%I
%S 3,18,3,3,2,78,3,4,3,118,2,146,3,3,3,164,2,44,2,2,3,53,2,3,3,4,3,53,2,
%T 403,3,18,3,3,2,957,3,3,2,99,2,369,3,3,3,533,2,8,3,18,3,164,2,3,3,7,3,
%U 381,2
%N Least base B >= 2 such that the repunit (B^n1)/(B1) 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^n1)/(x1) 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^n1)/(x1),easy)[2];
%p if max(seq(f[2],f=F)) >= 2
%p then return x
%p fi
%p od
%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^n1)/(b1)),b++);print1(b,", "))
%K nonn,more
%O 2,1
%A _Jeppe Stig Nielsen_, Jul 24 2014
