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A020492
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Balanced numbers: numbers k such that phi(k) (A000010) divides sigma(k) (A000203).
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88
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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
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1,2
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COMMENTS
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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, 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
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REFERENCES
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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.
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LINKS
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EXAMPLE
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sigma(35) = 1+5+7+35 = 48, phi(35) = 24, hence 35 is a term.
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MAPLE
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with(numtheory);
local n; for n from 1 to q do if (sigma(n) mod phi(n))=0 then print(n);
fi; od; end:
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MATHEMATICA
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Select[ Range[ 4000 ], IntegerQ[ DivisorSigma[ 1, # ]/EulerPhi[ # ] ]& ]
(* Second program: *)
Select[Range@ 4000, Divisible[DivisorSigma[1, #], EulerPhi@ #] &] (* Michael De Vlieger, Nov 28 2017 *)
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PROG
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(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
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CROSSREFS
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Cf. A000010, A000043, A000203, A000668, A011257, A023897, A133028, A291565, A291566, A292422, A351114 (characteristic function).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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