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%I #16 Sep 27 2014 16:04:41
%S 1,3,15,135,819,1365,1485,2295,9009,13923,63855,387387,397575,667275,
%T 14381055,16410735,99558459,271543725,3145425129,7096702977,
%U 741585912975,2148325363107,4847048133291,39206559148911,53164445037705,130468907286855,1229923663366167
%N Terms of A222263 such that 2n/sigma(n) - 1 = 1/2^k, for some integer k.
%C For all n>1, sigma(n)>n, therefore 2n/sigma(n)-1 is always less than 1, i.e., k>0.
%C For k=1 to 11, the smallest known numbers to give 1/2^k are: 3, 15, 135, 2295, 1485, 1365, 63855, 16410735, 397575, 667275, 271543725.
%C For k=12 to 19, they are: unknown, 741585912975, unknown, 39206559148911, 2569480266942180207, 1712973775775070501, unknown, 299364435975778645966263.
%H Hiroaki Yamanouchi, <a href="/A230166/b230166.txt">Table of n, a(n) for n = 1..39</a>
%e a(1)=1 since 2*1/sigma(1)-1 = 2-1 = 1 = 1/2^0 is of the required form with k=0.
%e For n=2, 2*2/sigma(2)-1 = 4/3-1 = 1/3 is not of the form 1/2^k.
%e a(2)=3 since 2*3/sigma(3)-1 = 6/4-1 = 1/2 = 1/2^1 is of that form with k=1.
%e For a(3)=15, 2*15/sigma(15)-1 = 30/(1+3+5+15)-1 = 30/24 - 1 = 6/24 = 1/2^2 is of this form with k=2.
%o (PARI) is_A230166(n)=(n=2*n/sigma(n)-1)>>valuation(n,2)==1 \\ - _M. F. Hasler_, Oct 12 2013
%Y Cf. A222263.
%K nonn
%O 1,2
%A _Michel Marcus_, Oct 11 2013
%E a(21) from _Donovan Johnson_, Dec 28 2013
%E a(22)-a(27) from _Hiroaki Yamanouchi_, Sep 27 2014
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