## MObius transform sum...

let be the sum (over all the divisors d of n):

$$f(n)= \sum_{d|n} \mu (n/d)g(d)$$ my question is if n=prime then you have only 2 numbers 1 and p that are divisors so you get:

$$f(p)= \mu (p)g(1) + \mu (1) g(p)$$ is that correct?...now the question is to know what's the value of mu(x) function for x=1 or p.

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 Quote by lokofer let be the sum (over all the divisors d of n): $$f(n)= \sum_{d|n} \mu (n/d)g(d)$$ my question is if n=prime then you have only 2 numbers 1 and p that are divisors so you get: $$f(p)= \mu (p)g(1) + \mu (1) g(p)$$ is that correct?
Correct.

 Quote by lokofer ...now the question is to know what's the value of mu(x) function for x=1 or p.
Step #1 when trying to learn about mobius inversion and such:

Look at the definition of the mobius function.

Complete this step and $$\mu(1)$$ and $$\mu(p)$$ will be apparant.