A230166 Terms of A222263 such that 2n/sigma(n) - 1 = 1/2^k, for some integer k.

1, 3, 15, 135, 819, 1365, 1485, 2295, 9009, 13923, 63855, 387387, 397575, 667275, 14381055, 16410735, 99558459, 271543725, 3145425129, 7096702977, 741585912975, 2148325363107, 4847048133291, 39206559148911, 53164445037705, 130468907286855, 1229923663366167

Offset: 1

Author

Michel Marcus, Oct 11 2013

Keywords

• nonn

Comments

For all n>1, sigma(n)>n, therefore 2n/sigma(n)-1 is always less than 1, i.e., k>0.

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.

For k=12 to 19, they are: unknown, 741585912975, unknown, 39206559148911, 2569480266942180207, 1712973775775070501, unknown, 299364435975778645966263.

Examples

			a(1)=1 since 2*1/sigma(1)-1 = 2-1 = 1 = 1/2^0 is of the required form with k=0.
For n=2, 2*2/sigma(2)-1 = 4/3-1 = 1/3 is not of the form 1/2^k.
a(2)=3 since 2*3/sigma(3)-1 = 6/4-1 = 1/2 = 1/2^1 is of that form with k=1.
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.
		

Links

• Hiroaki Yamanouchi, Table of n, a(n) for n = 1..39

Crossrefs

Cf. A222263.

Programs

PARI

is_A230166(n)=(n=2*n/sigma(n)-1)>>valuation(n,2)==1 \\ - M. F. Hasler, Oct 12 2013

Extensions

a(21) from Donovan Johnson, Dec 28 2013

a(22)-a(27) from Hiroaki Yamanouchi, Sep 27 2014