

A227882


Known number of n_multiperfect numbers that can produce an hemiperfect of abundancy (2*n1)/2.


0




OFFSET

2,2


COMMENTS

The hemiperfect that are obtained are coprime to p = 2*n1.
When p=2*n1 is prime, if m is a nmultiperfect 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*n1)/2.


LINKS

Table of n, a(n) for n=2..11.
Achim Flammenkamp, The Multiply Perfect Numbers Page
G.P. Michon, Multiperfect and hemiperfect numbers


EXAMPLE

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*n1 is not prime.
a(10) is also 0, since all known 10multiperfect 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 (halfinteger abundancy).
Cf. A006254.
Sequence in context: A001999 A157580 A101293 * A189799 A300946 A078096
Adjacent sequences: A227879 A227880 A227881 * A227883 A227884 A227885


KEYWORD

more,nonn


AUTHOR

Michel Marcus, Oct 25 2013


STATUS

approved



