%N Minimal common difference of maximal arithmetic progression starting at n such that each term k has tau(k)=tau(n)
%C a(1) is not well defined, since the maximal progression has only one term.
%e For n=6, A165500(n)=3, and the least difference d such that tau(6) = tau(6+d) = tau(6+2d) is d=2, so a(6)=2.
%Y Cf. A165499, A165500, A088430.
%A _Hugo van der Sanden_, Sep 21 2009, Oct 09 2009
%E Extended to n=22 taking advantage of A088430 for n=19, _Hugo van der Sanden_, Jun 02 2015