- #1
lokofer
- 106
- 0
let be the sum (over all the divisors d of n):
[tex] f(n)= \sum_{d|n} \mu (n/d)g(d) [/tex] my question is if n=prime then you have only 2 numbers 1 and p that are divisors so you get:
[tex] f(p)= \mu (p)g(1) + \mu (1) g(p) [/tex] is that correct?...now the question is to know what's the value of mu(x) function for x=1 or p.
[tex] f(n)= \sum_{d|n} \mu (n/d)g(d) [/tex] my question is if n=prime then you have only 2 numbers 1 and p that are divisors so you get:
[tex] f(p)= \mu (p)g(1) + \mu (1) g(p) [/tex] is that correct?...now the question is to know what's the value of mu(x) function for x=1 or p.