[seqfan] Re: Sum of omega(k) for k = 1..n

[Benoit CLOITRE]

It seems formula 19 at http://mathworld.wolfram.com/DistinctPrimeFactors.html gives the Mertens constant.
.

> Message du 12/01/12 05:30
> De : "Charles Greathouse" 
> A : "Sequence Fanatics Discussion list" 
> Copie à : 
> Objet : [seqfan] Re: Sum of omega(k) for k = 1..n
> 
> > a(n) - n*log(log(n)) <= c*O(n), for all n and for some constant c?
> 
> a(n) - n * log log n <= cn is true for all n > e and some c, yes.
> (The restriction on n is because of the double log.) But more is
> true:
> a(n) - n * log log n ~ cn
> for some c (which I think is around 0.25). I'm trying to find this constant.
> 
> This is actually pretty basic analytic number theory but for some
> reason I'm having trouble.
> 
> > Finally, could
> >
> > A013939(n) = n*log(log(n)) + O(n/log(n)) (for n large enough)
> >
> > be a better estimate?
> 
> No, that is false. It could be that
> 
> A013939(n) = n*log(log(n)) + cn + O(n/log(n))
> 
> for some c but I don't think so. All I know is that
> 
> A013939(n) = n*log(log(n)) + cn + o(n),
> 
> that is, the error in the approximation
> 
> A013939(n) ≈ n*log(log(n)) + cn
> 
> is less than eps * n for any eps > 0 and large enough n depending on
> the choice of eps.
> 
> Charles Greathouse
> Analyst/Programmer
> Case Western Reserve University
> 
> On Wed, Jan 11, 2012 at 11:06 PM, Ed Jeffery  wrote:
> > Charles Greathouse wrote:
> >
> >>Regarding A013939, what is the limit of
> >
> >>(A013939(n) - n log log n)/n?
> >
> >>Is this constant (roughly a quarter, I think) in the OEIS?
> >
> >>I wrote
> >
> >>a(n) = n log log n + O(n).
> >
> >>but it should really be
> >
> > ?a(n) = n log log n + kn + o(n).
> >
> >>for some k.
> >
> > Charles,
> >
> > Is the first expression the correct one since you should have the relation
> >
> > a(n) - n*log(log(n)) <= c*O(n), for all n and for some constant c?
> >
> > Otherwise, since the o(n) term is supposed to dominate, is the term
> > k*n (in the second expression) redundant? Finally, could
> >
> > A013939(n) = n*log(log(n)) + O(n/log(n)) (for n large enough)
> >
> > be a better estimate?
> >
> > Regards,
> >
> > Ed Jeffery
> >
> > _______________________________________________
> >
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> 
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