divisor function sigma

Calculate the divisor function sigma_k(n) for positive integer n

sigma(n,0) is equal to divisor_count(n)
sigma(n,1) is equal to sum(divisors(n))
sigma(n,k) is equal to sum([x**k for x in divisors(n)])

Default for k is 1.

References
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- http://en.wikipedia.org/wiki/Divisor_function
No.  I think, you should move this summation outside of the loop.