A141643 -id:A141643 - cp's OEIS frontend

A227882 Known number of n_multiperfect numbers that can produce an hemiperfect of abundancy (2*n-1)/2.

1, 3, 19, 0, 87, 117, 0, 30, 0, 0

Offset: 2

Michel Marcus, Oct 25 2013

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.

			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.

Achim Flammenkamp, The Multiply Perfect Numbers Page
G. P. Michon, Multiperfect and hemiperfect numbers

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.

cp's OEIS Frontend

A227882 Known number of n_multiperfect numbers that can produce an hemiperfect of abundancy (2*n-1)/2.

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