Sum of square free divisors of integer

Hi, is there any forumula to find the sum of square free (Square-free integer - Wikipedia, the free encyclopedia) divisors of an integer?

One way is to get the divisors then check for each if it is square free and take the sum. However is there any formula for calculating it?

I think mobius (Möbius function - Wikipedia, the free encyclopedia) function and Euler-Totient(Euler's totient function - Wikipedia, the free encyclopedia) functions can be used here but am not sure how?

E.g for n = 9 the divisors are 1,3,9 out of which the square free divisors are 1,3 hence sum = 4.

Thanks.

Re: Sum of square free divisors of integer

Isn't that just the sum of divisors of the radical of n? The radical of n is the product of the distinct prime divisors of n. For example, for $\displaystyle n=504=2^3\cdot3^2\cdot7$, $\displaystyle \text{rad}(504)=2\cdot3\cdot7=42$. The divisors of n are 1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 18, 21, 24, 28, 36, 42, 56, 63, 72, 84, 126, 168, 252, 504 of which 1, 2, 3, 6, 7, 14, 21, and 42 are square free. This is exactly the set of divisors of 42.

Re: Sum of square free divisors of integer

thanks for the reply but is there any more other efficient way?

Thanks

Re: Sum of square free divisors of integer

It would be the product of p+1 for all primes p dividing n. So in the case of 504, (2+1)(3+1)(7+1)=96.