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

Author

M. F. Hasler, Apr 25 2009

Keywords

• nonn

Comments

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

Examples

			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.
		

Links

• 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

Crossrefs

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

Programs

PARI

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

PARI

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

Python

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

Formula

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

Extensions

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