OFFSET
1,1
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
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
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 *)
PROG
(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
KEYWORD
nonn
AUTHOR
Yuval Dekel (dekelyuval(AT)hotmail.com), Oct 27 2003
EXTENSIONS
More terms from Benoit Cloitre and Ray Chandler, Oct 27 2003
STATUS
approved