[seqfan] Re: Asymptotic for A074753
franktaw at netscape.net
franktaw at netscape.net
Wed Jan 13 05:23:10 CET 2010
The density of the semiprimes themselves does go to zero; the number of
semiprimes up to n is about
n log log n / log n.
Looking at MathWorld,
http://mathworld.wolfram.com/DivisorFunction.html, we have formula 36
(from Hardy and Wright),
Sum(k=1..n, sigma(k)) = pi^2/12 n^2 + O(n log n)
which means that the average order of sigma(n) is pi^2/6 n. This
strongly suggests that the limit is non-zero, but I don't see how to
make that rigorous. It suggests 6/pi^2 = .6079... as the limit, but I
doubt that.
Franklin T. Adams-Watters
-----Original Message-----
From: Matthew Conroy <jumbo at madandmoonly.com>
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.
cheers,
Matt
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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>
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