities are measured by the people that stay behind, is a combination of the normal result t=sqrt(2d/A), and the speed of light result t=d/c, yielding
t=sqrt(2d/A + d^2/c^2). The speed seen back on Earth is v=At/sqrt(1+A^2*t^2/c^2). That's the speed you got montebianco.