A123926 - OEIS

%I #10 Jun 17 2013 18:14:54

%S 1,1,2,1,2,2,2,1,1,2,2,2,2,2,4,1,2,1,2,6,4,2,2,2,1,2,4,2,2,4,2,3,4,2,

%T 4,1,2,2,4,2,2,4,2,6,2,2,2,2,3,3,4,2,2,4,4,2,4,2,2,12,2,2,2,1,4,4,2,6,

%U 4,4,2,1,2,2,2,2,4,4,2,2,1,2,2,4,4,2,4,2,2,2,4,6,4,2,4,6,2,3,2,1,2,4,2,2,8

%N Greatest common divisor of sigma_k(n) for all k >= 1.

%C Has the property that if gcd(n,m) = 1, then a(n)*a(m) divides a(n*m). First inequality is a(4) = 1, a(5) = 2, but a(20) = 6. It appears that a(n) also always divides sigma_0(n) = tau(n).

%C Contribution from _Matthew Vandermast_, Feb 10 2010: (Start)

%C 1. If an integer m does not divide sigma_0(n), m will also not divide sigma_(totient m)(n). Therefore a(n) always divides sigma_0(n) = tau(n).

%C 2. a(n) is even iff sigma_1(n) is even. Cf. A028982, A028983.

%C 3. a(p)=2 for any odd prime p. If n is an odd integer with 2^e divisors, then a(n)=2^e.

%C 4. For any prime p and positive integer m, if p is congruent to 1 mod m, then a(p^(m-1))=m. It follows from Dirichlet's Theorem (see link) that every positive integer appears in the sequence infinitely often. (End)

%H Charles R Greathouse IV, <a href="/A123926/b123926.txt">Table of n, a(n) for n = 1..10000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DirichletsTheorem.html">Dirichlet's Theorem</a>

%e For n=4, sigma_1(n) = 7, sigma_2(n) = 21, both divisible by 7, but sigma_3(n) = 73, which is not, so a(4) = 1.

%t a[n_] :=  GCD @@ Table[DivisorSigma[k, n] , {k, 0, EulerPhi[n]}]; Table[a[n], {n, 1, 105}] (* _Jean-François Alcover_, May 21 2012 *)

%o (PARI) a(n)=my(d=divisors(n), g=#d); for(k=1, eulerphi(n), g=gcd(lift(sum(i=1,#d,Mod(d[i],g)^k)),g); if(g<3,return(g))); g \\ _Charles R Greathouse IV_, Jun 17 2013

%Y Cf. A109974, A000005, A000203, A001157.

%K nice,nonn

%O 1,3

%A _Franklin T. Adams-Watters_, Nov 21 2006