A088912 -id:A088912 - cp's OEIS frontend

A159907 Numbers m with half-integral abundancy index, sigma(m)/m = k+1/2 with integer k.

2, 24, 4320, 4680, 26208, 8910720, 17428320, 20427264, 91963648, 197064960, 8583644160, 10200236032, 21857648640, 57575890944, 57629644800, 206166804480, 17116004505600, 1416963251404800, 15338300494970880, 75462255348480000, 88898072401645056, 301183421949935616, 6219051710415667200

Offset: 1

M. F. Hasler, Apr 25 2009

nonn

Obviously, all terms must be even (cf. formula), but e.g. a(9) and a(12) are not divisible by 3. See A007691 for numbers with integral abundancy.
Odd numbers and higher powers of 2 cannot be in the sequence; 6 is in A000396 and thus in A007691, and n=10,12,14,18,20,22 don't have integral 2*sigma(n)/n.
Conjecture: with number 1, multiply-anti-perfect numbers m: m divides antisigma(m) = A024816(m). Sequence of fractions antisigma(m) / m: {0, 0, 10, 2157, 2337, 13101, 4455356, ...}. - Jaroslav Krizek, Jul 21 2011
The above conjecture is equivalent to the conjecture that there are no odd multiply perfect numbers (A007691) greater than 1. Proof: (sigma(n)+antisigma(n))/n = (n+1)/2 for all n. If n is even then sigma(n)/n is a half-integer if and only if antisigma(n)/n is an integer. Since all members of this sequence are known to be even, the only way the conjecture can fail is if antisigma(n)/n is an integer, in which case sigma(n)/n is an integer as well. - Nathaniel Johnston, Jul 23 2011
These numbers are called hemiperfect numbers. See Numericana & Wikipedia links. - Michel Marcus, Nov 19 2017

			a(1) = 2 since sigma(2)/2 = (1+2)/2 = 3/2 is of the form k+1/2 with integer k=1.
a(2) = 24 is in the sequence since sigma(24)/24 = (1+2+3+4+6+8+12+24)/24 = (24+12+24)/24 = k+1/2 with integer k=2.

Max Alekseyev, Table of n, a(n) for n = 1..130
G. P. Michon, Multiperfect Numbers & Hemiperfect Numbers, Numericana.
Walter Nissen, Abundancy : Some Resources .
Project Euler, Problem 241: Perfection Quotients.
Wikipedia, Hemiperfect number

Cf. A000203, A088912, A141643 (k=2), A055153 (k=3), A141645 (k=4), A159271 (k=5).

isok(n) = denominator(sigma(n,-1)) == 2; \\ Michel Marcus, Sep 19 2015

forfactored(n=1,10^7, if(denominator(sigma(n,-1))==2, print1(n[1]", "))) \\ Charles R Greathouse IV, May 09 2017

from fractions import Fraction
from sympy import divisor_sigma as sigma
def aupto(limit):
  for k in range(1, limit):
    if Fraction(int(sigma(k, 1)), k).denominator == 2:
      print(k, end=", ")
aupto(3*10**4) # Michael S. Branicky, Feb 24 2021

A159907 = { n | 2*A000203(n) is in n*A005408 } = { n | A054024(n) = n/2 }

Terms a(20) onward from Max Alekseyev, Jun 05 2025

A160678 Numbers n whose abundancy is equal to 13/2; sigma(n)/n = 13/2.

Original entry on oeis.org

170974031122008628879954060917200710847692800, 1893010442758976546037991125738431754692198400, 54361481238923605327597493185154939181072384000

Offset: 1

Gerard P. Michon, Jun 06 2009

nonn

This sequence includes many terms but it is conjectured to be finite.

			a(1) = 2^23 3^9 5^2 7^5 11^5 13^2 17 19^3 31 37 43 61^2 97 181 241.
As the "sum of divisors" function (sigma) is a multiplicative function, sigma(a(1)) is the product of the values of sigma at the above prime powers, respectively given as follows, in factorized form:
sigma(a(1)) = (3^2 5 7 13 17 241) (2^2 11^2 61) (31) (2^3 3 19 43) (2^2 3^2 7 19 37) (3 61) (2 3^2) (2^3 5 181) (2^5) (2 19) (2^2 11) (3 13 97) (2 7 13) (2 7^2) (2 11^2).
a(1) belongs to the sequence because the latter product boils down to 13/2 times the former.

Michel Marcus, Table of n, a(n) for n = 1..307
G. P. Michon, Multiperfect and hemiperfect integers
G. P. Michon, Multiplicative functions: Abundancy = sigma(n)/n
G. P. Michon and M. Marcus, Hemiperfect numbers of abundancy 13/2
Walter Nissen, Abundancy: Some Resources

Cf. A000203 (sigma function, sum of divisors), A141643 (abundancy = 5/2), A055153 (abundancy = 7/2), A141645 (abundancy = 9/2), A159271 (abundancy = 11/2), A159907 (half-integral abundancy, "hemiperfect numbers"), A088912 (least numbers of given half-integer abundancy). A007691 (multiperfect numbers, abundancy is an integer), A000396 (perfect numbers, abundancy = 2), A005101 (abundant numbers, abundancy is greater than 2), A005100 (deficient numbers, abundancy is less than 2).

is(n)=sigma(n,-1)==13/2 \\ Charles R Greathouse IV, Feb 21 2017

A317681 a(n) = smallest m such that sigma(m) = n*m/2.

Original entry on oeis.org

1, 2, 6, 24, 120, 4320, 30240, 8910720, 14182439040, 17116004505600, 154345556085770649600, 170974031122008628879954060917200710847692800, 141310897947438348259849402738485523264343544818565120000, 12749472205565550032020636281352368036406720997031277595140988449695952806020854579200000

Offset: 2

Jianing Song, Aug 04 2018

Interleaving of A007539 and A088912.
For even n, a(n) is a multiply perfect number; for odd n it is a hemiperfect number.
Note that 1 is the only number with abundancy 1, and 2 is the only number with abundancy 3/2 (in other words, 1 and 2 are solitary numbers; see A014567). For k >= 4 it is not known whether there are finitely many or infinitely many numbers with abundancy k/2. Also it is not known whether a(n) < a(n+1) always holds.
On the Riemann Hypothesis (RH), a(n) > exp(exp(n/(2*exp(gamma)))), where gamma = 0.5772156649... is the Euler-Mascheroni constant (A001620).

			a(7) = 4320 since sigma(4320) = 15120 = 7/2*4320 and 4320 is the smallest m such that sigma(m)/m = 7/2.

Max Alekseyev, Table of n, a(n) for n = 2..16
Achim Flammenkamp, The Multiply Perfect Numbers Page
Fred Helenius, A glossary of MPFN-related terms.
G. P. Michon, Multiplicative functions: Abundancy = sigma(n)/n
G. P. Michon, Multiperfect and hemiperfect numbers
Walter Nissen, Abundancy: Some Resources

Cf. A007539, A088912, A246454.

Numbers with abundancy k/2: A000396 (k=4), A141643 (k=5), A005820 (k=6), A055153 (k=7), A027687 (k=8), A141645 (k=9), A046060 (k=10), A159271 (k=11), A046061 (k=12), A160678 (k=13).

Nest[Append[#, Block[{m = #1[[-1]] + 1}, While[DivisorSigma[1, m] != #2 m/2, m++]; m]] & @@ {#, Length@ # + 2} &, {1}, 6] (* Michael De Vlieger, Aug 05 2018 *)

for(n=2, 10, for(m=1, 10^12, if(sigma(m)/m==n/2, print1(m, ", "); break())))

a(n) = my(k=1); while (sigma(k) != k*n/2, k++); k; \\ Michel Marcus, May 15 2025

a(2n) = A007539(n), a(2n+1) = A088912(n), n > 0.

a(15) = A088912(7) added by Max Alekseyev, Jun 05 2025

A216897 a(n) = smallest m such that sigma(m)/m = n + 1/3.

Original entry on oeis.org

3, 12, 1080, 18506880, 2198278051200, 28657168334177779210174055055360000, 30539655646420783598362725995379113771515252254037575464371359842304000000000

Offset: 1

Michel Marcus, Sep 19 2012

nonn

a(7) is a 77-digit number with factorization: 2^35 3^20 5^9 7^3 11 13^2 17 19 23 29 31 37^2 41 61 67 71 73 109 137 409 521 547 1093 36809 368089.

An upper bound for a(8) is a 165-digit number that can be found on given link where line begins with 25/3.

			a(1) = 3 because sigma(3)/3 = 4/3 = 1 + 1/3 and 3 is the earliest m such that sigma(m)/m = 1 + 1/3.

Walter Nissen and Michel Marcus, Extended Table of Abundancies

Cf. A160320, A088912.

a(7) confirmed and added by Max Alekseyev, Jun 06 2025

cp's OEIS Frontend

A159907 Numbers m with half-integral abundancy index, sigma(m)/m = k+1/2 with integer k.

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A160678 Numbers n whose abundancy is equal to 13/2; sigma(n)/n = 13/2.

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A317681 a(n) = smallest m such that sigma(m) = n*m/2.

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Mathematica

PARI

PARI

Formula

Extensions

A216897 a(n) = smallest m such that sigma(m)/m = n + 1/3.

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