sum of square-free divisors

[Pieter Moree]

Steven Finch wrote:

> How are you?  Given a positive integer n, let a(n) denote the sum of
> the
>  square-free divisors of n, including 1.  For example,
>
> a(1000) = 18 = 1+2+5+10.
>
> The sequence {a(n)} is A048250 in Sloane's database
>
> http://www.research.att.com/~njas/sequences/
>
> It's known that a(n) is the product of 1+p for each p|n, and that a(n)
> has  Dirichlet generating function:
>
> zeta(s)*zeta(s-1)/zeta(2*s-2)
Yes, I agree with that.

> Someone has suggested to me that sum_{n<=x) a(n) should be of the form
>  C*x^2. Is this true?  If yes, what exactly is the constant C?
C=1/2.

The proposition is

\sum_{n\le x}a(n)=x^2/2+O(x\sqrt{x}).
I checked it on maple. Seemed consistent.

My proof rest on the following lemma.

Lemma.
Let F(x)=\sum_{d<=x}\mu(d)^2 d.
Then F(x)=3x^2/pi^2+O(x\sqrt{x}).

Proof. Notice, as did the
very young Gauss, that \sum_{d<=x}d=[x]([x]+1)/2. We use this
and inclusion and exclusion. This yields

F(x)=\sum_{r^2<=x}\mu(r)\sum_{d<=x,
r^2|d}d=\sum_{r^2<=x}{\mu(r)/2}[x/r^2]([x/r^2]+1)r^2,

where we used that \sum_{d<=x, r^2|d}d=r^2[x/r^2]([x/r^2]+1)/2.

This yields F(x)={x^2/2}\sum_{r^2<=x}\mu(r)/r^2+O(x\sqrt{x)).
>From this the assertion easily follows.

Proof of proposition. The sum under consideration equals
\sum_{md<=x}\mu(d)^2 d =\sum_{m<=x}\sum_{d<=x/m}\mu(d)^2 d
=\sum_{m<=x}F(x/m)
with F(x)=\sum_{d<=x}\mu(d)^2 d.
Then apply the lemma and Euler's observation that
\sum 1/m^2=pi^2/6.


> A literature search has turned up nothing so far.  If you can answer
> my  questions, this would be marvelous!  I will certainly cite you in
> an  upcoming essay...
>
> Thank you very much.  I hope to hear from you.  Take care,
>
> Steve
>
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