

A159907


Numbers n with halfintegral abundancy index, sigma(n)/n = k+1/2 with integer k.


25



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

1,1


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, multiplyantiperfect 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 halfinteger 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


LINKS



FORMULA



EXAMPLE

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.


PROG

(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=", ")


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



