# Sum of divisor function

• November 15th 2010, 03:20 AM
I-Think
Sum of divisor function
I have to find a formula for the function
$f(n)=\sum_{d|n}\sigma(d)$
where
$n=p_1^{e_1}p_2^{e_2}***p_r^{e_r}$

And I want to ensure that I understand the function correctly, so I will illustrate my understanding by examples with numbers

$f(12)=\sum_{d|12}\sigma(d)$
$f(12)=\sum_{d|n}24$
$f(12)=60$

2nd example
$f(9)=\sum_{d|9}\sigma(d)$
$f(n)=\sum_{d|n}13$
$f(9)=14$

Am I correct?
• November 19th 2010, 06:12 PM
dwsmith
The sum of divisors function adds up the divisors: 1 + 12 + 2 + 6 + 3 + 4 = 28
• November 20th 2010, 01:06 AM
CaptainBlack
Quote:

Originally Posted by I-Think
I have to find a formula for the function
$f(n)=\sum_{d|n}\sigma(d)$
where
$n=p_1^{e_1}p_2^{e_2}***p_r^{e_r}$

And I want to ensure that I understand the function correctly, so I will illustrate my understanding by examples with numbers

$f(12)=\sum_{d|12}\sigma(d)$
$f(12)=\sum_{d|n}24$
$f(12)=60$

2nd example
$f(9)=\sum_{d|9}\sigma(d)$
$f(n)=\sum_{d|n}13$
$f(9)=14$

Am I correct?

$f(12)=\sigma(1)+\sigma(2)+\sigma(3)+\sigma(4)+\sig ma(6)+\sigma(12)$

CB