A245498 - OEIS

%I #27 Feb 09 2024 08:01:19

%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^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)[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^n-1)/(b-1)),b++);print1(b,", "))

%K nonn

%O 2,1

%A _Jeppe Stig Nielsen_, Jul 24 2014