%I A020492
%S A020492 1,2,3,6,12,14,15,30,35,42,56,70,78,105,140,168,190,210,248,264,270,357,
%T A020492 418,420,570,594,616,630,714,744,812,840,910,1045,1240,1254,1485,1672,
%U A020492 1848,2090,2214,2376,2436,2580,2730,2970,3080,3135,3339,3596,3720,3828
%N A020492 Balanced numbers: numbers n such that phi(n) (A000010) divides sigma(n) 
               (A000203).
%C A020492 The quotient A020492[n]/A002088[n] = SummatorySigma/SummatoryTotient 
               as n increases seems to approach Pi^2/36 or Zeta(2))^2 [~2.705808084277845]. 
               - Labos E. (labos(AT)ana.sote.hu), Sep 20 2004
%C A020492 If 2^p-1 is prime (a Mersenne prime) then m=2^(p-2)*(2^p-1) is in the 
               sequence because when p=2 we get m=3 and phi(3) divides sigma(3) 
               and for p>2, phi(m)=2^(p-2)*(2^(p-1)-1); sigma(m) =(2^(p-1)-1)*2^p 
               hence sigma(m)/phi(m)=4 is an integer. So for each n, 2^(A000043(n)-2)*(2^A000043(n)-1) 
               is in the sequence. - Farideh Firoozbakht (mymontain(AT)yahoo.com), 
               Nov 28 2005
%D A020492 D. Chiang, "N's for which phi(N) divides sigma(N)", Mathematical Buds, 
               Chap. VI pp. 53-70 Vol. 3 Ed. H. D. Ruderman, Mu Alpha Theta 1984.
%H A020492 T. D. Noe, <a href="b020492.txt">Table of n, a(n) for n=1..1000</a>
%e A020492 sigma(35) = 1+5+7+35 = 48, phi(35) = 24, hence 35 is a term.
%t A020492 Select[ Range[ 4000 ], IntegerQ[ DivisorSigma[ 1, # ]/EulerPhi[ # ] ]& 
               ]
%o A020492 (MAGMA) [ n: n in [1..3900] | SumOfDivisors(n) mod EulerPhi(n) eq 0 ]; 
               [From Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Nov 09 2008]
%Y A020492 Cf. A000010, A000203.
%Y A020492 Cf. A000043, A000668, A011257.
%Y A020492 Sequence in context: A015769 A015765 A015771 this_sequence A110590 A111271 
               A070926
%Y A020492 Adjacent sequences: A020489 A020490 A020491 this_sequence A020493 A020494 
               A020495
%K A020492 nonn
%O A020492 1,2
%A A020492 David W. Wilson (davidwwilson(AT)comcast.net)
%E A020492 More terms from Farideh Firoozbakht (mymontain(AT)yahoo.com), Nov 28 
               2005