Möbius function verification

Question

I am looking to verify my answer to the question $$F(n)=\sum_{d|n}{\mu(d)\sigma(d)}=(-1)^{\omega(n)}\prod_{j=1}^{\omega(n)}{p_j}$$ Where $\mu$ is the Möbius function, $\sigma$ is the sum of divisors of a number $n$, and $\omega$ is the number of distinct prime factors of $n$.

Let $n=\prod_{j=1}^kp_j^{\alpha_j}$. Now, since the function $\mu(n)\sigma{(n)}$ is multiplicative, so is $F(n)$ and thus I can look at $F(p_i^{\alpha_i})$ Now the divisors of $p_i^{\alpha_i}$ are $1, p_i, p_i^2, ..., p_i^{\alpha_i}$, so $\mu(p_i^{\alpha_i})=0$ if $\alpha_i>1$. Therefore, for each prime power factor of $n$, we get that $$F(p_i^{\alpha_i})=\mu(1)\sigma(1)+\mu(p_i)\sigma(p_i)$$ $$=1\cdot 1 + (-1)\cdot (p_i+1)=-p_i$$ Since there are $\omega(n)$ primes, the formula above follows.

Is this correct? My gut says yes...

Thank you for the verification – Lalaloopsy Jul 25 '14 at 12:49 — Lalaloopsy, Jul 25 '14 at 12:49
Related: http://math.stackexchange.com/questions/24362 – Bart Michels Mar 14 '15 at 11:08 — Bart Michels, Mar 14 '15 at 11:08

score 1 · Accepted Answer · answered Mar 10 '15 at 21:23

Your proof is correct.
If there is anything to remark, it is that your proof only covers the case where $n$ has at least one prime divisor, that is, $n>1$. This means one should verify the case $n=1$ separately. When $n=1$ the product is empty and hence (by convention) equal to $1$, which is the same as the LHS.

Möbius function verification

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