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A227882 Known number of n_multiperfect numbers that can produce an hemiperfect of abundancy (2*n-1)/2. 0
1, 3, 19, 0, 87, 117, 0, 30, 0, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET

2,2

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.

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*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.

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

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Last modified October 15 15:14 EDT 2019. Contains 328030 sequences. (Running on oeis4.)