A123926 Greatest common divisor of sigma_k(n) for all k >= 1.

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, 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, 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

Offset: 1

Author

Franklin T. Adams-Watters, Nov 21 2006

Keywords

• nice
• nonn

Comments

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).

Contribution from Matthew Vandermast, Feb 10 2010: (Start)

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).

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

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

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)

Examples

			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.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Dirichlet's Theorem

Crossrefs

Cf. A109974, A000005, A000203, A001157.

Programs

Mathematica

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

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