[seqfan] Re: Asymptotic for A074753

Matthew Conroy jumbo at madandmoonly.com
Wed Jan 13 03:39:14 CET 2010

hi Franklin,

To get a bound greater than 0, you should be able to restrict to
square-free integers.  The equivalent a(n) for square-free
integers gives a(100)=49, a(1000)=468, a(10000)=4709,
so it looks safe to assume that that will stay at a nicely
positive density.

Once restricted to square-free integers, you may more
easily be able to count.

Even just considering semiprimes, the density is pretty high, though
I think ultimately it may go to zero.

I'm interested in this question; I'd be happy to hear how you get along,
and I'll send you any results I can get, if you would like them.



On Jan 12, 2010, at 9:22 AM, franktaw at netscape.net wrote:

> Since my last asymptotic question got no response :-(, I thought I
> would try another one :-).
> For a(n) = http://www.research.att.com/~njas/sequences/A074753 (Number
> of integers k such that sigma(k) < n), what is the limit n->infinity
> a(n) / n?
> Up to 1 million, it appears to be about .673.
> Certainly the limit is less than 1; any 2k > 2n/3 will have sigma 
> (2k) >
> n, so the limit must be < 5/6.  Similar arguments with other factors
> can refine this; looking only at prime factors up to 100000, I get an
> upper bound of .7044..., and this clearly converges to something in
> this neighborhood.  I don't see any way to prove the actual limit is >
> 0, however.
> Franklin T. Adams-Watters
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