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A123926 Greatest common divisor of sigma_k(n) for all k >= 1. 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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
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)
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Dirichlet's Theorem
EXAMPLE
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.
MATHEMATICA
a[n_] := GCD @@ Table[DivisorSigma[k, n] , {k, 0, EulerPhi[n]}]; Table[a[n], {n, 1, 105}] (* Jean-François Alcover, May 21 2012 *)
PROG
(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
CROSSREFS
Sequence in context: A082070 A336648 A082902 * A336476 A082064 A082055
KEYWORD
nice,nonn
AUTHOR
STATUS
approved

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Last modified April 23 12:08 EDT 2024. Contains 371912 sequences. (Running on oeis4.)