# Simplifying expression Mobius Function

Can anybody help me simplify this expression using Mobius Inversion Formula or any other result in order to calculate F(3500) in a simple way?? $$F(n)=\sum_{d\mid n} \mu(d)d$$

• Hint: $\mu(n)n$ is multiplicative, hence so is $F(n).$ – Thomas Andrews Apr 8 at 17:48
• Thank you very much for your hint. This is the key of the exercise (It was not about Mobius Inversion) !! – Luis Gimeno Sotelo Apr 8 at 17:59

It is the product $$f(n)=(1-p_1)(1-p_2)\cdots(1-p_k)$$ where the $$p_i$$ are the distinct prime divisors of $$d.$$

This is because $$g(n)=\mu(n)n$$ is multiplicative, and hence so is $$F,$$ and $$F(p^k)=1-p$$ for prime $$p.$$

In your case, $$3500$$ has prime divisors $$2,5,$$ and $$7$$. So:

$$F(3500)=(1-2)(1-5)(1-7)=-24$$

You could write it as $$F(n)=n\sum_{d\mid n} \mu(d)\frac{1}{n/d}.$$ Then you get, by Möbius inversion: $$\frac{1}{n}=\sum_{d\mid n}\frac{F(d)}{d}$$

Or:

$$1=\sum_{d\mid n}\frac{n}{d}F(d)$$

That won't help you compute $$F$$ much however.