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

[Charles Greathouse]

> 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 <lejeffery7 at gmail.com> 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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