

A222263


Odd numbers n such that 2n/sigma(n)  1 = 1/x for some positive integer x.


6



1, 3, 15, 135, 315, 585, 819, 1155, 1365, 1485, 2295, 4455, 9009, 9405, 9945, 11115, 13923, 13965, 16965, 33705, 34335, 40365, 43785, 43875, 63063, 63855, 84825, 90675, 106425, 108927, 130815, 209385, 236115, 262845, 387387, 396495, 397575, 449295, 483945
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OFFSET

1,2


COMMENTS

Odd terms from A222264, see there for motivation and further links. In particular, the powers of 2 in A222264 which correspond to perfect numbers are excluded, so all n*x from this sequence are spoof perfect numbers, cf. A174292.
Here we do not exclude n if gcd(n,x) > 1 (the first such term is 1155), which would arguably be a "reasonable" additional condition to impose.
The first term with odd x is n=a(79)=9018009, x=22021, which yields Descarte's n*x=198585576189, see also A033870, A033871 and A222262.
No other term with odd x (and thus no other odd freestyle perfect number) is known as of today, to our best knowledge. See the paper by Banks et al. for some restrictions on such numbers.
One can note that when x is even, then sigma(n)/n is of the form (2k2)/k. For instance, for n=15 we have x=4, and sigma(n)/n = 8/5 with k=5. On the other hand, when x is odd, then sigma(n)/n is of the form (2k1)/k. For instance, for n=9018009 we have x=22021, and sigma(n)/n = 22021/11011 with k=11011.  Michel Marcus, Nov 24 2013


LINKS

M. F. Hasler and Donovan Johnson, Table of n, a(n) for n = 1..500 (first 143 terms from M. F. Hasler)
W. Banks, A. Guloglu, W. Nevans and F. Saidak, Descartes Numbers, in: Anatomy of Integers, 167174, American Mathematical Society, Providence R.I., 2008.


EXAMPLE

For a(3)=15, we have sigma(15)=sigma(3*5)=4*6=24, thus 2*15/sigma(15) = 30/24 = 1+1/4. Here, x=4, and if 4 was assumed to be prime, then sigma'(4)=1+4=5 (*) and sigma'(15*4) = sigma(15)*sigma'(4) = 24*5 = 120 = 2*(15*4), i.e. 15*4 is a spoof perfect number.
(*) Here, sigma' is the "freestyle" sumofdivisors function which assumes that 4 is prime, i.e., sigma'(n)=sigma(n/4^k)*(4^(k+1)1)/3 with 4^k  n, or k=valuation(n,4)=floor(A007814(n)/2).


PROG

(PARI) forstep(n=1, 9e9, 2, numerator(2*n/sigma(n)1)==1 & print1(n", "))


CROSSREFS

Cf. A000203 = sigma, the sumofdivisors function.
Sequence in context: A281186 A108210 A006717 * A246804 A230166 A059861
Adjacent sequences: A222260 A222261 A222262 * A222264 A222265 A222266


KEYWORD

nonn


AUTHOR

M. F. Hasler, Feb 20 2013


STATUS

approved



