cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

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

Views

Author

Yasutoshi Kohmoto, May 10 2008

Keywords

Comments

200286975596707184640 belongs to this sequence. - Gerard P. Michon, May 10 2009
81703797123392614369698250752 is in this sequence. - Gerard P. Michon, May 11 2009

Crossrefs

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

Programs

Extensions

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

Views

Author

Yasutoshi Kohmoto, May 10 2008

Keywords

Comments

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

Crossrefs

Programs

  • Mathematica
    Select[Range[100000], DivisorSigma[1, #]/# == 5/2 &] (* Robert Price, Apr 03 2019 *)
  • PARI
    isok(k) = sigma(k, -1) == 5/2; \\ Michel Marcus, Apr 04 2019

Extensions

Definition rewritten by M. F. Hasler, May 10 2008
First three terms confirmed by Ray Chandler, Sep 18 2008

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

Views

Author

Alex Ratushnyak, Jul 05 2013

Keywords

Comments

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

Crossrefs

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

Programs

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

Views

Author

Michel Marcus, Sep 16 2012

Keywords

Comments

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

Examples

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

Crossrefs

Programs

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

Views

Author

Farideh Firoozbakht, Nov 29 2003

Keywords

Comments

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

Examples

			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.
		

References

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

Crossrefs

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).

Programs

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

Extensions

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

Views

Author

Jianing Song, Aug 04 2018

Keywords

Comments

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).

Examples

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

Crossrefs

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).

Programs

  • Mathematica
    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 *)
  • PARI
    for(n=2, 10, for(m=1, 10^12, if(sigma(m)/m==n/2, print1(m, ", "); break())))
    
  • PARI
    a(n) = my(k=1); while (sigma(k) != k*n/2, k++); k; \\ Michel Marcus, May 15 2025

Formula

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

Extensions

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

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

Views

Author

Michel Marcus, Oct 25 2013

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

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.
Showing 1-7 of 7 results.