1

Let $f$ be a multiplicative function, i.e. for all coprime $a,b\in\mathbb{N}\quad f(ab)=f(a)f(b)$.

Consider: $$(f*f)(n)=\sum_{d\vert n}f(d)f\left(\frac{n}{d}\right),\quad\text{where * is Dirichlet convolution.}$$ Claim: $$(f*f)(n)=f(n)\sigma_0(n)\quad\text{where $\sigma_0(n)$ is the number of divisors function}.$$ Proof: $$\sum_{d\vert n}f(d)f\left(\frac{n}{d}\right)=\underbrace{\sum_{d\vert n}f(n)}_\text{by multiplicativity}=f(n)\sum_{d\vert n}1=f(n)\sigma_0(n).\square$$

I would just like to make sure I haven't made any silly mistakes.

  • Is your result valid for this particular case ? – Jean Marie Feb 16 '21 at 19:55
  • 1
    @JeanMarie Yes, I believe it is. Proof(hopefully as formal as I think): $\mu(n)\sigma_0(n)=0$ if $n$ has a squared factor, let alone cubed. So given $n$ essentially equal to its own radical, and using the fact that $\sigma_x(n)$ is multiplicative: $\mu(n)\sigma_0(n)=(-1)^k\sigma_0(p_1)\cdot..\cdot\sigma_0(p_k)=(-1)^k(2)^k=(-2)^k$ by definition of $\mu(n)$ and observing that $\sigma_0(p)=2.$ –  Feb 16 '21 at 20:37

0 Answers0