[seqfan] Re: Asymptotic for A074753
Matthew Conroy
jumbo at madandmoonly.com
Fri Jan 15 02:04:17 CET 2010
Thanks for that reference, Drew.
Using the full Table 1 in Wall's paper, I was able to calculate
lim inf a(n)/n > 0.6152954.
There is a lot of error in that table (there are large gaps between the
upper and lower bounds on A(x)). More recent results suggest
these could be much improved.
Note for instance, in the Wall paper
0.2441 < A(2) < 0.2909
but this was greatly improved by Marc Deleglise to
0.2474 < A(2) < 0.2480 in his paper
"Bounds for the density of abundant integers",Marc Deléglise,
Experiment.
Math. Volume 7, Issue 2 (1998), 137-143.
Unforunately, Deleglise's paper does not have results for other A(x).
cheers,
Matt
On Jan 13, 2010, at 6:44 AM, drew at math.mit.edu wrote:
> I believe you can obtain asymptotic lower (and upper) bounds on a
> (n)/n by
> applying the results of
>
> [1] "Density Bounds for the Sum of Divisors Function", Charles R.
> Wall,
> Phillip L. Crews and Donald B. Johnson, Mathematics of Computation,
> Vol.
> 26, No. 119, 1972, pages 773-777,
>
> which builds on earlier work by Behrend and Davenport (from 1933). The
> reference above is available online at http://www.jstor.org/stable/
> 2005106.
> The bound one can obtain is certainly not tight, but it is above 0.36.
>
> I realize that not everyone on this list has convenient access to
> JSTOR, so
> I will briefly summarize the key points that are relevant to A074753.
>
> Let pi_A(N,x) count the positive integers m <= N for which sigma(m)
> >= mx,
> where x is a real number in the interval [1,5].
>
> It was proven by Davenport (see reference in [1]) that
>
> A(x) = lim_{N\to\infty} pi_A(N,x)/N
>
> exists and is a continuous function of x. The reference [1] above
> gives a
> table of upper and lower bounds for A(x) for x = 1.0, 1.1 ....,
> 4.9, 5.0,
> for example, A(2) lies between 0.2441 and 0.2909.
>
> For a(n)=A074753(n), we may use this table to asymptotically bound a
> (n)/n.
> Indeed, for any real number x in [1,5] and positive integer n we have
>
> (*) floor(n/x) - pi_A(n/x,x) <= a(n)
>
> This is obtained by counting integers m <= n/x for which sigma(m) <
> mx <= n.
>
> Dividing (*) by n and taking limits as n tends to infinity yields
>
> (**) lim inf a(n)/n >= (1-A(x)) / x.
>
> From Table 1 of [1], we have A(2.1) <= .2372, which implies that
>
> (1) lim inf a(n)/n > 0.363238.
>
> This bound above can be likely be improved slightly by further
> subdividing
> the interval from [1,n] and applying multiple bounds on A(x).
>
> Regards,
>
> Drew
>
>> 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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