A087943 Numbers n such that 3 divides sigma(n).

2, 5, 6, 8, 10, 11, 14, 15, 17, 18, 20, 22, 23, 24, 26, 29, 30, 32, 33, 34, 35, 38, 40, 41, 42, 44, 45, 46, 47, 49, 50, 51, 53, 54, 55, 56, 58, 59, 60, 62, 65, 66, 68, 69, 70, 71, 72, 74, 77, 78, 80, 82, 83, 85, 86, 87, 88, 89, 90, 92, 94, 95, 96, 98, 99, 101, 102, 104, 105, 106

Offset: 1

Author

Yuval Dekel (dekelyuval(AT)hotmail.com), Oct 27 2003

Keywords

• nonn

Comments

Numbers n such that in the prime factorization n = Product_i p_i^e_i, there is some p_i == 1 (mod 3) with e_i == 2 (mod 3) or some p_i == 2 (mod 3) with e_i odd. - Robert Israel, Nov 09 2016

Links

• Enrique Pérez Herrero, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A000203, A059269, A066498, A034020, A028983, A074216, A329963 (complement).

Programs

Maple

select(n -> numtheory:-sigma(n) mod 3 = 0, [$1..1000]); # Robert Israel, Nov 09 2016

Mathematica

Select[Range[1000],Mod[DivisorSigma[1,#],3]==0&] (* Enrique Pérez Herrero, Sep 03 2013 *)

PARI

is(n)=sigma(n)%3==0 \\ Charles R Greathouse IV, Sep 04 2013

PARI

is(n)=forprime(p=2,997,my(e=valuation(n,p)); if(e && Mod(p,3*p-3)^(e+1)==1, return(1), n/=p^e)); sigma(n)%3==0 \\ Charles R Greathouse IV, Sep 04 2013

Formula

a(n) << n^k for any k > 1, where << is the Vinogradov symbol. - Charles R Greathouse IV, Sep 04 2013

a(n) ~ n as n -> infinity: since Sum_{primes p == 2 (mod 3)} 1/p diverges, asymptotically almost every number is divisible by some prime p == 2 (mod 3) but not by p^2. - Robert Israel, Nov 09 2016

Because sigma(n) and sigma(3n)=A144613(n) differ by a multiple of 3, these are also the numbers n such that n divides sigma(3n). - R. J. Mathar, May 19 2020

Extensions

More terms from Benoit Cloitre and Ray Chandler, Oct 27 2003