

A159907


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


22



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 a(k) 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

Table of n, a(n) for n=1..19.
G. P. Michon, Multiperfect Numbers & Hemiperfect Numbers, Numericana.
Walter Nissen, Abundancy : Some Resources .
Project Euler, Problem 241: Perfection Quotients.
Wikipedia, Hemiperfect number


FORMULA

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


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.


MAPLE

with(numtheory);
P:=proc(i)
local a, n;
for n from 2 to i do
a:=(n+1)/2sigma(n)/n; if a=trunc(a) then print(n); fi;
od;
end:
P(10000000000); # Paolo P. Lava, Dec 12 2011


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=", ")
aupto(3*10**4) # Michael S. Branicky, Feb 24 2021


CROSSREFS

Cf. A000203, A088912, A141643 (k=2), A055153 (k=3), A141645 (k=4), A159271 (k=5).
Sequence in context: A111430 A059332 A000794 * A242484 A088912 A342573
Adjacent sequences: A159904 A159905 A159906 * A159908 A159909 A159910


KEYWORD

nonn


AUTHOR

M. F. Hasler, Apr 25 2009


STATUS

approved



