A141643 -id:A141643 - 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

A055153 Numbers k such that sigma(k) = 7k/2.

Original entry on oeis.org

4320, 4680, 26208, 20427264, 197064960, 21857648640, 57575890944, 88898072401645056, 301183421949935616, 9083288595228991885541376, 22290964134962716779872256, 230361837156847526055247872

Offset: 1

Jud McCranie, Jun 16 2000

nonn

			Sigma(4320)=15120=7*4320/2, so 4320 is in the sequence.

G. P. Michon, Table of n, a(n) for n = 1..21
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. A007539, A000396, A005820, A027687, A000203, A141643, A141645.

Do[If[DivisorSigma[1, m]==3.5*m, Print[m]], {m, 2*10^8}]

is(k)=sigma(k,-1)==7/2 \\ Charles R Greathouse IV, Mar 09 2014

Terms confirmed through a(5) by Ray Chandler, Sep 18 2008
a(6) and a(7) found by Yasutoshi Kohmoto and confirmed by Washington Bomfim, Oct 19 2008
Edited by N. J. A. Sloane, Sep 19 2008, Apr 18 2009
a(9) from Avinoam Kalma, a(12) from Yasutoshi Kohmoto, and a(8), a(10), a(13)-a(21) from Michel Marcus, added by Gerard P. Michon, Jun 04 2009

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

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

A194472 Erdős-Nicolas numbers.

Original entry on oeis.org

24, 2016, 8190, 42336, 45864, 392448, 714240, 1571328, 61900800, 91963648, 211891200, 1931236608, 2013143040, 4428914688, 10200236032, 214204956672

Offset: 1

Alonso del Arte, Aug 24 2011

Abundant numbers m such that the sum of the first k divisors is equal to m for some k, thus this is a subsequence of A064510. k has to be less than tau(m) - 1 for this sequence, whereas in A064510 k = tau(m) - 1 is allowed (and thus perfect numbers are in that sequence).
a(17) > 5*10^11. 104828758917120, 916858574438400, 967609154764800, 93076753068441600, 215131015678525440 and 1371332329173024768 are also terms. - Donovan Johnson, Dec 26 2012
a(17) > 10^12. - Giovanni Resta, Apr 15 2017
Equivalently, numbers whose abundancy equals 1 + the sum of the reciprocals of its first k divisors for some k > 1. - Charlie Neder, Feb 08 2019
96892692739248881664, 41407449045801454927872, 101616496263816777695232, 1346571992706422996646631651147776, 3304572752464376776401640967110656 are also terms. - Michel Marcus, Feb 09 2019
All known terms of A141643 (abundancy 5/2) are terms. - Michel Marcus, Feb 11 2019
Named after the Hungarian mathematician Paul Erdős (1913-1996) and the French mathematician Jean-Louis Nicolas. - Amiram Eldar, Jun 23 2021
Are all terms in this sequence even? - Jenaro Tomaszewski, May 07 2023

			The divisors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24 and 1 + 2 + 3 + 4 + 6 + 8 = 24, hence 24 is in the list.
The divisors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. The first seven of these add up to 36, but the first eight add up to 52, therefore 48 is not on the list.

Jean-Marie De Koninck, Those Fascinating Numbers, Amer. Math. Soc., 2009, p. 141.

P. Erdős and J.-L. Nicolas, Répartition des nombres superabondants, Bull. Soc. Math. France, Vol. 103, No. 1 (1975), pp. 65-90.
Wikipedia, Erdős-Nicolas number.

Cf. A005835, A000396, A064510, A141643.

subtr = If[#1 < #2, Throw[#1], #1 - #2] &; selDivs[n_] := Catch@Fold[subtr, n, Drop[Divisors[n], -2]]; erdNickNums = {}; Do[If[selDivs[n] == 0, AppendTo[erdNickNums, n]], {n, 2, 10^5}]; erdNickNums (* Based on the program by Bobby R. Treat and Robert G. Wilson v for A064510 *)

isok(n) = {if (sigma(n) <= 2*n, return (0)); my(d = divisors(n), s = 0); for (k=1, #d-2, s += d[k]; if (s == n, return (1)); if (s > n, break);); return (0);} \\ Michel Marcus, Feb 09 2019

from itertools import accumulate, count, islice
from sympy import divisors
def A194472_gen(startvalue=1): # generator of terms >= startvalue
    return (n for n in count(max(startvalue,1)) if any(s == n for s in accumulate(divisors(n)[:-2])))
A194472_list = list(islice(A194472_gen(),5)) # Chai Wah Wu, Feb 18 2022

More terms from M. F. Hasler, Aug 24 2011

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

A383758 Least integer k for which sigma(k - x) + sigma(k + x) = n*k has at least one solution.

Original entry on oeis.org

1, 2, 6, 24, 93, 1952, 14412, 361881, 61824672

Offset: 2

Jean-Marc Rebert, May 09 2025

The corresponding x are : 0, 0, 0, 0, 87, 1828, 13308, 358839, ...
a(10) <= 61824672 via sigma(61824672 - 60697728) + sigma(61824672 + 60697728) = 10*61824672. - Michel Marcus, May 20 2025
a(11) <= 43293761280 via sigma(43293761280 - 40511560320) + sigma(43293761280 + 40511560320) == 11*43293761280. - Michel Marcus, May 25 2025
Note that for n=2,3,4,5,8,and 9, we have k+x = A383920(n). - Michel Marcus, Jun 09 2025
From David A. Corneth, Jun 13 2025: (Start)
a(10) = 61824672. We must have sigma(k-x) >= 5*(k-x) or sigma(k+x) >= 5 * (k+x).
The numbers <= 2*61824672 that have this property are 122522400. It has been checked that if k + x = 122522400 then k must be 61824672 to get the smallest such k. (End)

			a(4) = 6 because the equation sigma(6-x) + sigma(6+x) = 4*6 has the solution x = 0 and no smaller number possesses this property. See A000396, A383268, and A383269.
a(5) = 24 because the equation sigma(24-x) + sigma(24+x) = 5*24 has the solution x = 0. This is verified as follows: sigma(24-0) + sigma(24+0) = sigma(24) + sigma(24) = 60 + 60 = 120 = 5*24. Moreover, no smaller number possesses this property. See A141643.
a(6) = 93 because the equation sigma(93 - x) + sigma(93 + x) = 6 * 93 has the solution x = 87: sigma(93 - 87) + sigma(93 + 87) = sigma(6) + sigma(180) = 12 + 546 = 6*93. Moreover, no smaller number possesses this property.

Cf. A000203, A000396, A141643, A317681, A383268, A383269.

isok(k,n) = forstep(x=k-1, 0, -1, if (sigma(k - x) + sigma(k + x) == n*k, return(1)));
a(n) = my(k=1); while (!isok(k, n), k++); k; \\ Michel Marcus, May 10 2025

a(n) <= A317681(n).

a(10) from Michel Marcus and David A. Corneth, Jun 13 2025

A384838 Numbers k for which sigma(k - x) + sigma(k + x) = 5*k has at least one nonnegative solution.

Original entry on oeis.org

24, 53, 56, 63, 66, 74, 75, 79, 82, 84, 95, 112, 168, 192, 216, 227, 245, 252, 255, 274, 280, 282, 288, 308, 312, 347, 348, 351, 360, 365, 392, 395, 408, 420, 431, 432, 434, 458, 465, 466, 471, 476, 496, 528, 532, 560, 576, 579, 588, 624, 628, 644, 670, 694, 716, 720, 784

Offset: 1

Michel Marcus, Jun 10 2025

nonn

Michel Marcus, Table of n, a(n) for n = 1..5000

Cf. A000203 (sigma), A141643 (a subsequence).
Cf. A383268, A384839, A384840, A384841.
Cf. A383758.

isok(k) = for (x=0, k-1, if (sigma(k - x) + sigma(k + x) == 5*k, return(1)));

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

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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A055153 Numbers k such that sigma(k) = 7k/2.

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

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

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A194472 Erdős-Nicolas numbers.

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

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A383758 Least integer k for which sigma(k - x) + sigma(k + x) = n*k has at least one solution.

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