A292422 -id:A292422 - cp's OEIS frontend

A020492 Balanced numbers: numbers k such that phi(k) (A000010) divides sigma(k) (A000203).

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

Offset: 1

David W. Wilson

nonn

The quotient A020492(n)/A002088(n) = SummatorySigma/SummatoryTotient as n increases seems to approach Pi^4/36 or zeta(2)^2 [~2.705808084277845]. - Labos Elemer, Sep 20 2004, corrected by Charles R Greathouse IV, Jun 20 2012
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
The quotients sigma(n)/phi(n) are in A023897. - Bernard Schott, Jun 06 2017
There are 544768 balanced numbers < 10^14. - Jud McCranie, Sep 10 2017
a(975807) = 419998185095132. - Jud McCranie, Nov 28 2017

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

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.

Donovan Johnson, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
Jud McCranie, 670314 balanced numbers (first 1000 from T. D. Noe, first 10000 from Donovan Johnson)

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

Positions of 0's in A063514.

[ n: n in [1..3900] | SumOfDivisors(n) mod EulerPhi(n) eq 0 ]; // Klaus Brockhaus, Nov 09 2008

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

select(n->sigma(n)%eulerphi(n)==0,vector(10^4,i,i)) \\ Charles R Greathouse IV, Jun 20 2012

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

from math import prod
from itertools import count, islice
from sympy import factorint
def A020492_gen(startvalue=1): # generator of terms >= startvalue
    for m in count(max(startvalue,1)):
        f = factorint(m)
        if not prod(p**(e+2)-p for p,e in f.items())%(m*prod((p-1)**2 for p in f)):
            yield m
A020492_list = list(islice(A020492_gen(),20)) # Chai Wah Wu, Aug 12 2024

More terms from Farideh Firoozbakht, Nov 28 2005

A068390 Numbers k such that sigma(k) = 4*phi(k).

Original entry on oeis.org

14, 105, 248, 418, 1485, 3135, 3596, 3956, 4064, 5396, 8636, 20026, 23374, 25714, 35074, 35343, 39105, 41656, 55154, 56134, 56536, 71145, 74613, 87087, 124605, 150195, 175305, 192855, 263055, 393104, 413655, 421005, 474548, 604012, 697851, 711988, 819772

Offset: 1

Benoit Cloitre, Mar 03 2002

If 2^p-1 is a prime (Mersenne prime) greater than 3 then 2^(p-2)*(2^p-1) is in the sequence. So for n>1, 2^(A000043(n)-2)*(2^A000043(n)-1) is in the sequence. - Farideh Firoozbakht, Feb 23 2005

Theorem: If m>0, k is an integer and p=2^(m+2)+k-1 is a prime number then n=2^m*p is a solution to the equation sigma(x) = 4*phi(x)-k. The previous comment is the special case k=0. - Farideh Firoozbakht, Oct 01 2014

D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, p. 88.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (calculated using data from Jud McCranie, terms 1..1000 from Donovan Johnson)
Kevin A. Broughan and Daniel Delbourgo, On the Ratio of the Sum of Divisors and Euler’s Totient Function I, Journal of Integer Sequences, Vol. 16 (2013), Article 13.8.8.
Kevin A. Broughan and Qizhi Zhou, On the Ratio of the Sum of Divisors and Euler's Totient Function II, Journal of Integer Sequences, Vol. 17 (2014), Article 14.9.2.
Farideh Firoozbakht and M. F. Hasler, Variations on Euclid's formula for Perfect Numbers, JIS 13 (2010) #10.3.1

Cf. A000010, A000203, A079546, A000043, A292422.

Subsequence of A248150 (sigma(k) is divisible by 4).

[n: n in [1..10^6] | SumOfDivisors(n) eq 4*EulerPhi(n)]; // Vincenzo Librandi, Sep 25 2017

Select[Range[900000],DivisorSigma[1,#]==4EulerPhi[#]&] (* Harvey P. Dale, Nov 29 2013 *)

for(n=1,300000, if(sigma(n)==4*eulerphi(n),print1(n,",")))

More terms from Carl Najafi, Aug 16 2011

A293391 Integers n such that sigma(n)/phi(n) is a perfect square.

Original entry on oeis.org

1, 14, 30, 105, 248, 264, 418, 714, 1485, 3080, 3135, 3596, 3828, 3956, 4064, 5396, 6678, 8636, 10098, 12648, 20026, 20790, 21318, 22152, 23374, 24882, 25714, 26040, 35074, 35343, 39105, 41656, 43890, 44660, 49938, 55154, 56134, 56536, 61344, 71145, 74613, 86304, 87087, 94944

Offset: 1

Jose Arnaldo Bebita Dris, Oct 08 2017

nonn

From Robert Israel, Dec 12 2017: (Start)
Intersection of A011257 and A020492.
If x and y are coprime members of the sequence, then x*y is in the sequence.
Contains all members of A133028 except 3. (End)

			sigma(14)=3*8=24, phi(14)=14*(1/2)*(6/7)=6, sigma(14)/phi(14)=2^2, so 14 is in the list.

Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 349 terms from Robert Israel)
J. A. B. Dris and C. Leibovici, Why is this sequence not in the OEIS?, October 8 2017.
ProofWiki, Integers whose ratio between sigma and phi is square, misses the second term, 14 as of Dec 2017.

Cf. A000010, A000203, A011257, A020492, A133028, A289336, A289412, A292422.

for n from 1 to 100000 do
    r := numtheory[sigma](n)/numtheory[phi](n) ;
    if issqr(r) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Dec 07 2017

Select[Range[10^5], IntegerQ@ Sqrt[DivisorSigma[1, #]/EulerPhi[#]] &] (* Michael De Vlieger, Dec 08 2017 *)

isok(n) = my(q=sigma(n)/eulerphi(n)); issquare(q) && (denominator(q) == 1); \\ Michel Marcus, Dec 07 2017; corrected Sep 21 2019

a(n) = sigma(n)/phi(n) = m^2, for some integer m.

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A020492 Balanced numbers: numbers k such that phi(k) (A000010) divides sigma(k) (A000203).

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A068390 Numbers k such that sigma(k) = 4*phi(k).

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Magma

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A293391 Integers n such that sigma(n)/phi(n) is a perfect square.

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