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%I #17 Nov 14 2019 12:14:47
%S 1,3,19,0,87,117,0,30,0,0
%N Known number of n_multiperfect numbers that can produce an hemiperfect of abundancy (2*n-1)/2.
%C The hemiperfect that are obtained are coprime to p = 2*n-1.
%C 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.
%H Achim Flammenkamp, <a href="http://wwwhomes.uni-bielefeld.de/achim/mpn.html">The Multiply Perfect Numbers Page</a>
%H G. P. Michon, <a href="http://www.numericana.com/answer/numbers.htm#multiperfect">Multiperfect and hemiperfect numbers</a>
%e a(2) = 1, since the only perfect number multiple of 3 is 6, and 6/3=2 has abundancy 3/2.
%e a(3) = 3, since the 3 known hemiperfect of abundancy 5/2 are coprime to 5.
%e a(5) = a(8) = a(11) = 0, since for those n, 2*n-1 is not prime.
%e a(10) is also 0, since all known 10-multiperfect are at least divisible by 19^2.
%Y Cf. A000396 (2), A005820 (3), A027687 (4), A046060 (5), A046061 (6), A007691 (integer abundancy).
%Y Cf. A141643 (5/2), A055153 (7/2), A141645 (9/2), A159271 (11/2), A160678 (13/2), A159907 (half-integer abundancy).
%Y Cf. A006254.
%K nonn,more
%O 2,2
%A _Michel Marcus_, Oct 25 2013
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