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%I #23 May 30 2020 10:28:24
%S 2,5,6,8,10,11,14,15,17,18,20,22,23,24,26,29,30,32,33,34,35,38,40,41,
%T 42,44,45,46,47,49,50,51,53,54,55,56,58,59,60,62,65,66,68,69,70,71,72,
%U 74,77,78,80,82,83,85,86,87,88,89,90,92,94,95,96,98,99,101,102,104,105,106
%N Numbers n such that 3 divides sigma(n).
%C 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
%H Enrique Pérez Herrero, <a href="/A087943/b087943.txt">Table of n, a(n) for n = 1..5000</a>
%F a(n) << n^k for any k > 1, where << is the Vinogradov symbol. - _Charles R Greathouse IV_, Sep 04 2013
%F 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
%F 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
%p select(n -> numtheory:-sigma(n) mod 3 = 0, [$1..1000]); # _Robert Israel_, Nov 09 2016
%t Select[Range[1000],Mod[DivisorSigma[1,#],3]==0&] (* _Enrique Pérez Herrero_, Sep 03 2013 *)
%o (PARI) is(n)=sigma(n)%3==0 \\ _Charles R Greathouse IV_, Sep 04 2013
%o (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
%Y Cf. A000203, A059269, A066498, A034020, A028983, A074216, A329963 (complement).
%K nonn
%O 1,1
%A Yuval Dekel (dekelyuval(AT)hotmail.com), Oct 27 2003
%E More terms from _Benoit Cloitre_ and _Ray Chandler_, Oct 27 2003
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