%N Least base B >= 2 such that the repunit (B^n-1)/(B-1) of length n is not squarefree.
%C When n is prime, a(n) seems to be hard to determine.
%C Let p be a prime == 1 (mod n) (such a prime exists by Dirichlet's theorem). Since gcd(n, phi(p)) > 1 there exists b such that 1 < b < p and b^n == 1 (mod p). Then x = b + y*p for suitable y has x^n == 1 (mod p^2), and x == b (mod p), i.e., (x^n-1)/(x-1) is divisible by p^2. Therefore a(n) <= x < p^2. - _Robert Israel_, Jul 24 2014
%e a(17)=164 because (164^17 - 1)/163 is not squarefree (is multiple of 103^2), and 164 is the minimal number with that property.
%p A:= proc(n) local x,F;
%p for x from 2 do F:= ifactors((x^n-1)/(x-1),easy);
%p if max(seq(f,f=F)) >= 2
%p then return x
%p end proc;
%p seq(A(n), n=2..50); # _Robert Israel_, Jul 24 2014
%o (PARI) for(n=2,100,b=2;while(issquarefree((b^n-1)/(b-1)),b++);print1(b,", "))
%A _Jeppe Stig Nielsen_, Jul 24 2014