

A020492


Balanced numbers: numbers k such that phi(k) (A000010) divides sigma(k) (A000203).


93



1, 2, 3, 6, 12, 14, 15, 30, 35, 42, 56, 70, 78, 105, 140, 168, 190, 210, 248, 264, 270, 357, 418, 420, 570, 594, 616, 630, 714, 744, 812, 840, 910, 1045, 1240, 1254, 1485, 1672, 1848, 2090, 2214, 2376, 2436, 2580, 2730, 2970, 3080, 3135, 3339, 3596, 3720, 3828
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OFFSET

1,2


COMMENTS

If 2^p1 is prime (a Mersenne prime) then m = 2^(p2)*(2^p1) 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^(p2)*(2^(p1)1); sigma(m) = (2^(p1)1)*2^p hence sigma(m)/phi(m) = 4 is an integer. So for each n, A133028(n) = 2^(A000043(n)2)*(2^A000043(n)1) is in the sequence.  Farideh Firoozbakht, Nov 28 2005
Phi and sigma are both multiplicative functions and for this reason if m and n are coprime and included in this sequence then m*n is also in this sequence.  Enrique Pérez Herrero, Sep 05 2010
There are 544768 balanced numbers < 10^14.  Jud McCranie, Sep 10 2017


REFERENCES

D. Chiang, "N's for which phi(N) divides sigma(N)", Mathematical Buds, Chap. VI pp. 5370 Vol. 3 Ed. H. D. Ruderman, Mu Alpha Theta 1984.


LINKS



EXAMPLE

sigma(35) = 1+5+7+35 = 48, phi(35) = 24, hence 35 is a term.


MATHEMATICA

Select[ Range[ 4000 ], IntegerQ[ DivisorSigma[ 1, # ]/EulerPhi[ # ] ]& ]
(* Second program: *)
Select[Range@ 4000, Divisible[DivisorSigma[1, #], EulerPhi@ #] &] (* Michael De Vlieger, Nov 28 2017 *)


PROG

(Magma) [ n: n in [1..3900]  SumOfDivisors(n) mod EulerPhi(n) eq 0 ]; // Klaus Brockhaus, Nov 09 2008
(Python)
from sympy import totient, divisor_sigma
print([n for n in range(1, 4001) if divisor_sigma(n)%totient(n)==0]) # Indranil Ghosh, Jul 06 2017


CROSSREFS

Cf. A000010, A000043, A000203, A000668, A011257, A023897, A133028, A291565, A291566, A292422, A351114 (characteristic function).


KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



