A159271 -id:A159271 - 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

A141645 Numbers n such that sigma(n)/n = 9/2.

Original entry on oeis.org

8910720, 17428320, 8583644160, 57629644800, 206166804480, 1416963251404800, 15338300494970880, 6275163455171297280, 200286975596707184640, 215594611071909888000, 5997579964837140234240, 39887491844324122951680

Offset: 1

Yasutoshi Kohmoto, May 10 2008

nonn

200286975596707184640 belongs to this sequence. - Gerard P. Michon, May 10 2009

81703797123392614369698250752 is in this sequence. - Gerard P. Michon, May 11 2009

G. P. Michon, Table of n, a(n) for n = 1..27
G. P. Michon, Multiperfect and hemiperfect integers [From _Gerard P. Michon_, Jun 04 2009]
Walter Nissen, Abundancy : Some Resources

Cf. A055153, A141643, A159271, A160678.

Cf. A159907 = (half-integer abundancy). - Gerard P. Michon, May 10 2009

is(n)=sigma(n,-1)==9/2 \\ Charles R Greathouse IV, Mar 09 2014

Definition rewritten by M. F. Hasler, May 10 2008
Is it certain that there are no other terms below the highest value shown? - N. J. A. Sloane, Sep 07 2008
a(1) corrected and a(3) confirmed by Ray Chandler, Sep 18 2008
Missing term a(2)=17428320 communicated by Walter Nissen, Apr 17 2009. There are no further terms through 2^34.
Missing term a(4) inserted and a(6) and a(8) added by Grzegorz Lach (137.036(AT)gmail.com), Apr 18 2009. a(4) was also sent by Avinoam Kalma (a.kalma(AT)gmail.com), Apr 20 2009.
a(7) = 15338300494970880 added by Gerard P. Michon, May 10 2009.
a(10)=215594611071909888000 [and above] from Michel Marcus, added by Gerard P. Michon, Jun 04 2009

A141643 Numbers k such that sigma(k)/k = 5/2.

Original entry on oeis.org

24, 91963648, 10200236032

Offset: 1

Yasutoshi Kohmoto, May 10 2008

There are no other terms through 2^34. - Walter Nissen, Apr 17 2009
No more terms below 10^12. - Jud McCranie, Aug 30 2013
a(4) > 10^100 if it exists. - Max Alekseyev, Jun 05 2025

G. P. Michon, Multiperfect and hemiperfect numbers
Walter Nissen, Abundancy : Some Resources
Walter Nissen, Abundancy : Some Resources
Walter Nissen, Primitive Friendly Pairs with friends < 2^34 with denom < 20000

Cf. A055153, A141645, A159271, A160678.

Select[Range[100000], DivisorSigma[1, #]/# == 5/2 &] (* Robert Price, Apr 03 2019 *)

isok(k) = sigma(k, -1) == 5/2; \\ Michel Marcus, Apr 04 2019

Definition rewritten by M. F. Hasler, May 10 2008

First three terms confirmed by Ray Chandler, Sep 18 2008

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

A227302 Numbers m such that m divides sigma(2*m).

Original entry on oeis.org

1, 3, 12, 14, 60, 248, 336, 2160, 2340, 4064, 13104, 15120, 16380, 261888, 1089270, 4455360, 8714160, 10213632, 11784960, 16775168, 22766400, 45981824, 71495424, 98532480, 229909120, 689727360, 738152448, 4291822080, 4294934528, 5100118016, 7091219520

Offset: 1

Alex Ratushnyak, Jul 05 2013

nonn

If m belongs to the sequence, then sigma(2*m)/m is an integer, so sigma(2*m)/(2*m) is either an integer or half of an integer, so 2*m is either perfect, multiperfect or hemiperfect. - Michel Marcus, Jul 09 2013

Jinyuan Wang, Table of n, a(n) for n = 1..54

Cf. A000203, A000396, A007691, A159907, A232702.

Cf. A141643, A055153, A141645, A159271, A160678. (hemiperfect numbers)

is(n)=sigma(2*n)%n==0 \\ Charles R Greathouse IV, Nov 04 2016

A216781 Numbers such that numerator(sigma(n)/n) is odd and denominator(sigma(n)/n) is even.

Original entry on oeis.org

2, 4, 8, 16, 18, 20, 24, 32, 36, 40, 48, 50, 52, 64, 68, 72, 80, 88, 96, 98, 100, 104, 112, 116, 128, 136, 144, 148, 152, 160, 162, 164, 176, 180, 192, 196, 200, 208, 212, 224, 232, 240, 242, 244, 256, 272, 288, 292, 296, 304, 320, 324, 328, 338, 344, 352

Offset: 1

Michel Marcus, Sep 16 2012

nonn

a(n) contains powers of 2 (A000079 except 1), and hemiperfect numbers (A055153, A141645, A159271, A160678).

			sigma(2)/2 = 3/2 (odd/even).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A216780, A216782, A000079, A055153, A141645, A159271, A160678.

Select[Range[1000], OddQ[Numerator[DivisorSigma[1, #]/#]] && EvenQ[Denominator[DivisorSigma[1, #]/#]] &] (* Vincenzo Librandi, Jun 24 2014 *)

oeab(n) = {for (i=1, n, ab = sigma(i)/i; if ((numerator(ab) % 2 == 1) && (denominator(ab) % 2 == 0), print1(i, ", ")););}

A088912 a(n) = smallest m such that sigma(m) = (n+1/2)*m.

Original entry on oeis.org

2, 24, 4320, 8910720, 17116004505600, 170974031122008628879954060917200710847692800, 12749472205565550032020636281352368036406720997031277595140988449695952806020854579200000

Offset: 1

Farideh Firoozbakht, Nov 29 2003

2 is the only number m such that sigma(m)=1.5*m.
A direct consequence of Robin's theorem is that a(6)>5E16, a(7)>1.898E29, a(8)>2.144E51, a(9)>9.877E89 and a(10)>6.023E157. - Washington Bomfim, Oct 30 2008
If the Riemann hypothesis (RH) is true then Robin's theorem (Guy Robin, 1984) implies that the n-th term of this sequence is greater than exp(exp((n+1/2)/exp(gamma))) where gamma=0.5772156649... is the Euler-Mascheroni constant (A001620). For the 6th term (which is actually 1.7*10^44) this lower bound is 5.0*10^16. Similarly, if RH is true, the next term (7th term) is at least 1.9*10^29 (and is probably more than 10^90 or so). - Gerard P. Michon, Jun 10 2009
From Gerard P. Michon, Jul 04 2009: (Start)
An upper bound for a(7) is provided by a 97-digit integer of abundancy 15/2 (5.71379...10^96) discovered by Michel Marcus on July 4, 2009. The factorization of that number is: 2^53 3^15 5^6 7^6 11^3 13 17 19^3 23 29 31 37 41 43 61 73 79 97 181 193 199 257 263 4733 11939 19531 21803 87211 262657.
Similarly, an upper bound for a(8) is provided by a 286-digit integer of abundancy 17/2 (3.30181...10^285) equal to x/17, where x is the smallest known number of abundancy 9 (a 287-digit integer discovered by Fred W. Helenius in 1995). This is so because 17 happen to occur with multiplicity 1 in the factorization of x. (End)
A new upper bound for a(7) was found on Aug 15 2009 by Michel Marcus, who broke his own record by finding two "small" multiples of 2^35*3^20*5^5*7^6*11^2*13^2*17 that are of abundancy 15/2. The lower one (1.27494722...10^88) has only 89 digits. - Gerard P. Michon, Aug 15 2009
These are the least hemiperfects of abundancy n + 1/2. - Walter Nissen, Aug 17 2010
On Jul 24 2010, Michel Marcus found a 191-digit integer of abundancy 17/2 (2.7172904...10^190) whose factorization starts with 2^81 3^29 5^9 7^10 11^4 13^3 17^2 19 23^2... This is the best upper bound to a(8) known so far. - Gerard P. Michon, Aug 22 2010

			a(2)=24 because 1+2+3+4+6+8+12+24=2.5*24 and 24 is the earliest m such that sigma(m)=2.5*m.

Guy Robin, Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann, J. Math. Pures Appl. 63 (1984), 187-213.

G. P. Michon, Multiplicative functions: Abundancy = sigma(n)/n
G. P. Michon, Multiperfect and hemiperfect integers
G. P. Michon and M. Marcus, Hemiperfect numbers of abundancy 11/2
G. P. Michon and M. Marcus, Hemiperfect numbers of abundancy 13/2
G. P. Michon and M. Marcus, Hemiperfect numbers of abundancy 15/2
G. P. Michon and M. Marcus, Hemiperfect numbers of abundancy 17/2
Walter Nissen, Abundancy : Some Resources
Wikipedia, Hemiperfect number
Wikipedia, Riemann hypothesis

Cf. A007539, A000396, A005820, A027687, A046060, A046061.

Cf. A159907 (hemiperfect numbers: half-integral abundancy), A141643 (abundancy = 5/2), A055153 (abundancy = 7/2), A141645 (abundancy = 9/2), A159271 (abundancy = 11/2), A160678 (abundancy = 13/2).

a[n_] := (For[m=1, DivisorSigma[1, m]!=(n+1/2)m, m++ ];m); Do[Print[a[n]], {n, 4}]

a(5)-a(6) from Robert Gerbicz, Apr 19 2009
Cross-references from Gerard P. Michon, Jun 10 2009
Edited by M. F. Hasler, Mar 17 2013
a(7) from Michel Marcus confirmed and added by Max Alekseyev, Jun 05 2025

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

A335267 Composite numbers whose harmonic mean of their divisors that are larger than 1 is an integer.

Original entry on oeis.org

6, 15, 28, 30, 91, 117, 135, 252, 270, 496, 703, 864, 936, 1891, 1989, 2295, 2701, 4284, 4590, 5733, 8128, 8432, 12403, 18721, 19872, 21528, 38503, 41580, 49141, 51319, 56896, 79003, 88831, 104653, 121920, 146611, 188191, 218791, 226801, 235053, 269011, 286903

Offset: 1

Amiram Eldar, May 29 2020

nonn

The primes are excluded from this sequence since they are trivial terms.
The corresponding harmonic means are 3, 5, 5, 5, 13, 9, 9, 9, 9, 9, 37, ...
Equivalently, composite numbers m such that (sigma(m)-m) | m*(tau(m)-1), or A001065(m) | A168014(m).
The semiprimes terms of this sequence are of the form p*q where p and q = 2*p - 1 are primes (A129521).
If m is a k-perfect numbers, k = 2, 3, ... (i.e., sigma(m) = k*m), then sigma(m)-m = (k-1)*m. If (k-1)*m | m*(tau(m)-1) then (k-1) | (tau(m)-1). If k is odd then tau(m) is also odd, so m is a square, and sigma(m) is odd. Since m | sigma(m) this means that m is also odd. Since there is no known odd multiply-perfect number except for 1 (A007691), there are no known k-perfect numbers with odd k in this sequence.
The perfect numbers (k=2, A000396) are terms: if m is a perfect number then sigma(m)-m = m.
The 4-perfect number (k=4, A027687) m are terms if 3 | (tau(m)-1). Of the first 36 terms of A027687 there are 8 such terms, the first is A027687(26).
The 6-perfect number (k=6, A046061) m are terms if 5 | (tau(m)-1). Of the first 245 terms of A046061 there are 20 such terms, the first is A046061(19).
Hemiperfect numbers that are terms of this sequence include A055153(i) for i = 10, 18 and 20, A141645(21), and A159271(i) for i = 97 and 103.

			6 is a term since its divisors other than 1 are 2, 3 and 6, and their harmonic mean, 3/(1/2 + 1/3 + 1/6) = 3, is an integer.

Amiram Eldar, Table of n, a(n) for n = 1..1000

A000396 and A129521 are subsequences.
Similar sequences: A001599, A247077, A247078.
Cf. A000005 (tau), A000203 (sigma).
Cf. A001065, A032741, A168014,
Cf. A005382, A005383.
Cf. A027687, A046061, A055153, A141645, A159271.

Select[Range[10^6], CompositeQ[#] && Divisible[# * (DivisorSigma[0, #] - 1), DivisorSigma[1, #] - #] &]
Select[Range[287000],CompositeQ[#]&&IntegerQ[HarmonicMean[ Rest[ Divisors[ #]]]]&] (* Harvey P. Dale, Jan 21 2021 *)

A227882 Known number of n_multiperfect numbers that can produce an hemiperfect of abundancy (2*n-1)/2.

Original entry on oeis.org

1, 3, 19, 0, 87, 117, 0, 30, 0, 0

Offset: 2

Michel Marcus, Oct 25 2013

The hemiperfect that are obtained are coprime to p = 2*n-1.

When p=2*n-1 is prime, if m is a n-multiperfect is such that valuation(m, p) = 1, then let's define k = m/p, sigma(k) = sigma(m/p) = sigma(m)/sigma(p) = (n*m)/(p+1) = (n*m)/(2*n) = m/2. So sigma(k)/k = m/(2*k) = (k*p)/(2*k) = p/2 = (2*n-1)/2.

			a(2) = 1, since the only perfect number multiple of 3 is 6, and 6/3=2 has abundancy 3/2.
a(3) = 3, since the 3 known hemiperfect of abundancy 5/2 are coprime to 5.
a(5) = a(8) = a(11) = 0, since for those n, 2*n-1 is not prime.
a(10) is also 0, since all known 10-multiperfect are at least divisible by 19^2.

Achim Flammenkamp, The Multiply Perfect Numbers Page
G. P. Michon, Multiperfect and hemiperfect numbers

Cf. A000396 (2), A005820 (3), A027687 (4), A046060 (5), A046061 (6), A007691 (integer abundancy).
Cf. A141643 (5/2), A055153 (7/2), A141645 (9/2), A159271 (11/2), A160678 (13/2), A159907 (half-integer abundancy).
Cf. A006254.

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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A141645 Numbers n such that sigma(n)/n = 9/2.

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A141643 Numbers k such that sigma(k)/k = 5/2.

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

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A227302 Numbers m such that m divides sigma(2*m).

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A216781 Numbers such that numerator(sigma(n)/n) is odd and denominator(sigma(n)/n) is even.

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A088912 a(n) = smallest m such that sigma(m) = (n+1/2)*m.

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

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