View source for Prove the result for shifted HAPs instead of HAPs

For an explanation of the justification for considering shifted HAPs, or ''privileged'' APs (PAPs), see [http://gowers.wordpress.com/2010/01/19/edp1-the-official-start-of-polymath5/#comment-5415 this post]. The simplest way to generate bounded-discrepancy PAPs experimentally is to choose a very large number <math>N</math> at random (the shift), calculate <math>r_d</math> for all small <math>d</math> such that <math>d \mid N + r_d</math> where <math>0 \leq r_d < d</math>, and then construct a zero-based sequence <math>(x_n)</math> requiring that <math>| x_{r_d} + x_{r_d + d} + x_{r_d + 2d} + \ldots + x_{r_d + md} | \leq C</math> for all <math>d</math> and <math>m</math>.

From a theoretical point of view, the advantage of trying to prove EDP for PAPs rather than HAPs is that one will not be distracted by considerations that rely on all the sequences lining up at zero.