3
$\begingroup$

Similar to this problem, I am trying to find: $$f(n)=\frac1{n^3}\sum_{\substack{a=1\\(a,n)=1}}^{n}a^3$$ Similarly I did this: $$\begin{align}g(n)&=\sum_{d|n}f(d)\\ &=\color{gray}{\text{(see previous question)}}\\ &=\sum_{d|n}\sum_{d'|d}\mu(d')\frac{d'^3}{d^3}\sum_{m\le d/d'}m^3\\ &=\sum_{d|n}\sum_{d'|d}\mu(d')\frac{d'^3}{d^3}\left[\frac{\frac{d^2}{d'^2}\left(\frac{d}{d'}+1\right)^2}{4}\right]\\ &=\frac14\sum_{d|n}\sum_{d'|d}\left(d\frac{\mu(d')}{d'}+2\mu(d')+\frac1dd'\mu(d')\right)\\ &=\sum_{d|n}\left(\phi(d)+2\nu(d)+\frac1d\sum_{d'|d}d'\mu(d')\right)\\ &=n+2+\sum_{d|n}\frac1d\sum_{d'|d}d'\mu(d') \end{align}$$

Now how to resolve $$\sum_{d|n}\frac1d\sum_{d'|d}d'\mu(d')$$

  • 1
    As $(n,a)=(n,n-a)$ $$2n^3 f(n)=\sum_{a=1,(a,n)=1}^n\{a^3+(n-a)^3\}=2\sum_{a=1,(a,n)=1}^n(n^3-3n^2a+3a^2n)$$ $$=2n^3\phi(n)-6n^2\sum_{a=1,(a,n)=1}^na+6n\sum_{a=1,(a,n)=1}^na^2$$ Now $\sum_{a=1,(a,n)=1}^na=\dfrac{n\phi(n)}2$ – 2017-02-26

2 Answers 2

2

Completing the comment of @lab bhattacharjee :

$$\sum_{\substack{1\le a\le n\cr(a,n)=1}}a^2=\frac{n^2}{3}\varphi(n)+\frac{n}{6}\prod_{p\mid n}(1-p)$$

I guess you can now get the whole picture :)

  • 2
    how do you get those two (multiplicative) terms ? – 2017-02-26
1

$$\sum_{b\mid n}\frac{1}{b}\sum_{a\mid b}a\cdot\mu(a) =\sum_{a\mid n}a\cdot \mu(a)\sum_{c\mid\frac{n}{a}}\frac{1}{ac}=\sum_{a\mid n}\mu(a)\sum_{c\mid\frac{n}{a}}\frac{1}{c}$$ equals $$ \sum_{a\mid n}\mu(a)\sum_{c\mid\frac{n}{a}}\frac{ac}{n} = \frac{1}{n}\sum_{a\mid n} a\cdot\mu(a)\cdot\sigma_1\left(\frac{n}{a}\right) $$ that is $\frac{1}{n}$ times a multiplicative function (a Dirichlet convolution). If $n=p^k$, $$ \sum_{a\mid n} a\cdot\mu(a)\cdot\sigma_{1}\left(\frac{n}{a}\right)=\sigma_1(p^k)-p\cdot\sigma_1(p^{k-1})=1 $$ hence, simply, $$\sum_{b\mid n}\frac{1}{b}\sum_{a\mid b}a\cdot\mu(a) = \frac{1}{n}.$$